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Atoms and Molecules

Dr. S. S. Tripathy

AVOGADRO'S LAW AND ITS APPLICATIONS

Until 1811 Dalton's Atomic theory was held supreme. It was believed that gases aremonoatomic (H,N, O etc) and atoms combine in 1:1 ratio to give compound atom (what we callnow the molecule)

H + Cl --------------> HCl (compound atom)BERZELIUS HYPOTHESIS:Berzelius on the basis of Dalton's theory proposed:

" Equal volumes of all gases under identical conditions of temperature and pressure containequal number of atoms"Failure of Berzelius Hypothesis:According to Berzelius,1 atom of hydrogen(H) combines with 1 atom of chlorine(Cl) to produce 1 compound atom ofhydrogen cloride(HCl)(Note that there was no idea about molecule at that time and he believed that gases exist inmonoatomic state. One HCl molecule was called a "compound atom" at that time)So 'n' atoms of H should combine with 'n' atoms of Cl to produce 'n' compound atoms of HCl.Applying Berzelius hypothesis, if 1 volume of hydrogen gas contains 'n' atoms, then 1 volume ofother gases also would contain 'n' number of atoms. Hence1 volume of hydrogen should combine with 1 volume of chlorine to produce 1 volume of hydrogenchloride (temperature and pressure remaining constant).However this was found to be wrong by Gay Lussac's experimental study according to which,1 volume of hydrogen gas combines with 1 volume of chlorine gas to produce 2 volumes ofhydrogen chloride gas at the same temperature and pressure. Thus Berzelius hypothesis wasrejected and it was challenged by Avogadro who believed that elementary gases exist mostly asdiatomic molecules, i.e. O

2, N

2, H

2 and so on. The idea of compound atom was discarded and

replaced by the term molecule. The concept of MOLECULE was thus first introduced.Avogadro, in 1811, first gave a hypothesis which subsequently was turned into a law. He presumedthat the elementary gases like nitrogen, hydrogen, oxygen etc. exist as diatomic molecules. Hispresumption was later experimentally confirmed by Cannizzaro in 1950 after which the hypothesisturned into a law.Avogadro's LawEqual volumes of all gases under identical conditions of temperature and pressure containequal number of molecules.

Explanation:According to this law we can take any gas, say, hydrogen, chlorine, oxygen, hydrogen chloride,ammonia, carbon dioxide etc. and if the volumes of these gases are same while measured undersame temperature and pressure, then each of the gas must contain equal number of molecules.

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Atoms and Molecules

Dr. S. S. Tripathy

H2 N2 CO

Pressure= P mmTemp. = t0CVolume = V LNo. of molecules = n No. of molecules = n

Volume = V LTemp. = t0CPressure= P mm

No. of molecules = nVolume = V LTemp. = t0CPressure= P mm

Conversely if the gases contain equal number of molecules at the same temperature and pressurethen they possess equal volumes.

SAQ 1: Suppose you have been given three gas cylinders of equal volume (5 litres) containingthree gases N

2, O

2 and H

2. Each of them has a pressure of 2 atmospheres at temperature of

270C. If nitrogen cylinder contains 20, 00, 000 molecules, then find out the number of moleculesof oxygen and hydrogen present in the other cylinders.SAQ 2.There are 1 lack molecules each of NO and SO

2 in two different cylinders of V litre

capacity each at 300C. The pressure in NO cylinder is 860 mm of mercury. What is the pressurein the SO

2 cylinder?

Experimental validity:Avogadro's law explained satisfactorily the Gay Lussac's Law of combining gaseous volumes.In fact the success of the Avogadro's law is based on the Gay Lussac's experimental observations.Gay Lussac experimentally found that:Under identical conditions of temperature and pressure, if gaseous reactants completely reactto give gaseous products then there exists a whole number ratio always between the volumes ofthe reactants and products.(i) 1 volume of hydrogen gas combines with 1 volume of chlorine gas to give

2 volumes of hydrogen chloride gas(ii) 2 volumes of hydrogen gas combine with 1 volume of oxygen gas to give

2 volumes of water vapourAvogadro's law (the then hypothesis) could corroborate the Gay Lussac's findings stated above.Two assumptions made by Avogadro were that elementary gases like hydrogen, chlorineetc. are diatomic and one molecule of hydrogen chloride contains one H atom and oneCl atom.

H2 + Cl

2 → 2 HCl

1 molecule of hydrogen + 1 molecule of chlorine → 2 molecules of hydrogen chlorideSo x molecules of hydrogen + x molecules of chlorine → 2x molecules of hydrogen chlorideApplying Avogadro's hypothesis if 1 volume of hydrogen contains x molecules, 1 volume of

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Atoms and Molecules

Dr. S. S. Tripathy

other gases also would contain x molecules at the same temperature and pressure.1 volume of hydrogen + 1 volume of chlorine → 2 volumes of hydrogen chlorideThis is exactly what Gay Lussac observed experimentally.

SAQ 3: Indicate what volume of gaseous products would be formed from the following gaseousreactions. Assume complete reaction under constant temperature and pressure conditions.

(i)1 litre of nitrogen gas + 3 litres of hydrogen gas → ? litres of ammonia gas(ii)3 l of carbon monoxide gas + 1.5 l of oxygen gas → ? l of carbon dioxide gas(iii)10 l of hydrogen gas + 10 l of iodine vapours → ? l of hydrogen iodide vapours(iv)5 l of nitrogen monoxide gas + 2.5 l of oxygen gas →? l of nitrogen dioxide gas(v)150ml of methane gas + 300ml of oxygen gas →? ml of carbon dioxide +? ml of water vapour

Gram molar volume and Avogadro's LawJohann Josef Loschmidt(1865) and Jean Baptiste Jean Perrin(1909) were the scientists whoare remembered for the discovery of the famous number 6.023 X 1023 called Avogadro'sNumber i.e the number of species present in one mole of any substance.Again Cannizzaro experimentally found that one mole of any gas at NTP occupied 22.4 litres(grammolar volume). So from the concepts of Avogadro's number and gram molar volume, we can saythat:if the number of molecules of different gases taken in different vessels is each 6.023 X 1023 atNTP then each gas will occupy 22.4 litres. Conversely if the volume of each gas is 22.4 litres atNTP then each will contain 6.023 X 1023 molecules.

Pressure= 760 mmTemp. = 00CVolume = 22.4 LNo. of molecules = 6.023 X 10 No. of molecules = 6.023 X 10

Volume = 22.4 LTemp. = 00CPressure= 760 mm

No. of molecules = 6.023 X 10Volume = 22.4 LTemp. = 00CPressure= 760 mm

CO2 O2 Cl2

23 23 23

SAQ 4:(i) Supposing half mole of cows are grazing in a certain field. How many cows are therethen?

(ii)Supposing you purchased 12.046 X 1024 number of apples from the market. Howmany moles of apples you bought?

(iii)A 5 litre N2 gas cylinder was found to contain 1 lack molecules of N

2 and another

cylinder of unknown volume of oxygen was also found to contain 1 lack molecules at the sametemperature and pressure. What is the volume of the oxygen cylinder?

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Atoms and Molecules

Dr. S. S. Tripathy

(iv)Two gases X and Y kept in two different vessels at 270C and 1 atm. pressure. GasX occupied 1.5 litres and found contain 30,000,00 molecules. If the volume of gas B is 3 litres,how many molecules gas B might be having?

(v)How long it will take for a person to count Avogardo's number at the rate of fivecounts per second?

Alternative definition of Avogadro's Law:At constant temperature and pressure, if number of molecules(or moles) becomes more,

the volume of the gas becomes more, Avogadro's law can be stated as follows."The volume of any gas is directly proportional to the number of moles at constanttemperature and pressure".

V ∝ n, (at constant P and T) where 'n' is the number of moles

⇒ V = k n (where k is the proportionality constant)

⇒ Vn = k ⇒ V

n = nV1 2

1 2(at constant P and T)

SAQ 5: 1000 molecules were removed from a 200 ml of gas containing 5000 molecules at 270Cand 800mm fixed pressure. Will there be any change in volume? If so what will be the newvolume.

APPLICATION OF AVOGADRO'S LAW(i) To prove thatVapour Density or Relative Density of gas = (Molecular Mass)/2Before we establish the relationship between the vapour density and molecular mass of any gas,let us know what is meant by vapour density. You know that density of a solid or liquid whencompared with density of water, we call the ratio, specific gravity or relative density of thatsubstance. Since water has a density equals to 1gm per cm3, the specific gravity of any solid orliquid is numerically equal to density of that substance in gm per cm3. Say for example, thedensity of mercury is 13.6gm per cm3 i.e the density of mercury is 13.6 times greater thandensity of water(1gm/cm3). In other words mercury is 13.6 times heavier than water.Unlike liquids and solids, gases have very very low densities. Say for example the density ofhydrogen gas at NTP is 0.000089 gm per cm3, density of oxygen gas at NTP is 0.0014gm/cm3

and the like. Therefore the densities of gases are not compared with water, in stead, comparedwith density of the lightest gas hydrogen. HenceRelative density(R.D) or Vapour Density(V.D) of a gas:

=Density of a gas or vapour

Density of hydrogen gas(both measured under same temperature and pressure)

Can you calculate the relative density of oxygen gas from the densities of hydrogen and oxygengases given above. It is 0.0014/0.000089 = 16. Is it not half the molecular mass(32) of O

2 gas?

Now let us try to prove the relationship between vapour density and molecular mass of a gasusing Avogadro's law.

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Atoms and Molecules

Dr. S. S. Tripathy

=

mass of the gas or vapourvolume of the gas or vapour

mass of hydrogen gas

volume of hydrogen gas

V.D

If both the volumes are same, volume terms are cancelled out to give

=Mass of certain volume of gas or vapour

Mass of same volume of hydrogenV.D

(measured under same temperature and pressure)

=V.D Mass of V cc of the gas or vapourMass of V cc of hydrogen gas (let us take volume=V cc)

Let V cc of the gas contains 'n' molecules . Applying Avogadro's law, we have

V.D =Mass of 'n' molecules of the gas or vapou

Mass of 'n' molecules of hydrogen gasDividing by 'n' to both numerator and denominator, we have

=Mass of 1 molecule of the gas or vapour

Mass of 1 molecule of hydrogen gas

Since hydrogen molecule is diatomic(H2),

V.D =Mass of 1 molecule of the gas or vapour

Mass of 2 atoms of hydrogen gas

=Mass of 1 molecule of the gas or vapour

2 X Mass of 1 atom of hydrogen gas

= Molecular Mass2

Because we know that molecular mass = (mass of one molecule)/ (mass of one Hatom).IMPORTANT: Although density of a gas is dependent on temperature and pressure, itsvapour density is independent of any external conditions. Since molecular mass of asubstance is fixed, its V.D is also fixed. This is because V. D is actually relative density. Ifthe density of a gas changes due to change in external conditions, density of H

2 will also

change in the same proprotion and their ratio will always remain constant.SAQ 6: (i)What are the vapour densities of the following gases

(a) Carbon dioxide (b)nitrogen (c)sulphur dioxide(d)hydrogen chloride (e)nitric oxide(ii) How many times the density of ammonia is greater than the density of hydrogen ata certain fixed temperature and pressure.(iii)A certain unknown gas has a density equals to 0.00125gm/cc at NTP. What are itsvapour density and molecular mass?Can you identify the gas?(density of hydrogen gas

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Atoms and Molecules

Dr. S. S. Tripathy

at NTP=0.000089gm/cc)(iv)The vapour densities of a few gases are listed below. Identify the gases.(a)X= 40(it is gas of S and O) (b)Y= 24(a gas of O) (c)Z=15(a gas of N and O)(v)The vapour density of SO

2 gas at NTP is 32. What is its density?(density of hydrgen

is 0.000089gm/cc at NTP)(vi)The V.D of oxygen gas is 16 at 270C, what would be its VD at 1000C?

(ii) Atomicity of Hydrogen gas is 2Let us now prove on the basis of Avogadro's law that gaseous hydrogen molecule is diatomic i.eits atomicity is 2.Gay Lussac's experiment1 vol. of hydrogen gas + 1 vol. of chlorine gas → 2 vols. of hydrogen chloride gas(measured under same pressure and temperature)If 1 vol. of gas contains 'n' molecules, then according to Avogadro's law'n' molecules of hydrogen + 'n' molecules of chlorine → '2n' molecules of hydrogen chloride1 molecule of hydrogen + 1 molecule of chlorine → 2 molecules of hydrogen chlorideDividing by 2, we have½ molecule of hydrogen + ½ molecule of chlorine → 1 molecule of hydrogen chlorideOne molecule of hydrogen chloride contains only one H atom as it forms only one type of salt(e.g NaCl with NaOH)and not two types of salts as produced by sulphuric acid i.e NaHSO

4 and

Na2SO

4 with NaOH).This H atom of hydrogen chloride must have come from ½ molecule of

hydrogen.So, ½ molecule of hydrogen = 1 atom

1 molecule of hydrogen = 2 atomsAtomicity of hydrogen =2 and the hence formula of hydrogen gas is H

2

(iii)Atomicity of oxygen is 2The following Gay Lussac's experiment is taken2 vol. of hydrogen gas + 1 vol. of oxygen gas → 2 vol. of water vapour(under same temperature and pressure)Applying Avogadro's law'2n' molecules of hydrogen + 'n' molecules of oxygen → '2n' molecules of water vapour2 molecules of hydrogen + 1 molecule of oxygen → 2 molecules of water vapour1 molecule of hydrogen + ½ molecule of oxygen → 1 molecule of water vapourFrom electrolysis experiment, it was known that the number of atoms of H and O in watermolecule is 2 and 1 respectively as 18 gms of water gave 2 gms of hydrogen(correspond to 2 Hatoms) at cathode and 16gms of oxygen(correspond to one O atom) at anode.

So ½ molecule of oxygen = 1 atom1 molecule of oxygen = 2 atoms. Hence oxygen gas is diatomic(O

2)

(iv)Gram Molar Volume(GMV) = 22.4 liters at NTPYou know that one mole of any gas at NTP occupies 22.4 litres or 22400ml(cc). Let us see howit can be proved from Avogadro's law.

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Atoms and Molecules

Dr. S. S. Tripathy

Mass of 1 litre of gas or vapour at NTP Mass of 1 litre of hydrogen gas at NTP

2 XMolecular Mass = 2 X V.D=

(here we have taken the fixed volume to be 1 litre)

= 2 X mass of 1 litre of gas or vapour at NTP

0.089

(because the density of hydrogen at NTP=0.000089gm/ml=0.089gm/l) = 22.4 X Mass of 1 litre of gas or vapour at NTP = Mass of 22.4 litres of the gas or vapour at NTP

Hence the molecular mass is the mass in gm of 22.4 litres of the gas at NTP. In other words, onemole of any gas at NTP will occupy 22.4 litres.

SAQ 7: Experiments show that 1 volume of nitrogen gas reacts with 3 volumes of hydrogen gasto produce 2 volumes of ammonia at the same temperature and pressure. From this prove thatnitrogen molecule is diatomic (Use Avogadro's law). (The formula of ammonia is NH

3)

SAQ 8: What will be volumes of the following gases in cm3 at NTP(i) ½ mole of nitrogen gas (ii)5 moles of CO gas (iii)0.01 moles of SO

2

How Avogadro's Number(N0) was determined ?

It is humanly not possible to calculate this incredibly large number. Nobody has determinedon the basis of real counting. The calculation was made by indirect methods by several scientists.1. First Loschimdt in 1865 determined the number of molecules present in 1 cc of a gas atNTP by using the Kinetic Theory of gases(using molecular diameter and mean free path) to be2.6 × 1019. Subsequently Maxwell in 1873 determined this number to be 1.9 × 1019. It wasalready known by that time that 1 mole of a gas occupies 22400 mL at NTP. So the number ofmolecules present in 22400 mL was calculated to be 4.3 × 1023.2. Slightly later Kelvin determined this number of molecules in 22400 mL of a gas at NTPusing light scattering experiment to be 5 × 1023.3. In 1908 J. Perrin determined this number using his Brownian motion study in liquids to liebetween 6.5 × 1023 to 6.9 × 1023.4. Rutherford and Geiger used radioactive method(emission of alpha particles from radiumand uranium) to determine this number to be 6.2 × 1023.5. By 1933, more than 80 separate determinations of Avogadro's number were made whichgave nearly same value of the constant but there was no clear agreement on that constant. Thebest modern value of Avogadro's number which was universally accepted came from x-raydiffraction studies of crystals from the measurements of lattice parameters(details not discussedhere). From density of a metallic crystals and unit cell edge length avogadro's number wasdertermined. This value is 6.02214199 × 1023 which has been universally accepted.

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Atoms and Molecules

Dr. S. S. Tripathy

RESPONSE TO SAQsSAQ 1: O

2 = 20,00, 000 and H

2=20,00,000 (same in each case)

SAQ 2: 860 mm(same)SAQ 3: (i) N

2 + 3H

2 ------>2NH

3

First balance the equation. Look to the coefficient of the reactants and products. The volumes ofreactants and products are proportional to their coefficients. Note that for the Gay Lussac's lawto hold good it is required that all the reactants and produts are gaseous.So 1 litre of N

2 reacts completely with 3 litres of H

2 to give 2 litres of ammonia

(ii) CO+½ O2 ---------> CO

2

3 litres of CO gas reacts with 1.5 litres of O2 gas to form 3 litres of CO

2 gas

(iii) H2 + I

2 ------->2 HI

So 20 litres of hydrogen iodide gas will be produced.(iv) NO + ½ O

2 --------> NO

2

So 5 litres of NO2 will be formed

(v) CH4 + 2O

2 --------> CO

2 + 2H

2O

So 150ml of CO2 and 300ml of water vapour will be formed.

SAQ 4:(i)If one mole of any substance contains Avogadro's number of that substance then thenumber of cows present in half mole is ½ X 6.023X1023 = 3.0115 X 1023

(ii)6.023X1023 number of apples = 1 mole of applesSo 12.046 X1024 number apples = 20 moles of apples

(iii)If the number of molecules are same for two gases at equal temperature and pressure,then they will occupy the same volume. So the volume of oxygen cylinder is 5 litres.

(iv)At the identical temperature and pressure if the the volume of gas Y is two timesmore than that of X then the number of molecules of Y will be two times greater than X. Hencegas Y will contain 60,000,00 molecules.

(v)Time = (6.023 × 1023) /5 secs = 38.19 × 1014 years.SAQ 5: Yes, there will be change in volume as the pressure and temperature are kept constantin the two cases. According to Avogadro's law, the volumes of the gases will be directly proportionalto the number of molecules.

V1/V

2 = n

1/n

2

So 200ml/V2 = 5000/(5000-1000)= 5/4

⇒ V2 =(200X4)/5 = 160ml. So the new volume is 160ml.

SAQ 6: (i) (a)The Molecular Mass of CO = 12+16=28, hence its V.D =28/2=14(b)The molecular Mass of N

2= 28, So V.D=28/2=14

(c)SO2 = 32+32=64, So VD=32 (d)HCl= 1+35.5=36.5, V.D=18.25

(e)NO=14+16=30, VD= 15(ii)VD of NH

3 = 17/2 = 8.5, so ammonia gas has 8.5 times greater density than hydrogen.

(iii)VD= density of the gas/density of hydrogen = 0.00125/0.000089= 14. So its molecularmass will be 28. The gas could be either CO or N

2.

(iv) (a)M.M of gas X = 80, hence the gas is SO3(32+48)

(b)M.M of gas Y= 48, hence the gas is O3(ozone)

(c)M.M of gas Z= 30, hence the gas is NO(14+16)(v)V.D = (density of a gas) / (density of hydrgen) at the same temperature and pressure

So density of the gas = 0.000089X32=0.00284gm/cc

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Atoms and Molecules

Dr. S. S. Tripathy

(vi) V.D of oxygen will remain same(16). Note that VD is independent of temperatureand pressure. Although the densities of gases are different at different temperatures andpressures, but since VD is a ratio of two densities at the fixed temperature and pressure, it is aconstant quantity for a given gas. Since molecular mass of a gas is fixed, its VD is bound to befixed.SAQ 7: 1 vol. of nitrogen + 3 vols. of hydrogen ----------> 2 volumes of ammoniaApplying Avogadro's law we have,

n molecules of nitrogen + 3n molecules of hydrogen-----> 2n molecules ofammonia

1 molecule of nitrogen+ 3 molecules of hydrogen -----> 2 molecules of ammonia

½ molecule of nitrogen + 32 molecule of hydrogen -----> 1 molecule of ammonia

Since one molecule of ammonia contains one N atom(given), so½ molecule of nitrogen contains 1 atom of N1 molecule of nitrogen contains 2 atoms of N. Hence nitrogen gas is diatomic(N

2).

SAQ 8: (i)22.4/2 = 11.2 l= 11200cc (ii)22.4X5=112l=112000cc(iii)22.4X0.01=0 224 litres =224 ml.

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Atoms and Molecules

Dr. S. S. Tripathy

ATOMIC MASS

Atomic mass of an element is a relative mass of an atom which is expressed as compared to astandard element like, H, O or C.Hydrogen standard:

Atomic Mass = Mass of one H atom Mass of one atom of an element

In this scale the mass of one H atom is arbitrarily fixed at 1.000. Atomic mass of an elementmeans how many times the mass of one atom of an element is heavier than mass of one hydrogenatom.On H scale the atomic mass of oxygen was found to be 15.87 and that of carbon to be 12.02 i.eone atom of oxygen is 15.87 times heavier than one atom of hydrogen and one atom of carbon is12.02 times heavier than one atom of hydrogen. The atomic mass of oxygen is 15.87. This doesnot mean that mass of one oxygen atom is 15.87 gm. How can this be possible? The individualatom is so small and tiny that it cannot weigh 15.87 gm. Atomic mass hence is a relative massas compared to hydrogen or any other element.Carbon standard: Atomic Mass Unit(amu) or Carbon Unit(cu) or Unified Atomic MassUnit(u)From 1960 onwards, this standard is now universally accepted and comparison according tohydrogen standard is not being used nowadays.It is called the Atomic Mass Unit(AMU) or theUnified atomic mass unit(u) or carbon unit(cu) or dalton.

Atomic Mass = Mass of one atom of an element1 of the mass of one C atom12

a.m.u or c.u or u

In this definition the mass of one C atom(C-12 isotope) was presumed to be 12.00 a.m.u(i.enearest whole number of atomic mass of C obtained from H scale).If we make 12 equal slices (pieces) to a carbon atom whose atomic mass is 12 amu(each piecei.e 1/12 part of one carbon atom is called 1 amu), then how many times one atom of any otherelement is heavier than 1/12 part(one small slice) of carbon atom is atomic mass of the element.On the basis of carbon scale the atomic mass of H was found to be 1.0079 a.m.u(not 1.000taken in the hydrogen scale for making comparison with other atoms), and that of O is15.99491a.m.u and that of Na is 22.98977 a.m.u and so on.For the purpose of chemical calculations we approximate these atomic masses to their respectivenearest whole numbers, i.e 1 for hydrogen, 23 for sodium and 16 for oxygen and so on. Rememberthat the atomic mass of element is never a whole number excepting the carbon-12 isotopewhose atomic mass has been fixed arbitrarily at 12.0000. All other elements have non-whole number atomic masses (with figures after decimal point) but for simplicity in rememberingtheir values and for easy chemical calculations we use the value of atomic masses rounded offto the nearest whole number.SAQ 1: (i)In CU scale which element has a whole number atomic mass and what is that value?

(ii)In CU scale, the atomic mass of H is not 1, while in H scale, the atomic mass of C isnot 12. Justify.

(iii)Atomic mass of phosphorus is 31 and fluorine is 19. Is the statement correct? Justify

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Atoms and Molecules

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your answer.Mass Number: The number of protons and neutrons present in the nucleus is called the massnumber. Mass number of oxygen is 16, that of fluorine 19 and phosphorous 31 and so on. Notethat when atomic mass is rounded off to its nearest whole number we get its mass number.

SAQ 2: The atomic mass of oxygen is 15.99491a.m.u. What is its mass number? If its atomicnumber is 8, how many protons, neutrons and electrons it has?SAQ 3: Below given are the atomic masses of few elements. Find their mass numbers. Alsofind the number of neutrons present in each of them. The number of protons are indicated withinbrackets.

(i) 38.97377(Potassium: 19) (ii)126.9(Iodine: 53) (iii)107.89 (Ag: 47)SAQ 4: The atomic mass of oxygen is 16.0 amu.Is the statement correct? If not why? what isthe mass number of oxygen?

Gram Atomic Mass: Atomic mass when expressed in gm is called gram atomic mass. Whatdoes it actually imply? To understand this, I want to ask you a question. Could you tell the massof a sodium atom? Some of you may instantly answer it is 23gm. But it is wrong. You may say 23amu, which is correct as one atom of sodium is 23 times heavier than 1/12 part of a carbon atombut not 23 grams. How can a tiny atom weigh so much? Actually gm atomic mass is the massin gram containing Avogadro's number(N) of atoms. 23gms of Sodium contain 6.023X1023

atoms of sodium. Then you can calculate the mass of one atom of sodium in gram. We now stopfurther discussion on this and shall come back to it in the mole concept chapter. So we definegram atomic mass as the mass in gm. of Avogadro's number of atoms of the element.

SAQ 5: What is the value of one amu in gm.

ISOTOPES: (Average Atomic Mass)The atomic mass about which we studied before should in fact be called as isotopic masswhich is the relative mass of a particular isotope of the same element.Then the question arises what are isotopes? A simple example will make the point clear. Let ustake the case of chlorine. If you analyse chlorine gas obtained from any source, you will findtwo varieties of chlorine atoms, one having atomic mass 34.96885(mass number 35) and theother having atomic mass 36.96590(mass no. 37) in the ratio 3:1. This means that for every threeatoms of Cl35 atoms we shall find one Cl37 atom. These are called isotopes of chlorine. They arecalled Cl35 and Cl37 isotopes expressed in terms of their mass numbers. So isotopes are thesame elements (not different elements) i.e have the same identity but they differ only intheir atomic masses(Mass number). They differ in the number of neutrons present in theirnuclei while the number of protons is same in the isotopes. The atomic masses of individualisotopes are called isotopic masses. The isotopic mass of Cl35 is 34.96885 and that of Cl37

isotope is 36.96590. But in the periodic table and other places we find the atomic mass ofchlorine to be 35.5 not the isotopic masses any one of the two isotopes.This is average of the twoisotopic masses of the element. Do you know how the average atomic mass is calculated? It isnot the number average we often find i.e (35+37)/2=36. It is found in a different way and iscalled weight average. It is found as follows

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Atoms and Molecules

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Atomic mass of Cl = 3 X 34.96885 + 1 X 36.96590

3 + 1 ≈ 35.5

Or we can simplify the calculation by using their mass numbers instead of isotopicmasses.

Atomic mass of Cl = 3 X 35 + 1 X 37

3 + 1 ≈ 35.5

Since the Cl35 and Cl37 are always available in the ratio 3:1, the isotopic mass of Cl35 contributes3/4 and Cl37 contributes 1/4 to the average. Therefore we multiplied 3 with the isotopic mass ofCl35 and 1 with isotopic mass of Cl37 and took their sum and finally we divided this sum by thesum of 3 and 1(i.e 4). This gives the average atomic mass of 35.5 that we use in all chemicalcalculation involving chlorine.In terms of percentage we can also do the same thing by taking their relative abundance(availability) ratio as 75:25.

Atomic mass of Cl = 75 X 35 + 25 X 37

100 ≈ 35.5

Isotopes vary with respect to number of neutrons present in their nuclei. For example, Cl35

isotope has 18 neutrons(35-17) and Cl37 isotope has 20 neutrons(37-17), since the atomic numberof Cl is 17.The average atomic mass of a polyisotopic element can be determined from the followingrelationship.

Average atomic mass=(%)1 X (I.M)1 + (%)2 X (I.M)2 + and so on

100where (%)

1 is the percent composition of isotope 1 and (I.M)

1 is the isotopic mass of isotope 1

and so on.

Most of the elements remain as more than one naturally occurring isotopes. Isotopes of anelement have the same chemical identity i.e same atomic number but they differ in themass number i.e the number of neutrons. The isotopes vary in their atomic masses.

(i) OXYGEN: O exists as O16, O17 and O18 isotopes with a natural percent composition(abundance) as 99.78, 0.02 and and 0.2 % respectively. O16 is the most abundant (nearly100%) isotope in the mixture. So when we speak of O it is O16 having mass number 16. Althoughthe mass numbers of the isotopes are whole numbers 16,17 and 18 their atomic masses (isotopicmasses) are fractional i.e 15.99491, 16.99913 and 17.99916 respectively. But for simplicity weoften say that their atomic masses are 16, 17 and 18 respectively.

SAQ 6: How many protons and neutrons are available in the three isotopes of Oxygen.SAQ 7: Can the atomic mass of C13 isotope be exactly 13? If not for which isotope the atomicmass can be exactly a whole number?

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(ii) CARBON:Carbon has mainly two naturally occurring isotopes C12(98.88%), C13(1.12%)and the most abundant among the two is C12. This C12 isotope has been taken as the standard inthe amu scale for comparing the isotopic masses of all other isotopes. Note that carbon hasanother isotope C14 which is extremely scarce in nature and is radioactive and we shall studyabout it in other chapter.(iii) HYDROGEN: Hydrogen has two naturally occurring isotopes, H1(protium or hydrogen),H2 or D2(Deuterium or heavy hydrogen) with the percent composition of 99.9855 and 0.0145%and 10-15 % respectively. Another isotope of H is H3(or T3 called the tritium)also which isextremely scarce in nature and is radioactive. Excepting hydrogen isotopes of an element havethe same name. Deuterium and tritium are not to be mistaken as different elements.(v) BROMINE: Br has two isotopes Br79 and Br81 in the ratio 1:1(50:50)(vi) SULPHUR: S has two isotopes S32 and S34 in the ratio 95.6: 4.4Most of the elements that you find in the periodic table exists in more than one isotopic forms. Soin these cases we make use of the average atomic mass which we commonly say atomic mass.There are few elements which unfortunately remain in one isotopic form. For example F19 hasonly one isotope having atomic mass 18.99840. So in this case the atomic mass is same as itsisotopic mass.

Monoisotopic Elements: There are only 19 elements which do not have more than one isotope.These are called monoisotopic. Be9 , F19, Na23, Al27, P31, Mn55, Co59, I127, Cs133, Au197 are thecommon ones and Sc45, As75, Y89,Nb93, Rh103, Pr141, Tb159, Ho165, Tm169 are the uncommonones. Try to remember the common ones at least.The remaining elements have at least two or more than two isotopes. Sn(tin) has the largestnumber of naturally occurring stable isotopes of 10 and Xenon has the second highest 9).SAQ 8: Among the following elements which elements are monoisotopic and which arepolyisotopic? Br, O, F, Al, Na, Fe, Cl, S, Mg, Ca, K, I, C, N, Au, PSAQ 9: Si has three isotopes having isotopic masses 27.97693, 28.97650 and 29.97377. Whatare their mass numbers? The atomic number of Si is 14, find the number of neutrons in eachcase.SAQ 10: Is it wrong to take mass numbers in stead of isotopic masses for calculating theaverage atomic mass? Explain.SAQ 11: Br has two isotopes Br79 and Br81 in the ratio 1:1. What is its average atomicmass?SAQ 12: Oxygen has three isotopes: O16, O17 and O18 in the ratio 99.78:0.02:0.2. Is theaverage atomic mass of oxygen close to 16 or 18 ? What will be its average atomic mass?Calculate by using the mass numbers. What is this average called? How it is different from thecommon number average. Show by calculation.SAQ 13: The average mass of O is 15.9994 . What value should we use in chemicalcalculations?SAQ 14: Br has two isotopes Br79 and Br81. What atomic mass of Br we use in chemicalcalculations and why?SAQ 15: Hydrogen has two naturally occurring isotopes (H1 and H2) having isotopicmasses 1.008 and 2.014 respectively with a composition ratio, 99.985 : 0.015. Find the average

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atomic mass of H. Will the atomic mass be almost same as the atomic mass of H1 i.e 1.008(andnot close to the atomic mass deuterium)? If so why?SAQ 16: For every 10, 000 oxygen atom that you will count how many O16, O17 and O18

isotopes you will find?SAQ 17: Indicate which isotope is most abundant in the following element. Cl, O, S, H, C

Determination of Isotopic and Atomic Masses:In earlier times Cannizzaro and other workers like Dulong and Petit had used very crude methodsto find the atomic masses. They determined the atomic mass values on the basis of some physicaland chemical methods and these were merely average isotopic masses. For example, the averageatomic mass of Cl is 35.5 and Br is 80 and so on. The discussion on these methods will be madein the next chapter(equivalent mass). No method was known at that time to separate the differentisotopes of an element and get the individual isotopic masses. For example there are threedifferent varieties of O atoms having mass numbers 16, 17 and 18. These could not be found outexperimentally in earlier times. But after the discovery of an instrument called Massspectrometer by Aston in 1919, it was possible not only to separate the individual isotopesfrom each other but also find out their isotopic masses. The first success came in determiningthe isotopic masses of two isotopes of Ne as 20 and 22. At that time, Neon was known to havetwo isotopes. But later the third isotope Ne(21) was discovered. Today, powerful massspectrometers are available which give the isotopic masses correct upto many places (at least 7places) of decimal. Table below shows a list of the naturally occurring elements indicating theirnumber of stable and radioactive isotopes and their average atomic masses.Preliminary Idea about separation of isotopes and determination of isotopic masses ina mass spectrometer:

Unipositve ions are produced from an element in its gaseous state in a cathode ray tube. This isachieved by the bombardment of cathode rays which is nothing but a powerful electron beamonto the gaseous atoms. The knocking out of one electron from each atom converts them tounipoistive ions. Note that the masses of these ions are almost same as the masses of the neutralatoms as electron carries negligible mass. Since the element exists in more than one isotopicforms, their masses are different although their charges are same(+1). For instance, if we takechlorine gas, there would be two types of positve ions i.e 35Cl+ and 37Cl+ produced from the twoisotopes of chlorine. Similarly for Neon gas there would be three positive ions 20Ne+, 21Ne+ and22Ne+ from the three isotopes and so on. This mixture of positive ions of an element is thenallowed to passed through an electric field first and then through a magnetic field. The postiveions are deflected in an electric field towards the -ve plate of the field. But all the ions are notdeflected to the same extent. Depending on their masses, the deflections are different. Ionsfrom heavier isotopes are deflected to the lesser extent while the ions from lighter isotopes aredeflected to a larger extent. For instance, for chlorine, the two postive ions 35Cl+ and 37Cl+ getseparated into two beams, the 35 isotope being deflected more towards the positve plate ofelectric field. To achieve better separation and get sharper focus of ions, Aston used a magneticfield in succession to the electric field. The separated beam of isotopic ions get further separatedin the magnetic field and take a curved path. The magnetic field is applied perpendicular to thedirection of motion of the ion beam(current). So the mechanical effect is produced as per Fleming's

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Left hand rule perpendicular to the plane carrying the magnetic field and direction of motion ofthe beam. Thus all the isotopic ions take different curvatures, the lighter isotope turns to thegreatest extent while the heaviest to the minimun extent. These focussed ions coming at differentradii of curvature are allowed to fall on a calibrated photographic plate to produce sphericalspots at the appropriate isotopic masses. From position of the spot the isotopic masses could beknown and from the intensity of the spot, relative natural abundance could be determined. Chlorinefor instance produced two spots, one at mass number 35 and the other at 37 and intensity ratio ofthese spots were found to be 3:1 which is their relative abuandance. Note that as many numberof spots are observed for a particular element, same would be the number of isotopes.

N

S

IONISATION CHAMBER

Electron Beam

Cl(35) +

Cl(37) +

Cl(37) +

35

37

magnetic field

Nowadays, with the disovery of powerful electromagnets using superconducting materials,and the interfacing of computer, isotopic masses correct upto 7 places of decimal are availabe inthe form of lines in the computer monitor what is called the mass spectrum. The number of massspectral lines give the number of isotopes and their heights give their relative abundances.Elements with their stable and radioactive isotopes and average atomic mass:

Element At.No No. of stable No. of natural av. at. massisotopes radioisotopes

Ac 89 - 2 227Al 13 1 - 26.98Ar 18 3 - 39.948As 33 1 - 74.92At 85 - 3 210Ag 47 2 - 107.87Au 79 1 196.97Ba 56 7 - 137.34Be 4 1 - 9.012Bi 83 1 3 208.98B 5 2 - 10.81Br 35 2 - 79.909

3537

1

3

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Ca 20 6 - 40.08Cd 48 8 - 112.40C 6 2 1 12.011Ce 58 4 - 140.12Cs 55 1 - 132.905Cl 17 2 - 35.453Cr 24 4 - 52.00Co 27 1 - 58.93Cu 29 2 - 63.54Dy 66 7 - 162.54Er 68 6 - 167.26Eu 63 2 - 152.0F 9 1 - 19.0Fr 87 - 1 223Gd 64 7 - 157.2Ga 31 2 - 69.72Ge 32 5 - 72.59Hf 72 6 - 178.49He 2 2 - 4.003Ho 67 1 - 164.93H 1 2 - 1.00797Hg 80 7 - 200.59In 49 1 1 114.82I 53 1 - 126.9Ir 77 2 - 192.2Fe 26 4 - 55.85K 19 2 1 39.102Kr 36 6 - 83.8La 57 1 - 138.91Li 3 2 - 6.939Lu 71 1 1 174.97Mg 12 3 - 24.31Mn 25 1 - 54.94Mo 42 7 - 95.94Na 11 1 - 22.990Nd 60 6 1 144.24Ne 10 3 - 20.183Ni 28 5 - 58.71Nb 41 1 - 92.91N 7 2 - 14.007Os 76 7 7 190.2O 8 3 - 15.9994Pd 46 6 - 106.4P 15 1 - 30.974

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Po 84 - 7 209.0Pt 78 4 2 195.09Pr 59 1 - 140.91Pm 61 No naturally occuring isotopes 147

as it is a synthetic elementPb 82 4 4 207.19Pa 91 - 2 231.0Ra 88 - 4 226.0Rn 86 - 3 222Re 75 1 1 186.2Rh 45 1 - 102.905Rb 37 1 1 85.47Ru 44 7 - 101.1Sm 62 6 1 150.35Sc 21 1 - 44.96Se 34 6 - 78.96Si 14 3 - 28.09Sb 51 2 - 121.75Sr 38 4 - 87.62S 16 4 - 32.064Sn 50 10 - 118.69Ta 73 2 - 180.95Tc 43 No naturally occuring isotopes 98

as it is a synthetic elementTe 52 8 - 127.6Tb 65 1 - 158.92Tl 81 2 4 204.37Th 90 - 6 232.04Tm 69 1 - 168.93Ti 22 5 - 47.90U 92 - 3 238.03V 23 2 - 50.94W 74 5 - 183.85Xe 54 9 - 131.30Yb 70 7 - 173.04Y 39 1 - 88.904Zn 30 5 - 65.37Zr 40 5 - 91.22

In chemical calculations, the atomic mass that you shall use should be the nearest whole numberof the average atomic mass given in the above table. For K, for example, you should use 39instead of 39.12 given in the table. For elements having average atomic mass far away from thewhole number value(e.g Cl=35.5), you have to use as such. But from theoretical point of view,you should remember that the actual isotopic masses as well as the average atomic masses arenot whole numbers (excepting a C-12 isotope).

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DULONG'S AND PETIT'S LAW:(Study this law after completing the chapter 'Equivalent Mass')This is one of the oldest laws used to determine approximate average atomic masses of solidelements at room temperature and above.

In 1819 French scientists Pierre Dulong and Alexis Petit proposed this which can thefollowing alternative definitions.

(1) For solid element, the product of its specific heat capacity and atomic mass isapproximately equal to 6 cals K-1mol-1(25 JK-1mol-1).

Specific heat × atomic mass = 6 cals K-1mol-1(25 JK-1mol-1) (approximately)(2) The molar heat capacity is approximately constant for solid elements and is equal to

3R.(R = gas constant).(3) The amount of heat required to raise the temperature of a single atom of a solid is

independent of the type of atom. In other words the heat required to raise the temperature of 1gram atomic mass( one mole) of the solid element by 10C is approximately constant. This iscalled molar heat capacity which is approximately constant for all solid elements.Limitations of Dulong and Petit's Law:

1. This law is applicable to elements which are solids at room temperatures and mostlyfor metals. Large deviations were observed for solid nonmetals like sulphur, phosphorous, iodineetc.

2. Large deviations occured while measuring specific heats of S, Co, Te and Pt.3. The law is applicable only at ordinary temperature to elements having simple

crystal structures. At low temperatures and for complex crystal structures this law was notvalid.

4. The atomic mass obtained was approximate and not exact.Application of Dulong and Petit's Law:

Determinationa of exact atomic mass of elements:With the help of the equivalent mass of element which is always exact, the exact atomic masscan be determined.Step 1: Equivalent mass of the element is accurately determined by any experimental methodsuch as -

(a) Hydrogen displacement method (b) oxide method (c) chloridemethod (d) displacement method (e) double displacementmethod (f) electrolysis methodStep II: With the help of Dulong and Petit's law the approximate atomic mass is determined fromits specific heat data. Approximate atomic mass = 6/specific heatStep III : The approximate valency is obtained as follows.

Approximate valency = Approximate atomic mass/ equivalent massStep IV: Determination exact valency: Since valency is a whole number, the exact valency isobtained by rounding off the approximate valency to the nearest whole number.Step V : Correct atomic mass is obtained as follows : Exact atomic mass = Exact valency Xequivalent mass(note that the accuracy of this determination depends on the accurary of measurement ofequivalent mass by the experimental method)Example : A metallic element has specific heat 0.11 cal/g/K. Its oxide contains 22.27% oxygen.Calculate the correct atomic mass of the element.Solution: m(metal)/m(oxygen) = E(metal)/E(oxygen) 77.73/22.27 = x/8, So x = 27.92(EM)From Dulong and Petit's law : Approximate atomic mass = 6/0.11 = 58.182Approx. valency = 58.182/27.92 = 2.083

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So exact valency is 2Exact atomic mass = 27.92 X 2 = 55.92.SAQ 18: An element was found to have specific heat capacity 0.0276 cal/g0C. If 114.79 g of achloride of this element contained 79.34g of the metal element, calculate the exact atomic massof the element.

Do you know why the atomic mass standard was changed fromHydrogen to Carbon ?

Dalton experimentally found the atomic masses of some elements and hydrogen wastaken as the standard for comparision because it is the lightest element. H was taken as 1.0000amu and other elements have fractional atomic masses. Subsequently it was found H did notcombine with many elements, in stead oxygen combined with many more elements. ThenBerzelius suggested to adopt O(16) as standard for comparision. From 1850 oxygen standardwas used both by chemists and physicists. O was taken as 16.0000 and other elements havefractional atomic masses. In 1919, two other isotopes of O namely O(17) and O(18) werediscovered. Their relative abundance was O(16) = 99.78%, O(17) = 0.02% and O(18) = 0.2%.Then chemists took the average atomic mass of O which is slightly greater than 16 as standardfor comparision. One sixteenth of the average atomic mass of O was taken as 1 amu. Butphysicists continued with the same O-16 standard. The atomic masses of elements varied slightlyin Physicist and Chemist standards which was very unfortunate for the rest of the world. Chemistsand physicists did not agree for a common standard for long years until 1956. Chemists arguedthat by converting to physicist standard(O16), the atomic masses would change by 275 ppm. Toresolve the long standing dispute between the two groups, two scientists Alfred Nier and A.Ölander mediated and suggested to make C-12 as the standard. C-12 was already used as astandard in mass spectrometry. The physicists agreed to this proposal but chemists resistedbecause by taking C-12 as standard their atomic masses would change by 42 ppm. But this timetheir resistance was not high. E. Wichers lobbied the chemists and ultimately in 1961 both thegroups compromised to use C12 as standard. Then onwards C-12 standard is being usedinterntationally.That the isotopes of O are available in different proportions in different areas is an absurdthinking.

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RESPONSE to SAQs(Atomic Mass)

SAQ 1:(i)In amu scale carbon has a whole number of atomic mass which is 12.(ii)Since in amu scale the standard is C(12.0000), the atomic mass of H is 1.008. On the

other hand in hydrogen scale the standard is H(1.0000), the atomic mass of carbon is 12.02.(iii)Atomic masses of elements cannot be whole number values except

carbon(12.000).Hence the statement is wrong in the strictest sense. The atomic masses of Pand F are 18.99840 amu and 30.97377 respectively.SAQ 2: Mass Number=16. The number of protons= no. of electrons=8, the no. of neutrons=16- 8=8SAQ 3: (i)K: Mass No.= 39, No. of protons= no. of electrons=19 and no. of neutrons=39-19=20

(ii)I: Mass No.=127, No. of protons=no. of electrons=53, and no. of neutrons=127-53=74

(iii) Ag: Mass No.=108, No. of protons= no. of electrons=47, No of neutrons=108-47=61SAQ 4:No, the statement is incorrect. Because the atomic mass of an element cannot be awhole number in the amu scale except carbon(12). The Mass number of oxygen is 16.

SAQ 5: It is the mass of 1/12 part of one carbon atom in gm.We know that 6.203 X1023 atoms of carbon weigh 12 gm(one gm. atomic mass)

Hence 1 atom of carbon will weigh 12/(6.023X1023)gm = 1.9923 X 10-

23gmSo 1/12 part of this mass = (1/12) X 1.9923 X 10-23 = 1.66 X 10-24 gm. which is nearly equal tomass of one H atom.SAQ 6: O16: protons=8 and neutron=16-8=8, O17: protons=8, neutrons=17-8=9

O18: protons=8 and neutrons = 18-8=10SAQ 7: No, the atomic mass of C13 is a fraction. Only C12 isotope has the whole number atomicmass(12.00000amu).SAQ 8: Monoisotopic: F, Na, Al, I, P and Au; Polyisotopic: The restSAQ 9: The mass numbers are 28, 29 and 30. The number of neutrons in Si28 isotope is 14(28-14), in Si29 isotope 15(29-14) and Si30 isotope 16(30-14).SAQ 10: No, it is not wrong to use mass number in chemical calculations. Rather itsimplifies the calculation.SAQ 11: Average atomic mass of Br = (1X79 +1X81)/2 = 80SAQ 12: Average atomic mass of O = (99.78X16 + 0.2X17+0.02X18)/100= 16.002So the average atomic mass of O is close to mass number of O16 isotope. This happened becausethe other two isotopes O17 and O18 contribute almost insignificantly.This is called the weight average. The number average would have been (16+17+18)/3= 17. It isdifferent from the weight average value of 16 that we use.SAQ 13: We use a value of 16 for all chemical calculations because the average atomicmass is almost equal to 16.SAQ 14: We use the average atomic mass of 80(SAQ 11), not 79 or 81 which are thetwo isotopic masses.SAQ 15: The average atomic mass of H = (99. 985X1.008+0.015X2.041)/100= 1.009This value is almost same as the isotopic mass of protium(H1) of 1.008. It is not close to the

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isotopic mass of H2(2.041) because the contribution of H2 isotope is insignificant(0.015%). Forchemical calculation we often use an atomic mass of 1 for H instead of 1.008.SAQ 16: Since the percentage composition of O16, O17 and O18 isotopes are 99.78, 0.2and 0.02 respectively, for every 10,000 atoms we analyse, 9978 atoms will be of O16, 20 atomswill be of O17 and only 2 atoms will be O18.SAQ 17: The most abundant isotopes are

Cl: Cl35 (75%), O: O16(99.78%), S: S32(95.6%), H: H1(99.985%), C: C12(98.88%)SAQ 18: 79.34/35.45 = x/35.5 , So x = 79.228(EM), Approx. atomic mass = 6/0.0276 =217.391Approx. valency = 217.391/79.228 = 2.73 ; So exact valency = 3 and hence exact atomic mass= 79.228X3 = 237.684

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EQUIVALENT MASS

In many chemical reactions the reactants do not react in 1:1 mole ratio.

(i) Na + 1/2 Cl2 ---------> NaCl

(ii) Zn + Cl2 --------------> ZnCl

2

(iii) 2 Al + 3 Cl2 ----------> 2 AlCl

3

In the first example, one gram atomic mass of Na reacts with 1 gm. atomic mass of Cl(1/2 moleof Cl

2=one gm. atomic mass), in the second example one gm. atomic mass of Zn reacts with 2

gm. atomic mass of Cl2(one mole of Cl

2 = 2 gm. atomic masses), while in the third case 2 gm.

atomic mass of Al reacts with 6 gm. atomic mass of chlorine. In other words, one gm. atomicmass of Al reacts with 3 gm atomic mass of chlorine. So on the basis of balanced equation, it isnot always possible for the reactants to react in the 1:1 ratio.

In order to establish a 1:1 quantitative relationship between reactants a new mass scale, calledthe Equivalent Mass was introduced. One gm. equivalent mass of one reactant always reactswith one gm. equivalent mass of the other reactant.Then let us know what is equivalent mass of an element?

Atomic Mass= Change in ON of the element per atom

ValencyAtomic Mass

Equivalent Mass of an element =

Number of electrons gained or lost per a= Atomic Mass

In the example (i) given above,The equivalent mass of Na = Atomic Mass/1 =23/1=23, because here the

valency of Na is 1. We can also understand this in terms of change in ON and number ofelectrons lost.

Na → Na+ + e-

The change in ON when Na changes to Na+ =|0-1|=1. The number of electrons lost by one Naatom is also 1.

The equivalent mass of chlorine = Atomic mass/valency=35.5/1 =35.5Gm. Equivalent Mass: When equivalent mass expressed in gm. it is called gm. equivalent mass.From balanced equation, we know that 23 gms(1 gm. atomic mass) of Na reacts with 35.5gm(1gm. atomic mass) of chlorine. Now we can say that one gm. equivalent mass of Na reacts withone gm. equivalent mass of Cl, since the equivalent mass of Na and Cl are 23 and 35.5respectively.In the example(ii),

The equivalent Mass of Zn = Atomic mass/valency =65.5/2=32.75 as the valencyof Zn is 2(same as change in ON and number of electrons lost) and

The equivalent mass of chlorine is already known in example (i) to be 35.5.

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Looking to the balanced equation of (ii), we know that 65.5gm(one gm. atomic mass) of Znreacts with 71gms(2gm atomic mass) of Cl. So 32.75(65.5/2) gms of Zn must react with35.5(71)gms of chlorine. So we find here also that one gm. equivalent mass of Zn(32.75)reactswith one gm. equivalent mass of Cl(35.5).In the example(iii),

The equivalent mass of Al = Atomic mass/valency = 27/3=9(since the valencyof Al is 3 or change in the ON or number of elctrons lost is 3). From the balanced equation, wesee that 54gms(2gm atomic mass) of Al reacts with 6 X 35.5=210gms(6gm atomic mass) of Cl.So 9 gm of Al must react with (210/54) X 9=35.5gm of chlorine. Here again we found that onegm equivalent mass of Al(9gm) reacts with one gm. equivalent mass of Cl(35.5).Conclusion: One gm. equivalent mass of any element will react with one gm. equivalentof any other element.SAQ 1: Find the equivalent mass of the underlined elements from the following reactions. Theatomic masses of elements are given. Use symbol E for equivalent mass.

(i) Fe + Cl2 ------> FeCl

3(Atomic mass of Fe=56)

(ii) Cu + ½ O2 -------> CuO (Atomic mass of Cu =63.5 and O =16)

(iii) 2Cu + ½O2------> Cu

2O

(iv) Fe + H2SO

4 --------> FeSO

4 + H

2

(v) Mg + ½ O2 --------> MgO (Atomic mass of Mg =24)

SAQ 2: Does an element show more than one equivalent masses? Justify with example.SAQ 3: Find the equivalent masses for the following elements. Explain why you can correctlydetermine these even without knowing the actual chemical reactions.

Ca, H, Na, Al, Ag (At. Masses are 40, 1, 23, 27 and 108 respectively)

Note that in a reaction not only 1gm. equivalent mass of one reactant reacts with 1 gm. equivalentmass of the other reactant but also the product formed is 1 gm. equivalent mass. In otherwords there is a 1:1 relationship between reactants and products in terms of equivalent mass.Let us take one example.

Al + 3HCl -------> AlCl3 + 3/2 H

2

From the above balanced equation, 27gm(1gm atomic mass) of Al produces 3 gms of hydrogen(3gm atomic masse). So 9 gm of Al(1gm equivalent mass) will produce (3/27) X9 = 1 gm ofH(=one equivalent mass of H). So we found that the one gm. equivalent mass of Al producedone gm. equivalent mass of hydrogen. So it may be remembered here that one gm. equivalentmass of a reactant not only reacts with 1gm equivalent mass of the other reactant but also forms1 gm equivalent of each product.

That is why there is a famous saying in chemical sciences:Every chemical reaction takes place in equivalents.

Let us try to understand the concept of equivalent mass in a different way.Classical Definition of Equivalent Mass:Equivalent mass of an element is a number which shows how many parts by mass of the elementcan combine with or displace 1.008 parts by mass of hydrogen, 8 parts by mass of oxygen, 35.5parts by mass of chlorine or 1 equivalent mass of any other element.

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ALTERNATIVE DEFINITION OF EQUIVALENT MASS OF ELEMENT:It is the mass of the element which loses or accepts one mole of electrons (Avogadro'snumber of electrons) when oxidised from the neutral state to the desired oxidation state orreduced from the desired state to the neutral state.

O0 + 2e ---------------> O2-

One O atom requires 2 electrons to become O2-

Conversely, 2 electrons are accepted by 1 atom of OSo Avogadro's number(N)of electrons are accepted by N/2 atoms of O.Since the mass of Avogadro's number of atoms is the gm. atomic mass of the element,

mass of N/2 atoms of oxygen = 16/2=8 g = g. quivalent mass of oxygen.

Al --------> Al3+ + 3e,3 electrons are lost by 1 Al atom,N electrons are lost by N/3 atoms whose mass is equal to one third of its g. atomic

mass(27/3). So to be more precise, equivalent mass of an element is the atomic mass of theelement divided by the number of electrons lost/gained per atom or change in ON per atom inrespect of that element.

This is same as what we knew earlier. Equivalent mass is equal to atomic mass divided byvalency or change in ON.SAQ 4: Find the equivalent mass of the elements from the ion electron equations. The atomicmasses of the elements are given in the brackets.

(i) Co3+ + 3e --------> Co (59)(ii) Fe --------> Fe2+ + 2e (56)(iii) Co-------->Co2+ + 2e (59)(iv) Ba --------> Ba2+ 2e (137)(v) K---------> K+ +e (39)

SAQ 5: Using equivalent mass concept, answer the following(i) How many gms of hydrogen will react with 8 gms of oxygen to form water(ii)How many gms of chlorine will react with 18 gm of aluminium to form aluminium

chloride.(iii)How many gms of Mg will react with 24 gm of oxygen.(iv)How many gms of hydrogen gas will be evolved by 32.5gm of Zn when all the Znreacts with dilute acid. (Zn=65)(v)How many gms. of Ag will react with 5.35gm of Cl. (Ag=108)

SAQ 6: Find the mass in gm. in case of the following and calculate how much of chlorine eachwill need to form their chlorides.

(i) 0.5 gm equivalent of Mg (ii)2 gm equivalents of Na ( i i i ) 0 . 2 5 g mequivalent of Al

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Determination of Equivalent Mass of Unknown Metal:From the above discussion we found that the equivalent mass of an element can be calculated ifwe know the element i.e know its atomic mass and valency(ON) by using the relationE=Atomic Mass/valency. But if the identity of an element is not known, can we find its equivalentmass? Yes, we can do so by experimental methods. Let us see some simple experiments bywhich we can find the equivalent mass of unknown metals. .

(a)Displacement Method(b)Hydrogen Displacement Method(c)Oxide formation or Oxide reduction method(d)Chloride method(e)Double displacement method

(a) Displacement Method:A more active metal(metal lying in the upper position in the metal activity series) displaces a lessactive metal(metal lying at a lower position in the activity series) from the salt of the latter. Alsothe metals lying above hydrogen in the metal activity series can displace hydrogen from diluteacids. We know that one gram equivalent mass of one metal(or element) can combine or displaceone gram equivalent mass of the other metal or element. If the equivalent mass of one elementis known, then that of the other can be found out by measuring the masses of the displacingmetal(or element) and displaced metal or element.

Zinc(s) + copper sulphate(aq.) -----------> Zinc sulphate(aq.) + Copper(s)Zinc(s) + sulphuric acid(aq.) -------> zinc sulphate(aq.) + hydrogen(g)

Let, m1 = mass of Zn; m

2 = mass of Cu displaced(or mass of hydrogen displaced);

E1 = Eq. mass of Zn; E

2 = Eq. mass of Cu( eq. mass of hydrogen)

m1 gm of Zn displaces m

2 gm of Cu or hydrogen(from experiment)

E1 gm of Zn displaces (m

2/m

1)E

1 gm of Cu or hydrogen. According to the definition of

eq. mass,E

2 = (m

2/m

1)E

1

⇒m1m2

=E1

E2

This is called the Law of equivalents which states that masses of substances are directlyproportional to their equivalent masses. Law of equivalents is the basis of finding theequivalent mass by any method.

Example: 0.26gm of Al displaces 0.94gm of Copper from Copper Sulphate solution.If the equivalent mass of Aluminium is 9, calculate the equivalent mass of copper.Solution:We can find this by unitary method done before or from the law of equivalents discussed now.Let us apply the law of equivalents.

mCu

mAl

=ECu

AlE

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⇒ 0.94/0.26 = ECu

/9 ⇒ ECu

= 32. 53 .(a)Hydrogen Displacement Method:Active metals which lie above hydrogen in the metal activity series can displace hydrogen gasfrom dilute acids. This method is applicable only to those active metals which can displacehydrogen gas from dilute acids.

Equivalent Mass of an Active Metal:Metal + H

2SO

4 ----------------> Metal sulphate + H

2 (balancing cannot be done as we

do not know the valency of the metal).From the masses of the metal and displaced H2, we can

calculate the equivalent mass of the metal.Determination of equivalent mass of zinc by hydrogen displacement method:

Principle: A known mass of zinc is allowed to react with excess of dilute acid(e.g dil. H2SO4) toproduce its equivalent quantity of hydrogen gas. The mass of zinc which displaces 1.008 gm ofhydrogen is calculated which gives the equivalent mass of zinc.Working Procedure: A small piece of zinc metal is accurately weighed in an electronic balance.The weighed metal piece is put inside a porcelain basin containing water. An eudiometer tube( along narrow calibrated glass tube closed at one end) completely filled with dil. H2SO4 is invertedover the metal piece carefully using the thumb. It is fixed in a vertical position with the help of aclamp and stand. The reaction between zinc and acid immediately starts producing bubbles ofhydrogen gas which is collectd by the downward displacement of water. When no more gasevolves(i.e when the reaction is complete), the eudiometer tube is carefully removed with thehelp of thumb and transferred to a tall jar containing water.

Levelling: Keeping the tube in vertical position inside the tall jar and by making upward anddownward movement the tube is brought to a position at which the level of water inside theeudiometer tube is same as the level of water outside the tube i.e inside the tall jar. This is calledlevelling. At this position, the pressure of the gas inside the tube is equal to atmoshperic peressure.At this position the volume of the hydrogen gas is recorded from the calibrations of the tube. Theatmospheric pressure(=gas pressure) is recorded from the barometer. The room temperature isalso noted.

AirGas Jar

Tall Jar

atm

27

Atoms and Molecules

Dr. S. S. Tripathy

Calculation: Let the mass of the metal= W gThe volume of hydrogen gas collected in the eudiometer tube= V

1 mL

Pressure of the moist gas = atmospheric pressure = Pmm of Hg (recorded from barometer)Room temperature = t0C = (273+t) KAqueous tension at t0C = f mm of HgPressure of dry gas = (P-f) mm of Hg(Since hydrogen gas is collected over water, it is mixed with some water vapour which has afixed vapour pressure at t0C. This is called aqueous tension at t0C. This has to be substractedfrom the pressure of the moist gas(P) to find the pressure of dry hydrogen gas).Applying combined gas equation, the volume of hydrogen gas at NTP(V

2) is calculated.

P V1 2 21

T1=

TP V

2⇒

1(P-f) V

(273+t)=

760 V

2732 ⇒ 1(P-f)

(273+t)

273V =

X XV

760 X = y mL(say)

We know that density of hydrogen gas at NTP is 0.000089 g/mL ( 2.016g/224000cc)1 mL of hydrogen gas at NTP weighs 0.000089 gso, y mL of hydrogen at NTP weighs 0.000089 X y gNow 0.000089 X y g of hydrogen is displaced by W g of the metal

So, 1.008 g of hydrogen is diplaced by W

0.000089 X yX 1.008 g = z g (say) of metal

So, equivalent mass of the metal = zAlternative method(Equivalent volume method):2.016 g of hydrogen gas occupies 22.4 L at NTP1.008 g (i.e 1 g. eq. mass) of hydrogen gas occupies 11.2 L at NTPSo in stead of converting the volume of hydrogen gas NTP into mass using the density data, wecan directly use the equivalent volume of hydrogen(11.2L at NTP) to calculate equivalent massof the metal.y mL of hydrogen gas at NTP is displaced by W g of metalSo, 11200 mL(11.2L) of hydrogen gas is displaced by (W/y) X 11200 g = z g (say)of metalHence equivalent mass of the metal = z

Example: 1.47gm of a metal was treated with excess of dil H2SO

4 and 541.8cc of hydrogen

gas was collected over water at 150C and 752.5mm of Hg pressure. Calculate the equivalentmass of the metal.(Aqueous tension at 150C = 12.5mm and density of hydrogen gas at NTP=0.000089 gm/cc)Solution: Let us first convert the experimental volume into NTP conditions by applying the combinedgas equation;

Experimental: P1= (P-f)= 752.5-12.5)mm V

1=541.8cc

T1=(273+15)K

NTP : P2=760mm V

2=?

T2=273K

⇒ (752.55 - 12.5)X 541.8(273 + 15)

=760 X V

2732

⇒ =V2 500.1 cc

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Atoms and Molecules

Dr. S. S. Tripathy

Mass of 1 cc hydrogen is 0.000089gm at NTP (density of H2 gas =0.00009gm/cc)

So mass 500.1cc of hydrogen = 500.1 X 0.000089 = 0.0445 gm0.0445 gm of hydrogen gas is displaced by 1.47gm of the metal

1.008 gm of hydrogen is displaced by 1.47 X 1.008= g

0.044533.29 of the metal

Equivalent Mass of the metal = 33.29Alternatively, we can use law of equivalents to solve the above problem.

mmH

=M EM

EH(Where m

M is the mass of metal and m

H is the mass of hydrogen while

EM and EH are the equivalent masses of metal and hydrogen respectively.In the above problem, we have

1.470.0445

E

1.008= M ⇒ E

M = 33.29

For simplicity, we shall hencforth be using the law of equivalents to calculate the equivalent massof an element.For calculation you can take equivalent mass of hydrogen to be 1 instead of 1.008.There might a slight difference in the answer but this can be ignored.EQUIVALENT VOLUME METHOD:500.1 cc of hydrogen gas is produced by 1.47 gm of the metalSo, 11,200 cc of hydrogen(1 eq. volume) is produced by (1.47/500.1) X 11,400 = 32.9 gmEquivalent mass of the metal = 32.9Can you guess why there has been a difference between the answers obtained in two differentmethods? The reason, the experimental density of hydrogen gas at NTP has been taken as0.000089g/cc in the first method while the theoretical density is 0.00009g/cc which can be obtainedby using gram molar volume(22400cc) for 2.016g of hydrogen gas at NTP. The second methodhas made use of equivalent volume(half of molar volume) and that is why the answers aredifferent.

SAQ 7: 0.139gm of a metal when dissolved in a current of hydrochloric acid evolved 29.5ml ofhydrogen gas collected over water at 130C and 741mm pressure. Find the equivalent mass of themetal. If the atomic mass of the element is 59, what will be its valency.(Aq. Tension at 130C =11.2mm)(b) Oxide Formation Method:

(i)Direct oxide formation:We know that many metals when heated in presence of oxygen or air produces their correspondingoxides

Metal + O2 --------------> Metallic oxide (We cannot balance this

equation since the valency of the metal is not known)Example: Al + O

2 --------> Al

2O

3 (equation not balanced)

If the mass of the metal is measured before the reaction and the mass of the metal oxide formedis also measured after the reaction, we can calculate the mass of oxygen that has combined withthe given mass of the metal . From this, the mass of the metal that would combine with 8 gm ofoxygen can be calculated. This is the equivalent mass of the metal since the equivalent mass ofoxygen is 8. Look to the following example.

29

Atoms and Molecules

Dr. S. S. Tripathy

Example: 50 mg of a metal is heated strongly in air until constant mass of 83.3 mg isobtained. Find the equivalent mass of the metal.Solution:

Mass of the metal = 50 mg = 0.05 gMass of the metal oxide = 83.3 mg = 0.0833gmSo mass of oxygen = 0.0833-0.05 = 0.0333 gm0.0333 g of oxygen has combined with 0.05gm of metal8g of oxygen has combined with 12 gm of the metalHence Eq. Mass of the metal = 12.

Law of Equivalent method:E M=0.05

0.0333 8 ⇒ EM

= 12

(Can you guess, which metal has equivalent mass equal to 12 ? It is magnesium).(ii)Indirect Oxide Formation:

In many cases metal oxides are not obtained directly from the metal. In such cases oxides areprepared in two steps. This is called indirect oxide formation method. The equivalent mass ofcopper can be obtained by this method.

Step-I : Metal + HNO3 -----------> Metal Nitrate + NO

2 + H

2O

Step-II: Metal Nitrate-----heat--------> metal oxide + NO2 + O

2

We know the mass of metal in the beginning and also know the mass of the metallic oxide at theend. Then we can know the mass of oxygen from which the equivalent mass of the element canbe calculated by the same method discussed before.Example: A sample of copper metal weighing 0.8g was first dissolved in conc. HNO3 to form itsnitrate. Cupric nitrate thus formed is then strongly heated until constant dry mass is obtained.Reddish brown gas is evolved during ignition. The mass of copper oxide formed was 1.0 g.Calculate the equivalent mass of copper.Solution: Mass of copper metal = 0.8 gMass of oxygen = 1.0 - 0.8 =0.2 g, Using law of equivalents, we have,

E M=8

0.80.2 ⇒ E

M = 32

(c) Oxide Reduction Method:In this case the metal oxide is reduced by H

2 or any other reducing agent to the corresponding

metal.CuO + H

2 -----------> Cu + H

2O

This method is also similar to the oxide formation method. When the masses of the metal oxideand metal are known, it is then easy to calculate equivalent mass of the metal in the same wayas done before.Example: On heating 1.127 gm of a metallic oxide in a current of hydrogen, 0.9g ofmetal was formed. Calculate the equivalent mass of the metal.Solution:

Mass of metallic oxide= 1.127gm Mass of the metal=0.9gmSo mass of oxygen= 1.127- 0.9= 0.227gmUsing law of equivalents,

30

Atoms and Molecules

Dr. S. S. Tripathy

E M=8

0.90.227 ⇒ E

M = 31.71

(d) Chloride Method:In this case the element reacts with chlorine to produce the chloride of the element. The mass ofthe element and chlorine that reacted are experimentally known. The mass of element whichreacted with 35.5 gm of chlorine is calculated. This gives the equivalent mass of the element asthe equivalent mass of chlorine is 35.5. Look to this example.Example: Chloride of a metal M contains 47.23% of the metal.Find the equivalentmass of the metal.Solution: Mass of the metal=47.23gm, Mass of chlorine =100-47.23= 52.77gm

52.77gm of chlorine reacted with 47.23gm of the metal, so35.5gm of chlorine must react with (47.23/52.77)X35.5=31.77gmHence the equivalent mass of the metal is 31.77.

Alternatively, using law of equivalents we have,E M=47.23

52.77 35.5⇒ E

M = 31.77

SAQ 8: 0.475 gm of a metal chloride was formed by the reaction of certain mass of the metalwith excess of chlorine. If the equivalent mass of the metal is 12, what is the mass of the metalreacted?

(e)Double Displacement Method:A known mass of a reactant(A) is allowed to react with excess of another reactant(B) in aqueoussolution to produce a precipitate containing the metal present in A. The precipitate is dried andweighed. From the dry masses of the reactant A and the corresponding product, we can find theequivalent mass of the metal present in A.Example: BaCl

2 + Na

2SO

4 -------> BaSO

4 ↓+ NaCl

This is a double displacement reaction under the main category of metathesis reactions. Bariumchloride reacts with sodium sulphate to produce a white precipitate of barium sulphate andaquous solution of sodium chloride.EQUIVALENT MASS OF A COMPOUND:The equivalent mass(EM) of a compound involved in a metathesis reaction(reaction in whichON remains unchanged) is determined from the following relationship.

EM of a compound = EM of the basic radica(cation) + EM of acid radical(anion)

EM of a radical(ion) =radical or ionic masvalency

Example: EM of barium sulphate(BaSO4) = EM of Ba2+ + EM of SO4

2-

= 137/2 + 96/2 = 68.5 + 48 = 116.5EM of barium chlorideBaCl

2) = EM of Ba2+ + EM of Cl- = 137/2 + 35.5/1= 68.5 + 35.5=104

Now let us apply law of equivalents to barium chloride and barium sulphate.

mBaCl2

BaSO4

m =E

BaCl2E EBaSO4

=E

Ba Cl

SO4

+ E

EBa + =

x 35.5/1x 96/2

+

+

31

Atoms and Molecules

Dr. S. S. Tripathy

(where x is the EM of Ba) From this relationship, the equivalent mass of Ba can be calculated.Example 1: 0.925 g of anhydrous barium chloride was dissolved in water and treated with excesssulphuric acid. The weight of the dry barium sulphate obtained was 1.036. Find the equivalentmass of the barium. Also find the percentage of error with respect to the theoretical value.Solution:

mBaCl2

BaSO4

m =E

BaCl2E EBaSO4

=E

Ba Cl

SO4

+ E

EBa + =

x 35.5/1x 96/2

+

+

⇒ =x 35.5/1x 96/2

+

+0.9251.036

⇒ x = 68.12

Theoretical equivalent mass = Atomic mass/valency = 137/2= 68.5Percentage of experimental error = [(68.5-68.12)/68.5] X 100 =0.554%Examle 2: On heating 1.127 gm of a metallic oxide in a current of hydrogen, 0.9g of metalwas formed. Calculate the equivalent mass of the metal.We can solve this problem by comparing mass of metal oxide and mass of metal with equivalentmass of the metal oxide and that of metal. There is no necessity of using the mass of oxygen.

=mass of metal

mass of metal oxideEM of metal

EM of metal oxide = x

x + 8⇒ 0.9/1.127 = x/(x+8) ⇒ x = 31.71In other words we can compare the masses with the equivalent masses of any pair of substances,may it be elements or compounds.

SAQ 9: 3.31 g of pure and dry lead nitrate was dissolved in water and treated with excess ofpotassium chromate solution. A yellow precipiate of lead chromate was obtained which wasseparated and dried and weighed to 3.23 g. Calculate the equivalent mass of lead.d(Cr=52)SAQ 10: 1.53 g of metal hydroxide on strong heating produced 0.995 g of its oxide. Calculatethe equivalent mass of the metal.

PRACTICE QUESTIONS1. 24gm of a metal produced 22.4litres of H

2 gas at NTP from an acid. What is the equivalent

mass of the metal? Guess which is the metal?2. Enough steam was passed over 5gm of red hot iron till all the iron reacted completely toproduce 2.67 litres of H

2 gas at NTP. Calculate the equivalent mass of iron. Justify this by

writing the equation.3. 0.13gm of a metal combines with 56ml of O

2 gas at NTP. Calculate the equivalent mass

of the metal.4. 0.640gm of an unknown metal gave 0.851gm of its chloride. Calculate the equivalentmass of the metal. Can you guess which is the metal?5. 1.314gm of an unknown metal displaced 2.158gm of sliver from AgNO3 solution. Findthe equivalent mass of the unknown metal. Could you guess the metal?

32

Atoms and Molecules

Dr. S. S. Tripathy

RESPONSE TO SAQsSAQ 1:(i) E= 56/3= 18.66(since the valency of Fe here is 3 i.e same as the change in ON of Fe

=|0-3|=3)(ii) E(Cu) =63.5/2 =31.75(since Cu forms Cupric oxide in which the valency of Cu is 2) E(O)=16/2=8 (since the valency of O is 2 i.e the change in ON in O =|0-2|=2(iii)E(Cu) = 63.5/1 =63.5(since Cu forms cuprous oxide in which the valency of Cu is 1)(iv)E= 56/2 =28(since Fe forms ferrous sulphate in which Fe show a valency of 2)(v)E= 24/2=12(since the valency of Mg is 2)

SAQ 2: Yes, some elements which show variable valency(oxidation state) has more than oneequivalent masses. For example Fe shows two equivalent masses, 56/3 and 56/2 for ferric andferrous respectively. Cu shows two equivlent massse, 63.5/2 and 63.5/1 for cupric and cuprousstates respectively.SAQ 3: E(Ca)=40/2=20 E(H) =1/1=1 E(Na)=23/1=23,

E(Al) =27/3=9, E(Ag)=108/1=108This is because all the elements given in this SAQ show one valency i.e one ON(not variablevalency). Therefore, they have one equivalent mass each.SAQ 4: (i)E(Co)= 59/3 (ii)E(Fe)=56/2 (iii)E(Co)=59/2

(iv)E(Ba)=137/2 (v)E(K)=39/1SAQ 5: (i) E(H)=1 and E(O)=8, So 8gms of O will react with 1 gm of H. This is becausewe know that one gm equivalent mass of any element will react with one gm. equivalent mass ofany other element.

(ii) E(Al)=27/3=9 and E(Cl)=35.5.9gms of Al reacts with 35.5 gms of Cl, so18gm of Al reacts with (35.5/9) X 18=71gm Cl.

(iii) E(Mg)=24/2=12, E(O)16/2=8, So 8gms of O reacts with 12gm of Mg.Therefore, 24gm of O must react with (12/8) X 24=36gms of Mg.

(iv) E(Zn)= 65/2=32.5, E(H)=1, So 32.5gms of Zn will displace 1gm of hydrogen.(v) E(Ag)=108/1=108, E(Cl)35.5, So 35.5 gms of Cl will react with 108gm of Ag,

So 5.35gm of Cl will react with (108/35.5) X 5.35 = 10.8gms of Ag.SAQ 6:

(i) E(Mg)=24/2=12, So 1gm. equivalent =12gms0.5 gm. equivalent =12X0.5=6gmWe know that 12gm of Mg will react with 35.5gm of Cl(E).So 6 gm of of Mg will react with (35.5/12) X6 =17.75gms of Cl.

(ii) E(Na)=23/1=23, So 1 gm. equivalent of Na =23gmSo 2 gm. equivalents of Na =23X2=46 gms.We know that 23gms of Na(E) will react with 35.5gms of Cl.So 46gms of Na will react with (35.5/23) X 46=71gms of Cl.

(iii) E(Al)=27/3=9, E(Cl)=35.5, So, 1 gm. equivalent of Al =9gmSo 0.25 gm equivalent = 9X0.25 = 2.25gmWe know that 9gm of Al will react with 35.5gm of Cl.So 2.25gm of Al will react with 35.5/9X2.25= 8.75gms of Cl.

SAQ 7:

33

Atoms and Molecules

Dr. S. S. Tripathy

Let us first convert the volume from the experimental conditions to NTP condition byusing the combined gas equation.P

1 =(P-f)= (741-11.2)mm, V

1=29.5cc T

1=(273+13)K

NTP: P2=760mm V

2=? T

2=273K

⇒ 2

273760 X V

=(273 + 13)

(741 - 11.2)X 29.5 ⇒ V2 = 27.04cc

1cc of hydrogen gas at NTP weighs 0.000089gm27.04cc of hydrogen gas at NTP weights 27.04 X 0.000089= 0.0024gmSo 0.0024gm of hydrogen is displaced by 0.139gm of the metalSo 1gm of hydrogen will be displaced by (0.139/0.0024) X 1 = 57.9gmSo the equivalent mass of the metal is 57.91.Valency = Atomic Mass/equivalent mass = 59/57.91= 1.018Since valency is always a whole number, the true valency cannot be 1.018, it is 1 i.e thenearest whole number of calculated valency.So the valency of the element =1.

SAQ 8: Let the mass of the metal = x gm; Mass of metal chloride = 0.475gm(given)So mass of chlorine = (0.475 - x)gm

(0.475-x)gm of Cl reacted with x gm of the metal

So 35.5gms of Cl must react with x0.475-x

X 35.5 gm (equivalent mass of

metal)

According to the data: x0.475-x

X 35.5 = 12 ⇒ x = 0.119gm.

So the mass of the metal = 0.119gm.

SAQ 9 : m(metal nitrate)

m(metal chromate) =E(metal nitrate)

E(metal chromate) =E(metal E(NO3-)+

+E(metal E(CrO42-)=

x + 62/1

x + 116/2

x = 103.5

SAQ 10:1.530.995

= E + 17E + 8 ⇒ E = 8.73

ANSWER TO PRACTICE QUESTIONS1. 22.4litres of H

2 at NTP = 2gms (1 mole)

2gms of H2 is displaced by 24gm of the metal(given)

So 1 gm of H2 must be displaced by (24/2) X 1 = 12gm.

So the equivalent mass of the metal is 12. The metal must be Mg.2. 22.4 litres of H

2 at NTP = 2gms

So 2.67 litres of H2 at NTP = (2/22.4) X 2.67 = 0.238gm

0.238gm of H2 is displaced by 5 gm of iron

1 gm of H2 must be displaced by (5/0.238) X 1 = 21.0 gm.

So the equivalent mass of iron = 21The reaction is: Fe + H

2O --------> Fe

3O

4(magnetic oxide) + H

2

34

Atoms and Molecules

Dr. S. S. Tripathy

Here ON of Fe goes from 0 to +8/3 = 2.66(Fe3O

4). The atomic mass of Fe= 56,

So equivalent mass = Atomic mass/ change in ON = 56/2.66 = 21.0

3. 22400ml of O2 gas at NTP = 32gms

So 56ml of O2 at NTP = 0.08gm

0.08gm of Oxygen combines with 0.12 gm of the metalSo 8gm of Oxygen must combine with (0.12/0.08) X 8 = 12. The metal is Mg.

4. Mass of metal chloride = 0.851gm, mass of metal = 0.6490 gmSo mass of chlorine = 0.851 - 0.640 = 0.211gm0.211gm of the chlorine combines with 0.649gm of metalSo 35.5gm of chlorine combines with (0.640/0.211) X 35.5 = 107.6gmSo the equivalent mass of the unknown metal is 107.6 and the meal would be Ag(sliver).

5. We can apply law of equivalents here. In all the questions from 1-3 given before, wecould have used the law of equivalents in stead of solving by unitary method.

m1m2

=E1

E2

where m1 = mass of unknown metal, m

2=mass of of Ag,

E1=Eq.Mass of unnown metal and E

2=equivalent mass of Ag.

⇒ (1.314/2.158) = E1/108 (Since Eq. mass of Ag = 108/

1=108)⇒ E

1 = 65. 76(Eq. mass of the unknown metal).

35

Atoms and Molecules

Dr. S. S. Tripathy

MOLECULAR MASS

When you are asked to calculate the molecular mass of CO2, you immediately add the atomic

mass of one carbon atom(12) with 2 times the atomic mass of oxgyen atom(2X16=32) and theresult is 12+32=44. So the molecular mass of CO

2 is 44. What does it mean? It means that one

molecule of carbon dioxde is 44 times heavier than one atom of hydrogen or in the cu scale onemolecule of carbon dioxide is 44 times heavier than 1/12 part of the mass of a carbon-12isotope. So we write,

Molecular Mass = Mass of one molecule

1/12 part of one atom of C(12) amu

Gm. Molecular Mass is the molecular mass expressed in gm. It is often called one gm. mole orsimply one mole. Can you say how many molecules of the substance is present in one gm.molecular mass(one mole) of the substance? The answer is Avogadro's number of molecules.More about mole concept will be discussed in the next chapter, Mole Concept.Determination of Molecular Mass of a substance:Add the atomic masses of all the atoms present in the molecule. You will get its molecular mass.SAQ 1: Find the molecular masses of the following:

K2Cr

2O

7, CaCO

3, MgS

2O

3, K

2SO

4, Al

2(SO

4)

3.18H

2O, FeSO

4.(NH

4)

2SO

4.6H

2O, SO

3,

H2C

2O

4, CH

3COCH

3, Fe

4[Fe(CN)

6]

3, Ca

3(PO

4)

2, C

6H

12O

6

DETERMINATION OF MOLECULAR MASS

1. Gram Molar Volume(GMV) Method:We know that one mole of any gas or vapour at NTP condition will occupy 22.4l(22400ml). Thisis called gram molar volume(GMV). If the volume of a fixed mass of an unknown gas or vapouris known at certain temperature and pressure, then we can first convert the volume to NTPconditions and then find out the molecular mass by finding out the mass of the vapour occupying22.4 litres at NTP. Look to the following example.Example: 80 mg of liquid on vaporization occupied 24.9 ml at 270C and 740mm pressure.Calculate the molecular mass of the subtance.Solution:

First the volume is changed to NTP conditon.

⇒ =760 X V2

273740 X 24.9

273+27 ⇒ V2 = 22.06 ml

22.06ml of the vapour at NTP weighs 0.08gm22400 ml of the vapour at NTP weighs 81.23 gmHenc molecular mass of the substance is 81.23.

SAQ 2: At STP, 5 litres of a gas weighs 14.4gm. What is its molecular Mass? If the gas is madeup of S and O, could you guess what is the gas?SAQ 3: 380ml of an unknown gas at 270C and 800mm of Hg pressure weighed 0.455gm. Calculate

36

Atoms and Molecules

Dr. S. S. Tripathy

the molecular mass of the gas. If the gas is made up of a single element, could you guess whatthe gas is? Supposing the gas consists of two elements C and O, could you guess what it is?

2. Vapour Density Method (for volatile liquids):Victer Meyer's Method:This method is applicable for volatile substances(usually low boiling liquids) which vapouriseseasily without decomposition.Principle: A known mass of the volatile liquid is vapourised in the Victor Meyer's apparatuswhich dispaces equal volume of air. The displaced air is collected by downward displacementof water. The vapour density of the vapour is determined from the following relationship.

=Mass of certain volume of gas or vapour

Mass of same volume of hydrogenV.D

(at fixed temp. and pr.)Then molecular mass of the the substance is determined from the relationship

Molecular Mass = 2 X V.DDescription of the the Apparatus:(a)Victor Meyer's tube: This is the main component of the apparatus. It is a long (nearly60cm)narrow glass tube with an elongated bulb at the end. The tube is connected with a sidedelivery tube at th top which is inserted inside the beehive shelf immersed in a trough of water.Thereis glass wool at the bottom of the tube.(b)Outer Jacket: The Victor Meyer's tube is kept inside a jacket made of copper or glass. Aliquid which boils at least 20-300C higher than the boiling point of the volatile liquid is taken insidethe outer jacket. Water is usually taken inside this jacket. The outer jacket is connected to anoutlet tube at the upper end for the exit of water vapour.(c)Hoffmann's bottle: A very small glass bottle having stopper is used to contain volatile liquid.

Functioning: A small quantity of the volatile liquid whose V.D is tobe determined is taken in the stoppered Hoffmann's bottle and isweighed accurately. The water inside the outer jacket is boiled sothat within a period of time Victer Meyer tube attaines a temperatureof 1000C. Excess water vapour produced in the outer jacket iscontinously removed through the upper outlet. The air inside theVicter Meyer's tube expands while boiling of water in the outerjacket continues and the excess air passes out of the tube over thetrough of water in the form of gas bubbles. When the tube attainsequilibrium at 1000C, there is no more expansion of air and the airbubbles stops appearing in the trough of water.Then a graduated measuring tube which is completely filled withwater is inverted over the beehive shelf. Then the Hoffmann'sbottle(which is loosely stoppered) containing the volatile liquid iscarefully and quickly dropped into the Victor Meyer's tube by

37

Atoms and Molecules

Dr. S. S. Tripathy

opening its mouth and quickly closing it. While falling down, the loosely stopperd Hoffmannbottle opens and the liquid instantly vapourises inside the tube as the temperature is 1000C whichis greater than the boiling point of the liquid Since the vapour is heavier than air, it remains insidethe tube and displaces equal volume of air from the tube which is collected in the gas tube by thedownward displacement of water. The Hoffmann bottle settles down over the glass wool at thebottom of Victor Meyer's bute without breaking.Levelling and volume measurement: Then the gas tube contining air is carefully taken out andimmersed vertically in tall jar containing water. The tube is brought to a position at which thelevel of water inside the tube is same with the level of water outside in the tall jar. At this position,the volume of the air is read from the calibration. At this position the pressure of the moist airinside the tube is equal to one atmosphere. This is called levelling. The pressure is noted from thebarometer. Room temperature(temperature of water in the tall jar) is recorded.

Calculation:Let the mass of the volatile liquid = xgm.(Note that this mass will remain unchanged in both the liquid and vapour states)Let the volume of air dislaced by the vapour of the volatile liquid= VmlTemperature=t0C = (273+t)0CPressure = Pmm of Hg, and Aqueous Tension = f mm at t0C.So pressure of dry air = (P-f)mmFirst the volume of air displaced(Vml) is converted from the experimental conditions to NTPconditions by using the combined gas equation.

273760 X V2=

273+t

(P-f) X V

⇒ V2 =173 X (P-f) X V(273+t) X 760

= y ml (say (volume at NTP)

Note that the volume of displaced air = volume of the vapour of the volatile liquidNow we have to find out the mass of y ml of hydrogen gas at NTP. We know thatthe mass of 1ml of hydrogen gas at NTP =0.000089gm (since density = 0.000089g/cc at NTP)Hence the mass of y ml of hydrogen gas at NTP = 0.000089 X y gm

Vapour Density = Mass of y ml of unknown gas at NTPMass of y ml of hydrogen gas at NTP

=x

0.000089 X y(note that the mass of volatile liqiud will not change when vapourised)

Molecular Mass = 2 X V.D

Example: 2.0 gm of a certain unknown gas occupies 418 ml at 270C and 755mm Hgpressure. Find out the molecular mass of the gas.(aqueous tension at 270C =15mm)

38

Atoms and Molecules

Dr. S. S. Tripathy

Solution:V

1= 418ml, P

1= 740mm, T

1= 273+27

Converting to the NTP condition,

273+27 273760 X V2

=(755-15) X 418

V2 = 370 .37 ml(volume at NTP)

V.D = Mass of 370.37ml of the gas at NTP

Mass of 370.37ml of hydrogen gas at NP

=0.000089 X 370.37

2

(Note that the initial mass of the unknown gas(here it is 2gm) cannot change despite itsvapourisation)

V.D = 60.67, Hence Molecular Mass = 2 X 60.67= 121.34

Note that the molecular mass could also be calculated by Gram Molar Volume Methodwithout determining vapour density. You can adopt any method you like. But experimentalmethod(Victor Meyer) demands that you must first find out the VD and then MM.

SAQ 4: A certain vapour at a certain temperature and pressure was found to be 29 timesheavier than hydrogen gas at the same temperature and pressure. What is the molecular mass ofthe substance.SAQ 5: 0.45 g of a volatile liquid in Victer Meyer's apparatus displaced 98.75 mL of moist air at200C and 718 mm of Hg pressure. Calculate the molecular mass of the liquid. (aqeous tension at200C = 17.4 mm of Hg)SAQ 6: The density of some gases are given below at NTP. Find their molecular masses. Thedensity of hydrogen gas at NTP=0.000089gm/cc. If all the gases are elemental in nature. Identifythem.

(i)0.0014 gm/cc (ii)0.00316gm/cc (iii)0.00089gm/ccSAQ 7:In a Victor Meyer's experiment, 0.168gm of a volatile liquid dislaced 49.4ml of air measuredover water at 200C and 740mm of pressure. Calculate the vapour density and molecular mass ofof the compound. (Aqeous tension at 200C= 18mm)SAQ 8: Find the molecular mass of the volatile liquid given in SAQ 7 by gram molar volumemethod. Do you find any difference?

PRACTICE QUESTIONS1. When 3.2 gm of sulphur is vapourized at 4500C and 723 mm pressure, the vapour occupiesa volume of 780ml. What is the molecular formula of sulphur vapour under these conditions?2. 250ml of ozonised oxygen(ozone + oxygen)at NTP weighed 0.393gm. On passing thesample through turpentine oil there was contraction in volume by 50ml. Find the molecular massof ozone.3. If a gas has a density of 0.5 gm per litre at NTP, then find the mass of one mole of the

39

Atoms and Molecules

Dr. S. S. Tripathy

gas.4. The molecular mass of a compound is 46. What will be the volume of air displaced by0.1665g of the substance at 150C and 773.3 mm pressure in Victor Meyer's apparatus? (AqueousTension at 150C =13.3 mm)5. What is the mass of one mole of a gas, one gm. of which occupies 0.9822 litres at1000C and 740mm pressure?

RESPONSE TO SAQs(Molecular Mass)

SAQ 1: K2Cr

2O

7 = 2X 39 + 2X52 + 7 X 16= 294, CaCO3 = 40 + 12 + 3X16=100

K2SO

4 = 2X39 + 32+4X16 = 174; MgS

2O

3(136), Al

2(SO

4)

3.18H

2O(666),

FeSO4.(NH

4)

2SO

4.6H

2O(392), SO

3(80), H

2C

2O

4(90), CH

3COCH

3(58) Fe

4[Fe(CN)

6]

3(860)

Ca3(PO

4)

2(310), C

6H

12O

6 (180)

SAQ 2:If 5 litres of the gas weighs 14.4gm at STPThen 22.4 litres of the gas will weigh (14.4/5) X 22.4= 64.5 (Molecular mass)If the gas consist of S and O , it has to be SO

2. Its molecular mass is 64 which is close

to the experimental value. Note that there would definitely ramain some error in any experimentalmethod.SAQ 3: First we have to convert the volume into STP condtions.

273+27 273760 X V2

=800 X 380

⇒ V2 = 364ml

365ml of the gas weighs 0.455gm at STP22400 ml of the gas will weigh (0.455/365) X 22400 = 27.9So the molecular mass of the gas is 27.9 and it is undoubtedly N

2(2X14=28) as it consists

of only one element. If it consists of C and O, it has to be CO(12+16=28).SAQ 4: The vapour density of the vapour =29, because the density of the vapour 29 timesgreater than density of hydrogen at the same temperature. So its Molecular Mass

=2X29=58SAQ 5:NTP volume(V2) = 84.8 mL, V.D = 0.45/(0.000089X84.8) = 59.62, So MM=119.24SAQ 6:(i) VD = (density of the gas or vapour)/(density of hydrogen) (at same temperatureand pressure, here it is NTP)

VD= (0.0014/0.000089)= 15.73, So MM= 2X15.73=31.46 ≈ 32. So the gas isO

2

(ii) VD= 0.00316/0.000089= 35.5, MM=71, so the gas is Cl2

(iii) VD=0.00089/0.000089=10, MM=20, so the gas is Ne.

SAQ 7: V1= 49.4ml, T

1=(273+20), P

1 = (740-18)mm

Let us first convert the volume to NTP conditions.

273760 X V2

=(740-18) X 49.4

273+20⇒ V

2= 43.72ml

=mass of 43.72ml of the vapour at NTPmass of 43.72ml of hydrogen at NTP

0.168

43.72X0.000089= 43.175V.D =

40

Atoms and Molecules

Dr. S. S. Tripathy

M.M = 2XV.D = 2X43.175= 86.35SAQ 8: Let us take the NTP volume V

2 from the SAQ 7 = 43.72ml

43.72ml of the vapour weighs 0.168gmSo 22400ml of the vapour will weigh (0.168/43.72)X22400 = 86.075gm (M.M)

We noticed that the M.M found by two methods are little different. This is obvious as thedifference is due to the difference in methods adopted. In VD method we made use of thedensity of hydrogen gas as 0.000089gm/cc which is not very exact. That is why this difference.But you are advised to adopt any method unless otherwise asked for specifically.

ANSWERS TO PRACTICE QUESTIONS1. Convert the volume to NTP condition first. The NTP volume V

2=280.18ml(see for

yourself)280.18ml of the vapour at NTP weighs 3.2gm.22400ml of the vapour at NTP must weigh (3.2/280.18) X 22400 = 255.83gm.So the the molecular formula (S)

n can be known: n=255.83/32 ≈ 8 : S

8

2. Turpentine oil absorbs ozone only. Since the contraction of volume was held by 50ml, thevolume of ozone =50ml. So the volume of oxgyen = 250-50=200mlV.D of the mixture = 0.393/(250X0.000089) = 17.66, So MM = 2X 17.66=35.32

Av. M.M = 35.32 = 200 X 32 + 50 X x250 ⇒ x = 48.6 (where x=M.M of ozone)

Note that the actual molecular mass of ozone is 48 which is close to the result.3. 1litre weighs 0.5gm at NTP, So 22.4litres will weigh 11.2gms. This is the mass of one

mole(MM).4. Let the volume of air displaced = x ml at 150C and (773.3-13.3)mm pressure. Let us

convert this volume to NTP condition. The NTP volume V2 =0.957x ml.

V.D = 0.1665/(0.957x X 0.000089) = 46/2=23 ⇒ x = 85.5mlSo the volume of air displaced in the Victor Meyer's experiment is 85.5ml

5. Similar to No.3. Ans: 32g (Try yourself)

41

Atoms and Molecules

Dr. S. S. Tripathy

Mole Concept and Avogadro lasw and i

MOLE CONCEPT

Mole is a chemical unit( like a dozen, a gross etc. used as family units) which can be defined intwo ways.(i) Number wise: one mole of anything contains Avogadro's Number (6.023 X 1023)of species of that thing.(ii) Mass wise: It is the mass of the substance which contains Avogadro's number ofspecies.Mole can be determined for atoms, molecules as well as for ions.(a) For atoms : e.g C, Na, H, Fe etc. (b)For molecules :e.g N

2, H

2, CO

2, H

2SO

4 etc.

(c) For ions : e.g SO42-, NO

3- etc.

(a) For atoms:The atomic mass expressed in gm(gm atomic mass) is the mass of Avogadro's number ofatoms. So one mole of atoms weigh one gm. atomic mass(e.g : one mole of Na atoms weighs 23gms , one mole of H atoms weigh 1.008 gm and one mole of C weigh 12 gms. and so on)N.B: Some authors write one mole of atoms as one gm. atom and according to them theterm mole should not be used for atoms. However, we have used both gm. atom and molefor atoms.SAQ 1:(i)How many bananas are there in one mole of banana?

(ii)You bought 1/10 mole of cricket balls for your team. How many balls you bought. Ifyou require only 20 balls per year, how long your team can play cricket?(iii)A cigarette smoker smoked 12.046 X1023 cigarettes before his death in lung cancer.How many moles of cigarettes did he smoke during his life time?

SAQ 2: Find the mass of 6.203 X 1023 atoms of (a)Phosphorus (b)Calcium(c)Helium (d)Boron

SAQ 3: Find the number of atoms present in (i) 19 gms of Fluorine (ii)39 gms of Potassium(iii)63.5 gms of Cu

NO. OF ATOMS AND MOLES(GM. ATOMS)PRESENT IN A GIVEN MASS OFAN ELEMENTExample: Calculate the number of atoms present in 2.3 gms of NaSolution:

23 gms(gm. atomic mass) of Na contain 6.203 X 1023 atoms of Na

2.3 gms of Na therefore contain 6.023 X 1023

23X 2.3 = 106.023 X

22atoms of Na

SAQ 4:Calculate the number of atoms present in the following:(i) 0.12 gm of C (ii)40 gms of He (iii)4 gms of Oxygen (iv)0.001 gm of

42

Atoms and Molecules

Dr. S. S. Tripathy

Sulphur (v)40mg of Calcium (vi)6.35 kgs of CopperThe atomic masses are as follows: C=12, He=4, O=16, S=32, Ca=40, Cu=63.5

SAQ 5: Calculate the mass of the following(i) 1 million carbon atoms (iii) 12X 1030 of K atomsAtomic Masses: C =12 K=39

No. of moles(gm. atoms) of atoms of an element present in a given mass of element:

You already know that 1 mole of atoms (also called 1 gm. atom of an element) contains Avogadro'snumber of atoms which weigh one gm atomic mass. For examle, the atomic mass of Na is 23. So23 gms of Na = 1 mole of Na atoms =1 gm atom of Na which contains 6.023 X 1023 atoms ofNa. So if you are asked to find the number of moles of atoms or gm. atoms of Na present in 2.3gm of Na, it is 1/10 of a mole =0.1 mole of atoms(gm. atom). Also if you are asked to find thenumber of atoms present in it, you can find out by the method described earlier. See this example.

Example: Find the number of moles(gm. atoms) of sodium atoms present in 0.023 gm of NaSolution:

23 gms of Na = 1 mole of Na atoms(1gm atom of Na)

0.023gm of Na = 23X =

10.023 0.001 mole of Na atoms or 0.001 gm atom of Na

SAQ 6: Calculate the number of moles(gm. atoms) of atoms present in the following(i)1.27 gms of Iodine (ii)400 mgs of Oxygen (iii)5.04 gms of H (iv)0.28gm

of Silicon (Atomic Masses: I =127, O=16, H=1, Si=28)

(b) For molecule and Ions:

Molecular mass expressed in gms(gm. molecular mass) is the mass of Avogadro's numberof molecules. Similarly ionic mass expressed in gm. is the mass of Avogadro's number ofions. Hence one mole of molecules will weigh one gm. molecular mass and one mole of ions willweigh one gm. ionic mass. One mole of N

2 gas will weigh 14X2= 28gms, one mole of CO

2 will

weigh 44 gms and one mole of SO42- will weigh 32+4X16=96 gms and so on.

SAQ 7:Calculate the mass of 6.023X1023 molecules of the following:(i)CaCO

3(ii)S

8(iii)HCl (iv)H

2SO

4

( Atomic masses: Ca=40 , C=12, S=32, Cl=35.5, S=32, O=16, H=1)SAQ 8: Calculate the mass of 6.023X1023 number of ions in case of the following

(i)NO3

- (ii)Cr2O

7 2- (Atomic masses: N=14, Cr=52)

NUMBER OF MOLECULES PRESENT IN GIVEN MASS OF A SUBSTANCEAND VICE VERSA

If 36.5 gms (gm molecular mass) of HCl contains 6.023X1023 molecules of HCl, then x gm of

43

Atoms and Molecules

Dr. S. S. Tripathy

HCl will contain (6.023X1023/36.5)X x molecules. So if the mass of a substance is given, we cancalculate the number of molecules. We can also get the mass of a substance which will containa fixed number of molecules. See the following examples.Example: Calculate the number of molecules present in 7 gms of N

2 gas.

Solution:The molecular mass of N

2 is 28.

28 gms of N2

contains 6.023X1023 molecules

7 gms of N2 contains 6.023 X 10

23

X =23

1028

7 1.505 X molecules.

SAQ 9:Find the mass of the following.(i)2.0076 X 1020 molecules of Na

2CO

3(ii)3.0115 X 1040 molecules of H

2SO

4

NO. OF MOLES PRESENT IN A GIVEN MASS OF A SUBSTANCESince the gm. molecular mass is the mass of 1 mole of a substance, we can find out the numberof moles present in given mass of a substance. Look to the following example.Example: Find the number of moles of sulphuric acid molecules present in 490 gmsof H

2SO

4 .

Solution:Molecular Mass of H

2SO

4 = 98

98 gms of H2SO

4 = 1 mole

490 gms of H2SO

4= 490/98=5 moles

SAQ 10: Find the number of moles in case of the following:(i)18.6 gms of Phosphorous(P

4) (ii)1.47kg of H

2SO

4

(iii)3.55gm of Cl2 gas

SAQ 11: Calculate the number of moles and molecules present in(i)0.106 gmsNa

2CO

3(ii)48gms of Oxygen (iii)15.5 gms of P

4

(iv)0.49 gms of H2SO

4(v)2.8 kg of Nitrogen gas

SAQ 12: Find the mass of the following(i) 1023 molecules of hydrogen (ii)100 million molecules of K

2SO

4

(iii) 1.2 X 10 69 molecules of NitrogenSAQ 13: Find the mass of the following.

(i)0.001mole of Ca3(PO

4)

2(ii)2.5moles of H

2SO

4

(iii)1/20 mole of MgCO3

MASS OF ONE ATOM OF AN ELEMENTSince the mass of Avogadro's number of atoms is gm. atomic mass. So from this the absolutemass of one atom of an element can be calculated. Note that this is the actual mass of an atomwhich is incredibly minute and small. When you are asked, what is the mass of one oxygen atom,you often say 16gm. But just think how the mass of one tiny oxygen atom could be 16gm?? With16gm of sugar you can prepare a cup of tea and could one atom of oxygen can weigh anincredibly large 16gm?? No, what we say 16gm; is its gm. atomic mass, which is the mass ofAvogadro's number of atoms. So from this the actual mass of one atom in gm can be calculated.See this example.

44

Atoms and Molecules

Dr. S. S. Tripathy

Example: Find the mass of one oxygen atom:Solution:

6.023X1023 atoms of oxygen weigh 16 gms1 atom of oxygen weighs 16/( 6.023X1023)= 2.66 X 10-23 gm.Do you notice how small the mass of one atom is!!!!

SAQ 14: Find out the mass of one atom of each of the following.The atomic masses are givenwithin brackets.

(i)C(12) (ii)Al(27) (iii)H(1) (iv)S(32)(v)Cl(35.5) (vi)Ag(108)

SAQ 15: What is the mass of 1 amu in gm? How this is related with the mass of hydrogen atom?

MASS OF ONE MOLECULE OF A SUBSTANCESince the gm. molecular mass is the mass of Avogadro's number of molecules in the same wayas gm atomic mass is the mass of Avogadro's number of atoms, we can calculate the actualmass of one molecule of a substance. See this example.Example: Calculate mass of one molecule of CO

2.

Solution:Molecular mass of CO

2 = 12 +16X2 = 44

6.023X1023 molecules of CO2 weigh 44 gms

So 1 molecule of CO2 weighs 44/(6.023X1023) = 7.305 X 10-23 gm.

You also found that like the mass of one atom, the mass of one molecule is also very small.

SAQ 16:.Find the mass of one molecule in gm for the following(i)NH

3(ii)CaCO

3(iii)(NH

4)

2SO

4(iv)CH

3COOH (v)NaCl

NUMBER OF MOLECULES AND MOLES OF A GAS PRESENT IN A GIVENVOLUME OF GAS AT A PARTICULAR TEMPERATURE AND PRESSURE

Gram Molar Volume(GMV) at NTP = 22.4 litresone mole of any gas occupies 22.4 litres at NTP.e.g 32 gms of O

2 gas, 2 gms of H

2 gas, 44 gms of CO

2 gas, 28 gms of N

2 gas etc. each occupies

22.4 litres at NTP (00C and 760mm Hg Pressure). So the number of moles and molecules of agas present in a given volume of gas can be easily calculated. Also we can do the opposite thing.We can calculate the volume of gas provided we know the number of molecules or moles ormass. The examples below will make these more clear.

Example : Calculate the number of molecules, moles and mass of oxygen gas present in224cc of oxygen gas at NTP.Solution:

22400 cc of O2 gas at NTP contain 6.023X1023 molecules of O

2

224 cc of O2 gas at NTP contain 6.023X1021 molecules of O

2

Again 22400 ml of O2 gas contains 1 mole

45

Atoms and Molecules

Dr. S. S. Tripathy

224 ml of O2 contains 224/22400 = 0.01mole

1 mole of oxygen weigh 32 gms0.01mole of oxygen weigh 32X0.01 = 0.32 gm

N.B : For gas present at conditions other than NTP, the combined gas equation(P

1V

I/T

1=P

2V

2/T

2 )is to be used to convert the given volume to NTP or volume at NTP to

volume at given conditions depending on the requirement. . If the volume data is given,theconversion is made first and if the volume is to be found out, then the conversion is done atthe end.Example : Calculate the volume in cc of nitrogen gas containing 2.4 X 1020 molecules at270C and 800mm pressure.Solution:Here volume is to be calculated at some other temperature and pressure. So the conversionis to be made at the end.

6.023X1023 molecules of N2 at NTP occupies 22400 cc

2.4 X 1020 molecules of N2 occupies 8.96cc at NTP

But we need to find the volume at the conditions given in the question. Let us apply gaslaw.

=760 X 8.96273

800 X V2(273+23) ⇒ V

2= 9.22cc

Example 3: Calculate the number of molecules and mass in gm. of CO2 present in 200 ml

of the gas at 270C and 800mm pressure.Solution: Since the volume data is given here, the gas equation is to be used first to get thevolume at NTP from the volume at given conditions.

273760 X V2

273+27800 X 200 = ⇒ V

2 = 191.6 ml(Volume at NTP)

22400 ml of CO2 gas at NTP weighs 44 gms

191.6 ml of CO2 weighs (44/22400)191.4 = 0.376gm =376 mg.

Again 22400 ml of the gas at NTP contains 6.023 X1023 molecules at 191.6 ml of the gas contains 5.1 X1021 molecules .

SAQ 17:(i) Find the volume of 4gms of CO

2 gas at NTP. How many molecules are present in it?

(ii) Calculate the mass of N2 gas present in 560ml of it at NTP. Also find out the number of

moles and molecules present in it.(iii) Find the number of molecules present in 680ml of O

2 gas at 270C and 900mm of Hg

pressure. Also calculate the mass in gm and number of moles.(iv) Calculate the volume of H

2 at 270C and 900 mm pressure if it contains 0.5 mole of

hydrogen gas.

46

Atoms and Molecules

Dr. S. S. Tripathy

PRACTICE QUESTIONS(Mole Concept)

SET-I1. Find the number of gm. atoms(mole of atoms) present in the following. Also calculate thenumber of atoms present in it. The atomic masses are given inside brackets.

(i)4Kg of Ca (40) (ii) 32.7gm of Zn(65.4) (iii)7.09gm ofCl(35.45)

(iv)95.4gm of Cu(63.55) (v)8.62gm of Fe(55.85)2. How many H

2 molecules are present in 8.5 gm of H

2? How many H atoms are in it?

3. Find the mass of one S atom in gm.4. Find the mass of the following:

(i)12.046 X 1024 atoms of H(1) (ii)3.0115 X 1030 atoms of Zn(65.5)(iii) 3.6138 X 1022 atoms of Ag(108)

5. How many moles and how many molecules of H2O are present in 48gms of H

2O?

6. What is the mass of 0.004mole of glucose(C6H

12O

6)? What is the mass of one glucose

molecule?7. What is the mass of the following. Also find their volumes at NTP.( N stands for Avogadro'snumber)

(i) 3N molecules of CO2(ii)0.5N molecules of N

2(iii)0.0002 N molecules of O

2

8. How many molecules of water and atoms of H and O are present in 9 gms of H2O?

9. Calculate the number of moles in each of the following. Also find the number of molecules ineach case.

(i)10gms of CaCO3

(ii)3.5kg of H2SO

4(iii)3.16gm of KMnO

4

10. Find the number of NO3- ions present in 62gm of NO

3- . What is the mass of one NO

3- ion?

11. Calculate the number of H atoms, S atoms and O atoms present in 7gms of H2SO

4.

12. Calculate the gm. atoms(mole of atoms) of carbon and oxygen present in 22gms of CO2 gas

at NTP. Also find the number of carbon and oxygen atoms present in it.13.Calculate the number of oxygen atoms present in 25gm of CaCO

3. Also find the number of

gm. atoms(mole of atoms) of oxgyen present.14. How many H atoms are present in 25gm of NH

4Cl? Also find the the number of moles of

NH4Cl present.

SET-II

1. Calculate(i)The number of boron atoms in 594gms of Boron(B=11)(ii)the number of calcium atoms in 10gms of calcium(Ca=40)(iii)the number of iron atoms in 8.37gms of iron(Fe=56)(iv)the number of gm. atoms in 1.0gm of Lithium(Li =7)(v)the number of Uranium atoms in 1 gm of U238 isotope( U = 238)

2. A sample of pure element that had a mass of 1.0g was found to contain 1.5 X 1022

atoms. Find out the atomic mass of the element.3. How many gm. atoms and atoms of S are present in 20gm of sulphur?(S=32)

47

Atoms and Molecules

Dr. S. S. Tripathy

4. How many moles are there in each of the following:(i)0.14gm of nitrogen gas (ii)9.2gm of NO

2 gas (iii)3.5 X1022 molecules of

N2O

5. Calculate the number of moles in 2.2gm of CO2.

6. Calculate the number of molecules in 2.4gm of CO.7. Calculate the mass of

(i)one CO2 molecule (ii)One Au atom (At. mass of Au = 197)

(iii)One H2SO

4 molecule (iv)One hydrogen atom

8. An atom of Hg weighs 33.3 X 10-23 gm. Calculate the atomic mass of Hg.9. One atom of an element(X) weighs 6.6 X 10-23 gm. Calculate the number of gm. atomsin 20kg of it.10. Calculate the number of methane molecules and the number of hydrogen and carbon

atoms present in 20gms of methane?11. How many atoms of each kind are present in 2.56gms of sucrose(C

12H

22O

11)?

12. Which of the following contains the greatest number of molecules and which containsthe least.

(i)8 gm of CO2

(ii)8gm of O2

(iii)8gm of N2

(iv)8gm of H2

13. How many moles and molecules are present in the following.(i)19.2 gm of H

2SO

4(ii)4.4 Kg of CO

2(iii)1 kg of ammonium

dichromate(Cr=52)14. Calculate the number of atoms of each kind in

(i)4.4gm of CO2

(ii)2.7gm of Al (iii)10 gm of CaCO3

15. A piece of Zn weighs 0.65gm. How many atoms of Zn does it contain?(Zn =65)16. Calculate the volume at NTP occupied by 2.4gm of SO

2.

17. The volume of a gas in a tube is 1.15 X 10-7 ml at NTP. Calculate the number of H2

molecules of the gas present in the tube.18. What is the number of atoms in 2gm of He and also find its volume at STP.(He=4)19. How many molecules of water are present in one cc of water?(density of water=1gm/cc)20. Calculate the number of molecules present in

(i)100 ml of CO2 at NTP (ii)1 litre of HCl at NTP

(iii)250 ml of H2 at 270C and 800 mm pressure

21. How many gm atoms are contained in(atomic masses are given within brackets)(a)32.7g of Zn(65.5) (b)7.09g of Chlorine(35.5) (c)95.4g of Cu(63.5)(d)4.31gm of Fe(56) (e)0.328g of S(32)

22. What is the volume of :(i)CO gas containing 3 X 1021 molecules at STP(ii)N

2 gas containing 2.4 X 1025 molecules at 270C and 800mm

pressure.23. Calculate the number of moles of carbon atoms present in

(a)1.0gm of C (ii)12 gm of carbon (iii)5.66 X 1020 carbon atoms24. Calculate the mass in gm for the following

(i)3 g atom of Ag (ii)0.5 mole of Ne (iii)2.5 moles of CO2

( i v ) 0 . 0 1mole of K

2CO

3

48

Atoms and Molecules

Dr. S. S. Tripathy

25. Find the number of g. atoms and mass of an element having 3 X 1024 atoms. Atomicmass of the element is 32.26. How many atoms of Na, S, O are present in 0.25 mole of Na

2SO

4?

27. Find the molecular mass of a gas, 7.525 X 1023 molecules weigh 35 gms. Also find thevolume of the 5gm of the gas at NTP.28. Calculate the number of oxygen atoms present in 17.6 g of CO

2.

29. Calculate the mass in gm of HNO3 present in 0.005moles of it. Also calculate the number

of molecules of HNO3 present in it.

30. Calculate the number of SO4

2- ions present 2.5gm of Na2SO

4.

RESPONSE TO SAQs(Mole Concept)

SAQ 1:(i) One mole of banana contains 6.023 X 1023 number of bananas. What a huge number!!!(ii) one mole of cricket balls = 6.023 X 1023 , so 1/10 mole of it will contain ten times less

i.e 6.023 X 1022 number of balls. If you need 20 balls per annum, then the number ofyears required to exhaust the stock = 6.023 X 1022 /20= 3.1 X 1021 years, which meansthousands of generations will play cricket with that stock of balls!!!!

(iii) 6.023 X 1023 number of cigarettes =1 mole of cigarettesSo 1 cigarette = 1/(6.023 X 1023)mole12.046 X 1025 cigarettes = 12.046 X 1023 X(1/6.023 X 1023)=2 molesSo the man smoked 2 moles of cigarettes during his life time.

SAQ 2: The mass of Avogadro's number of atoms= one gm. atomic mass(a)31gms (b)40gms (c)4gms (d)11gms.All these are the atomic masses of the elements.

SAQ 3: (i)6.023 X 1023 atoms of F (ii)6.023 X 1023 atoms of K(iii)6.023 X 1023 atoms of Cu

Note that in all these cases, mass of the substance given were their respective gm. atomicmasses. So the amount must contain Avogadro's number of atoms.SAQ 4: (i) 12 gms(gm. atomic mass) of C contain 6.023 X 1023 number of C atoms

0.12gms of C must contain (6.023 X 1023/12)X0.12= 6.023 X 1021 atoms(ii) 4 gms(gm atomic mass) of He contains 6.023 X 1023 atoms of He

40gms of He must contain (6.023 X 1023/4)X40= 6.023 X 1024 atoms of He(iii) 16gms(gm. atomic mass) of oxygen contains 6.023 X 1023 atoms of O

4gms of oxgyen must contain (6.023 X 1023/16)X4= 1.5 X 1023 atoms(iv) 32gms(gm. atomic mass)of S contains 6.023 X 1023 number of S atoms

0.001gms of S must contain (6.023 X 1023/32)X0.001= 1.88 X 1019 atoms of S(v) 40gms(gm atomic mass) of Ca contain 6.023 X 1023 atoms of Ca

40mg i.e 0.04gm of Ca must contain (6.023 X 1023/40)X0.04=6.023 X 1020

atoms of S(vi) 63.5gm(gm atomic mass) of Cu contains 6.023 X 1023 atoms of Cu

6.35Kg i.e 6350 gm of Cu must contain (6.023 X 1023/63.5)X6350=6.023 X 1025

atoms

49

Atoms and Molecules

Dr. S. S. Tripathy

SAQ 5:(i) 6.023 X 1023 number of carbon atoms weigh 12gms(gm atomic mass)1 million i.e 106 atoms will weigh [12/ (6.023 X 1023)]X106≈ 2X12-17 gms.

(How much small the mass is!!!!)(ii) 6.023 X 1023 atoms of K weigh 39gms (gm atomic mass)

12 X 1030 atoms of K will weighs [39/( 6.023 X 1023)]X12X1030 =78X107 gms= 7.8 X 108gms

SAQ 6:(i) 127gm(gm atomic mass) of I = 1 mole of I atoms=1 gm atom So 1.27gm of I = (1/127)X1.27= 1/100=0.01 mole of I atoms =0.001gm atom

(ii) 16 gms(gm atomic mass) of O = 1 mole of O atoms= 1gm atom400mg i.e 0.4gm of O = (1/16) X 0.4= 4/160 = 0.025 mole of atoms(gm atom)

(iii) 1 gm(gm atomic mass) of H = 1 mole of H atoms =1gm atom5.04gms of H = (1/1) X 5.04= 5.04 moles of H atoms =5.04 gm atoms

(iv) 28gms(gm atomic mass) of Si = 1 mole of Si atoms(1gm atom)0.28gm of Si = (1/28)X0.28= 1/100 =0.01 mole of Si atoms(0.01gm atom)

SAQ 7: (i) Molecular Mass of CaCO3 = 40 + 12 + 3X16= 100

The mass of 6.023X1023 molecules is gm molecular mass =100gms(ii) Molecular Mass of S

8= 8X32=256, So the mass of 6.023X1023 molecules of S

8

is 256gms.(iii) Molecular Mass of HCl = 1+35.5 = 36.5, So the mass of 6.023X1023 molecules

of HCl = 36.5gms.(iv) Molecular Mass of H

2SO

4 = 2+32 +64 =98, So the mass of 6.023X1023 molecules

of H2SO

4 is 98 gms.

SAQ 8:(i) Ionic Mass of NO3

- = 14 + 48 = 62, So 6.023X1023 number of NO3- ions wil

weigh 62gms.(ii) Ionic Mass of Cr

2O

7 2- = 2X52+ 7X16 = 216, So the mass of 6.023X1023 number

of Cr2O

7 2- is 216gms.

SAQ 9:(i) Molecular Mass of Na2CO

3 = 2X23+12+3X16=106,

So 6.023X1023 molecules of Na2CO

3 weighs 106 gms.

2.0076 X 1020 molecules of Na2CO

3 must weigh (106/6.023 X 1023)X2.0076 X

1020 =35.33 X10-3gm = 0.03533 gm = 35.33 mg(ii) Molecular Mass of H

2SO

4 = 2+32+64=98,

6.023X1023 molecules of H2SO

4 weighs 98 gms.

3.0115 X 1040 molecules of H2SO

4 will weigh (98/6.023 X 1023)X3.0115 X 1040

= 490 X 10 16gm = 490 X 10 13 kg = 490 X 10 10 quintals(!!!!).SAQ 10:(i) Molecular Mass of P

4 = 4X31=124

124 gms of P4 = 1 mole ⇒ 18.6gms of P

4 = (1/124)X18.6= 0.15 mole

(ii) Molecular Mass of H2SO

4 = 98

98 gms of H2SO

4 =1 mole

1.47kg i.e 1470gm of H2SO

4 = (1/98)X1470 = 15 moles.

(iii) Molecular Mass of Cl2 = 2X35.5=71

71gm of Cl2 = 1 mole ⇒ 3.55gm of Cl

2 = (1/71)X3.55= 1/2 =0.05 mole.

Note that in this case we had to find out the number of molecules(not atoms) as chrorine is adiatomic molecule.

50

Atoms and Molecules

Dr. S. S. Tripathy

SAQ 11:(i) Molecular Mass of Na2CO

3 = 46 + 12+48=106,

106gm of Na2CO

3 = 1 mole

0.106 gm of Na2CO

3 = (1/106)X0.106=0.001 mole

1 mole of Na2CO

3 contains 6.023 X 1023 molecules

So 0.001 mole of Na2CO

3 must contain 0.001X6.023 X 1023= 6.023 X 1020 molecules

Alternatively:106gm of Na

2CO

3 contains 6.023 X 1023 molecules

0.106 gm of Na2CO

3 contains (6.023 X 1023/106)X0.106 = 6.023 X 1020

moleculesYou are advised to adopt any method you like.

(ii) Molecular Mass of O2 =32,

32 gms of Oxgyen = 1 mole of O2 .

48gms of oxygen = (1/32)X 48 = 1.5 moles of O2

1 mole of O2 conatins 6.023 X 1023 molecules

1.5 moles of O2 must contain 2 X6.023 X 1023= 9.0345 X 1023 molecules.

(iii) Molecular Mass of P4 = 4X31=124

124gms of P4 = 1 mole

15.5 gms of P4 = 1/124 X 15.5 = 0.125 mole.

1 mole of P4 contains 6.023 X 1023 molecules of P

4

0.125 mole of P4 must contain 0.125 X6.023 X 1023 = 7.528 X 1022 molecules.

(iv) 98gms(gm molecular mass) of H2SO

4 = 1mole

0.49gm of H2SO

4 = 1/98 X 0.49 = 0.005 mole

1 mole of H2SO

4 contains 6.023 X 1023 molecules

0.005 mole of H2SO

4 must contain 0.005 X 6.023 X 1023 = 3.01 X 1022 molecules.

(v) Mol. Mass of N2 = 28,

28 gm of N2 = 1 mole

2.8kg i.e 2800gm of N2 = 1/28 X 2800 = 100 moles.

1 mole of N2 conains 6.023 X 1023 molecules of N

2

100 moles of N2 must cotain 100X6.023 X 1023= 6.023 X 1025 molecules.

SAQ 12:(i) Mol. Mass of H2 = 2,

6.023 X 1023 number of H2 molecules 2gm (gm. molecular mass)

1023 number of molecules will weigh (2/6.023 X 1023) X 1023 = 0.332gm(ii) M.M of K

2SO

4 = 2X39+32+4X16=174

6.023 X 1023 molecules of K2SO

4 weigh 174gm

So 100 million i.e 100 X106 molecules will weigh (174/6.023 X 1023 )X(100X106)=29 X10-15 gm

(iii) M.M of N2 =28

6.023 X 1023 molecules of N2 weigh 28 gm

So 1.2 X 10 69molecules of N2 will weigh

(28/6.023 X1023)X(1.2X1069)=5.578X1046gmSAQ 13:(i)M.M of Ca

3(PO

4)

2 = 3X40+2X31+8X16= 310

1 mole of Ca3(PO

4)

2= 310gms

0.001mole = 310X0.001=0.31gm.

51

Atoms and Molecules

Dr. S. S. Tripathy

(ii) M.M of H2SO

4= 98,

1 mole of H2SO

4 = 98gms ⇒ 2.5moles of H

2SO

4= 2.5X98=245gms.

(iii) M.M of MgCO3 =24+12+48=84

1mole of MgCO3 = 84gms

1/20 mole of MgCO3 = 1/20X84 = 4.2 gms.

SAQ 14: (i) Mass of 6.023X1023 atoms of Carbon = 12gmSo the mass of 1 atom of carbon = 12/(6.023X1023 ) = 2 X10-23gm.

(ii) Similarly mass of one Al atom = 27/(6.023X1023 )=4.5 X10-23 gm(iii) Mass of 1 H atom = 1/(6.023X1023 )= 1.66 X 10-24gm.(iv) Mass of 1 S atom = 32/(6.023X1023 )= 5.31 X10-23gm.(v) Massof 1 Cl atom = 35.5/(6.023X1023 )= 5.9X10-23g.(vi) Mass of 1 Ag atom = 108/(6.023X1023 )= 17.93 X 10-23gm=1.79X10-22gm

SAQ 15: 1 amu is the 1/12 part of one atom of C-12 isotope.First let us find out the mass of 1 C atom which is equal to 12/(6.023X1023 )=1.992X10-23gm.

This the mass of 12 amu.So the mass of 1 amu = (1.992X10-23)/12 = 1.66 X 10-24 gmLook to the response of previous SAQ 14 (iii), the mass of H atom was found to be 1.66X 10-24 gm. Hence mass of 1 amu and 1 H atom are same.

SAQ 16:(i) M.M of NH3 = 14+3=17

6.023X1023 molecules of NH3 weigh 17gm

So, 1 molecule of NH3 will weigh 17/(6.023X1023 )= 2.82 X10-23gm.

(ii) M.M of CaCO3 = 40 + 12 + 48 =100,

So the mass of 1 molecule of CaCO3 = 100/(6.023X1023 )= 1.66X10-22gm

(iii) M.M of (NH4)

2SO

4 = 2(14+4)+32+64=132.

So the mass of 1 molecule of (NH4)

2SO

4 = 132/(6.023X1023)= 2.19X10-22gm

(iv) M.M of CH3COOH = 2X12+4+32=60

So the mass of 1 molecule of CH3COOH= 60/(6.023X1023) = 9.96X10-23gm

(v) Formula Mass of NaCl = 23+35.5 = 58.5So the mass of 1 NaCl = 58.5/(6.023X1023)=9.71X10-23gm

SAQ 17:(i) M. M of CO

2= 44

So 44gms of CO2 occupy 22.4 litres at NTP

So 4 gm of CO2 will occupy (22.4/44)X4 = 2.0363 litres= 2036.3 mls.

44gms. of CO2 contains 6.023X1023 molecules

4 gm will contain (6.023X1023 /44)X4 = 5.47X1022 molecules.(ii) M.M of N

2 =28

22400ml of N2 gas at NTP will weigh 28gm

560ml of N2 gas at NTP will weigh (28/22400)X560 = 0.7gm.

For finding the number of moles, we can either start from gm. molecular massor gm. molar volume.

22400 ml of N2 at NTP = 1mole

560 ml of N2 at NTP = (1/22400)X560 = 1/40 =0.025 mole

Alternatively,28gms of N

2 = 1mole

52

Atoms and Molecules

Dr. S. S. Tripathy

0.7gm of N2 = 0.7/28= 0.025 mole

22400 ml of N2 at NTP contains 6.023X1023 molecules

560 ml of N2 at NTP must contain (6.023X1023 /22400)X560 = 1.5X1022

molecules.You can also calculate the number of molecules from the mass of N

2(0.7gm)

(iii) Since volume data is given, we have to convert it to NTP condition at once.The NTP volume = V

2 = 732.79 ml

22400ml of O2 at NTP contains 6.023X1023 molecules

732.79ml of O2 at NTP must contain (6.023X1023/22400) X 732.79 =

1.97X1022molecules(iv) Since the volume has been asked, the conversion is to be done at the end.

1 mole of H2 occupies 22.4 litres at NTP

So 0.5 mole of H2 must occupy 11.2 litres at NTP.

Now convert this volume to the volume at required conditions:(11.2 l X 760mm)/273K = (V

2 X 900)/300K ⇒ V

2 = 10.393litres.

ANSWERS TO PRACTICE QUESTIONS

1. (i) 40 gms of Ca = 1gm atom(1 mole of atoms)So 4kg i.e 4000gm of Ca = (1/40)X4000= 100 gm atoms.1 gm. atom contains 6.023X1023 of Ca atoms100gm atoms must contain 6.023X1023 X100 = 6.023X1025 atoms of Ca

(Note that you can also find the number of atoms from gm. atomic mass, i.e 40 gms ofCa contains 6.023X1023 of atoms and so 4000 gms will contain how many atoms?)

(ii) 65.4 gms of Zn = 1gm atom(mole of atoms)So 32.7 gms of Zn = (1/65.4)X32.7 = 0.5 gm atom65.4 gms of Zn contains 6.023X1023 of atoms32.7gms of Zn must contain (6.023X1023 /65.4)X32.7= 3.0115 X1023 atoms.

[Note that you can also find the number of atoms from the number of gm. atoms calculatedas in (i)](iii) 35.45gms of Cl= 1 gm atom

7.09 gms of Cl =(1/35.45)X7.09= 0.2 gm atom1 gm atom of Cl contains 6.023X1023 atoms of Cl0.2 gm atom of Cl must contain 6.023X1023 X 0.2= 1.2046 X 1023 atoms

(iv) 63.55 gms of Cu = 1 gm atom(mole of atoms)95.4gms of Cu= (1/63.55)X95.4=1.5 gm atoms.1 gm. atom of Cu contains 6.023X1023 of atomsSo 1.5 gm atoms of Cu must contain 6.023X1023 X1.5= 9.03 X 1023 atoms.

(v) 55.85gms of Fe= 1gm atom(mole of atoms)8.62gms of Fe = (1/55.85)X8.62=0.154 gm atom1 gm atom of Fe contains 6.023X1023 atoms0.154 gm. atom of Fe must contain 6.023X1023 X0.154=0.927 X 1023 atoms.

2. M.M of H2= 2

2gms of H2 contains 6.023X1023 molecules of H

2

53

Atoms and Molecules

Dr. S. S. Tripathy

8.5 gms of H2 must contain (6.023X1023 /2)X8.5=25.597X 1023 molecules

1 H2 molecule contains 2 H atoms(since hydrogen molecule is diatomic)

25.597X 1023 molecules of H2contains 25.597X 1023 X 2 =51.195 X1023 atoms

of H3. Atomic Mass of S = 32

6.023X 1023 atoms of S weighs 32 gms1 atom of S must weigh 32/(6.023X 1023)= 5.313 X 10-23 gm.

4. (i) 6.023X1023 atoms of H weigh 1gm (gm atomic mass)12.046 X 1024 atoms of H must weigh (1/6.023X1023 )X12.046 X 1024 =20gms

(ii) 6.023X1023 atoms of of Zn weighs 65.5gm(gm atomic mass)3.0115 X 1030 atoms of Zn must weigh (65.5/6.023X1023 )X3.0115 X 1030

=32.75X107gm.(iii) 6.023X1023 atoms of of Ag weighs 108gms(gm atomic mass)

3.6138 X 1022 atoms of Ag must contain (108/6.023X1023 )X3.6138 X 1022

=6.48gm.

5. M.M of H2O=18

18gms of H2O = 1 mole ⇒ 48 gms of H

2O = (1/18)X48= 2.67 moles

1 mole of H2O contains 6.023X1023 number of molecules

2.67 moles will contain 6.023X1023 X2.67 =16.08 X 1023 molecules.

6. M.M of glucose(C6H

12O

6) = 6X12 + 12X1+ 6X16=180

1 mole of glucose = 180 gms0.004 mole of glucose = 180X0.004 = 0.72gm6.023X1023 molecules of glucose weigh 180 gms.1 molecule of glucose must weigh 180/6.023X1023 =29.88 X10-23=2.98X10-22gm.

7. (i) N molecules of CO2 weigh 44gms ( N = 6.023X1023 )

3N molecules of CO2 must weigh 44X3=132 gms.

N molecules at NTP occupy 22.4 litres3N molecules at NTP must occuply 22.4X3= 67.2litres.

Alternatively:44 gms. of CO

2 occupies 22.4 litres at NTP

132gms of CO2 occupies (22.4/44)X132 = 67.2 litres.

(ii) N molecules of nitrogen gas weighs 28 gm(N =6.023X1023 )0.5 N molecules of nitrogen gas must weigh 0.5X28=14gms.N molecules of nitrogen gas at NTP occupy 22.4 litres0.5N molecules of nitrogen must occupy 0.5X22.4=11.2 litres at NPT.

(iii) N molecules of O2 weighs 32 gms.

0.0002N molecules of O2 must weigh 32X0.0002 = 0.0064gm

N molecules of O2 gas occupy 22.4litres at NTP

0.0002N molecules of O2 gas must occupy 22.4X 0.0002=0.00448litres=4.48ml

8. M.M of H2O= 18

18gms of water contains 6.023X1023 molecules.

54

Atoms and Molecules

Dr. S. S. Tripathy

9gms of water must contain 3.0115X 1023 molecules1 H

2O molecules contain 2 H atoms and 1 O atom

3.0115X1023 molecules contain 2X3.0115X1023 H atoms and 3.0115X1023 Oatoms

Hence the number of hydrogen atoms= 6.023X1023 and number oxygenatoms=3.0115X1023

9. (i) M.M of CaCO3 =40+12+48=100

100gms of CaCO3 = 1 mole

10gms of CaCO3 = 1/10=0.1 mole.

1 mole of CaCO3 contains 6.023X1023 molecules

0.1 mole of CaCO3 must contain 6.023X1023 X0.1= 6.023X1022 molecules

Alternatively:100gms of CaCO

3 cotanins 6.023X1023 molecules

10gms of CaCO3 must contain (6.023X1023 /100) X 10 = 6.023X1022 molecules.

(ii) M.M of H2SO

4 = 98,

98gms of H2SO

4 = 1 mole

3.5kg i.e 3500gm of H2SO

4 = 3500/98=35.71 moles

1 mole contains 6.023X1023 molecules35.71 moles must contain 6.023X1023 X35.71= 2.15 X1025 molecules.

(iii) M.M of KMnO4= 39+55+64=158

158gms of KMnO4 = 1 mole

3.16gms of KMnO4 = 3.16/158= 0.02 mole

1 mole contains 6.023X1023 molecules0.02 mole must contain 6.023X1023 X 0.02= 1.2046X1022 molecules.

10. Ionic Mass of NO3

- = 14+48=6262gms of NO

3- contains 6.023X1023 ions. This is what the question asks.

6.023X1023 number of NO3

- ions weigh 62gm1 ion of NO

3- must weigh 62/(6.023X1023 ) = 1.029X10-22 gm.

11. M.M of H2SO

4 = 98

98 gms of H2SO

4 contains 6.023X1023 molecules

7gm of H2SO

4 must contain (6.023X1023 /98)X7= 4.3X 1022 molecules

1 molecule of H2SO

4 contains 2 H atoms, 1 S atom and 4 O atoms.

So 4.3X 1022 molecules must contain 2X4.3X 1022 H atoms, 1X4.3X 1022 Satoms and 4X 4.3 X 1022 O atoms.

So the number of H atoms = 8.6X1022, number of S atoms= 4.3X 1022

and the number of O atoms =1.72 X 1023

12. 44gms of CO2 contains 1gm atom of C and 2 gm atoms of O.

22gms of CO2 contains 0.5gm atom of C and 1gm atom of O.

No. of C atoms = 0.5X 6.023X1023 and no. of O atoms = 1X 6.023X1023

13. 100gms of CaCO3 contains 3X 6.023X1023 number of O atoms

25gms of CaCO3 must contain (3/4) X 6.023X1023 number of O atoms.

100gms of CaCO3 contains 3 gm. atoms(moles of atoms) of O.

25gms of CaCO3 must contaian 3/4=0.75 gm atoms of O.

55

Atoms and Molecules

Dr. S. S. Tripathy

14. M. M of NH4Cl = 14 + 4 + 35.5= 53.553.5gms of NH

4Cl contains 4 X 6.023X1023 atoms of H

25gms of NH4Cl must contain 11.26 X1023 atoms of H

53.5gms of NH4Cl contains 1 mole of it.

25gms of NH4Cl contains 25/53.5= 0.467 mole.

SET-II1 (i) 11gms of B contains 6.023X1023 atoms.

So 594gms of B contains (6.023X1023)/11 X 594= 3.252X 1025 atoms.(ii) No of Ca atoms = (40/10)X 6.023X1023 = 2.409 X1024 atoms.(iii) No. of Fe atoms = (8.37/56)X 6.023X1023 = 9.0 X1022 atoms(iv) 7gms of Li = 1gm. atom of Li

1 gm of Li = 1/7 = 0.143 gm atom.(v) No. of U atoms = (1/238)X6.023X1023 = 2.53X1021 atoms.

2. 1.5 X1022 atoms of an element weighs 1gmSo 6.023X1023 atoms of the element must weigh (6.023X1023)/(1.56 X1022)=38.6So the atomic mass of the element= 38.6.

3. 32gms of S contains 6.023X1023 atoms.20gms of S must contain (20/32) X6.023X1023 = 3.76 X1023 atoms.32gms of S = 1 gm atom ⇒ 20gms of S= 20/32=5/8 gm.

atom.4. (i) 0.14/28=0.005 mole (ii)M.M of NO

2=14+32=46,

The number of moles= 9.2/46= 0.2(iii) 6.023X1023 molecules of N

2O weighs 1 mole

3.5 X1022 molecules of N2O weights (3.5X1022)/(6.023X1023)= 0.058 mole.

5. 44gms of CO2 = 1mole, So 2.2 gm= 2.2/44= 0.05 mole.

6. M.M of CO=12+16=28; 28gms contain 6.023X1023 molecules,So 2.4 gms of CO contains (2.4/28)X6.023X1023= 5.16 X1022 molecules.

7. (i) 6.023X1023 number of molecules weigh 44gms So 1 molecule must weigh 44/(6.023X1023) = 7.3 X 10-23gm.

(ii) 197/(6.023X1023)=3.27X10-22gm. (iii) 98/(6.023X1023) = 1.627X10-

22 gm.(iv) 1/(6.023X1023)= 1.66 X10-24gm.

8. 1 atom of Hg weighs 33.3 X 10-23gmsSo 6.023X1023 atoms of Hg must weigh (6.023X1023)X(33.3 X10-23)= 200.56(atomicmass)

9. 1 atom of the element weighs 6.6 X 10-23 gm6.023X1023 atoms of the element must weigh 6.023X1023 X 6.6 X 10-23 =39.7539.75gms = 1 gm atom ⇒ 20kg i.e 20,000gms = 20000/39.75= 503.144 gm. atoms.

10. M.M of CH4=16.

16gms of CH4 contains 6.0.23 X 1023 molecules.

20gms of CH4 must contain (20/16) X6.0.23 X 1023 =7.528 X1023 molecules.

1 CH4 molecule contains 1 C atom

7.528 X1023 molecules must contain 7.528 X1023 C atoms

56

Atoms and Molecules

Dr. S. S. Tripathy

1 CH4 molecule contains 4 H atoms

7.528 X1023 molecules must contain 4X7.528 X1023 =3.011X1024 H atoms.11. M.M of C

12H

22O

11= 12X12+22+11X16=342

342gms of sucrose contains 6.023 X1023 molecules2.56gms of sucrose must contain (2.56/342)X6.023 X1023 =4.5X1021 molecules1 molecule of sucrose contains 12 C atoms.4.5X1021 molecules of sucrose must contain 12 X 4.5X1021 C atoms.1 molecule of sucrose contains 22 H atoms4.5X1021 molecules of sucrose must contain 22 X4.5X1021 H atoms1 molecule of sucrose contains 12 O atoms4.5X1021 molecules of sucrose must contain 12 X4.5X1021 O atoms.

12. (i)8/44=0.18mole (ii)8/32=0.25mole (iii)8/28=0.285mole(iv)8/2=4moles

Since the number of molecules is directly proportional to the number of moles, 8gms ofH

2(iv)contains highest number of molecules and 8gms of CO

2(i) contains the lowest

number of molecules.13. (i) No. of moles = 19.2/98=0.196; No. of molecules =0.196 X6.023 X 1023

(ii) No. of moles = 4400/44=100; Hence no of molecules =100X.6.23 X 1023

(iii) M.M of (NH4)

2Cr

2O

7 = 2(18)+2X52 + 7X16= 252

So the number of moles = 1000/252=3.97; No. of molecules=3.97X6.023 X 1023

14. (i) 44gms contain 6.023 X 1023 molecules of CO

2

4gms of CO2 contains 6.023 X 1022

molecules

1 CO2 molecule contains 1 C atom

6.023 X 1022 molecules must contain 6.023 X 1022

C atoms.

1 CO2 molecules cotains 2 O atoms

So 6.023 X 1022 molecules must contain 2X6.023 X 1022

O atoms.

(ii) 27gms of Al contains 6.023 X 1023 atoms

So 2.7gms of Al must contain 6.023 X 1022 atoms.(iii) M.M of CaCO

3 =100,

100 gms of CaCO3 cotains 6.023 X 1023

molecules

10gms of CaCO3 must contain 6.023 X 1022

molecules

So the number of Ca atoms= 6.023 X 1022

The number of C atoms =6.023 X 1022 and number of O atoms = 3X6.023 X

1022

15. 65gms of Zn contains 6.023 X 1023 atoms of Zn

0.65gm of Zn must contain 6.023 X 1021 atoms of Zn

16. 64gms of SO2 occupy 22.4litres, So 2.4gm must occupy 0.84 litre at NTP.

17. 22400ml of H2 gas at NTP contains 6.023X1023molecules, So 1.15 X 10-7ml will contain

3.1 X 1012 molecules.18. 4gms of He contains N atoms, so 2gms of He must contain N/2 atoms(N=6.023X1023)

4gms occupy 22.4litres, so 2gms must occupy 11.2 litres.(note that He monoatomic)19. density = m/v ⇒ mass of 1cc of water = 1gm.. We know that 18gms of H

2O contains

N molecules, So 1 gm must contain N/18 molecules of H2O. (N=6.023X1023)

20. (i)N/224 molecules (ii)N/22.4 molecules

57

Atoms and Molecules

Dr. S. S. Tripathy

(iii) First the volume is converted to the NTP conditions. V2= 239. 47ml.

22400ml at NTP contains N molecules, so 239.47ml must contain 0.01 N molecules.(N=6.023X1023)

21. (a)32.7/65.5= 0.5 (b)7.09/35.5= 0.2 (c)95.4/63.5=1.5(d)4.31/56=0.0769 (e)0.328/32= 0.01

22. (i) 6.023 X 1023 molecules of CO occupy 22.4 litres3 X 1021 molecules of CO must occupy 0.112 litre= 112ml

(ii) First let us find the volume at NTP.Volume at NTP = 892.5 litres. Then convert to given conditions.(760mm X 892.5 l)/273 = (800mm XV

2)/300 ⇒ V

2 = 931.73litres.

23. (a) 1/12 mole(gm atom) (b)1 mole(1gm atom)(c) 5.66 X 1020/N=9.4X10-4 mole(gm atom) (N=6.023X1023)

24. (a) 3X108gms (ii)0.5 X20gms (iii)2.5X44gms (iv)0.01X 138gms25. N atoms = 1gm atom, So 3X1024 atoms = 5 gm atoms. The mass of 5gm.

atoms=5X32=160 gms.26. 1 mole of Na

2SO

4 contains 2 N atoms of Na and N atoms S and 4N atoms of O.

So 0.25 mole of Na2SO

4 must contain 0.5 N atoms of Na and 0.25atoms of S and N

atoms of O. (N=6.023X1023)27. If 7.525 X 1023 molecules weigh 35 gms, N molecules must weigh 27.9gms(MM).

27.9gms of the gas occupy 22.4 litres, So 5gms will occupy (22.4/29.9)X5= 3.74 litres28. 44gms of CO

2 contains 2X N atoms of O. So 17.6gms will contain 0.8 X N atoms of O.

29. 1 mole of HNO3 weighs 63gms, So 0.005mole must weigh 0.315gm. The no. of moleucles

present is 0.005 X N. (N=6.023X1023)30. MM of Na

2SO

4 = 142; 142gms of Na

2SO

4 contain N SO

42- ions,

So 2.5 gms of Na2SO

4 must contain 0.0176N sulphate ions.(N=6.023X1023).

1

Stoichiometry-I

Dr. S. S.Tripathy

STOICHIOMETRIC CALCULATIONS

The term stoichiometry is derived from two Greek words, 'stoicheon' which means element and 'metron' whichmeans measure. Stoichiometry essentially deals with the study of quantitative relationship among elements in acompound or in a chemical reaction. Stoichiometry hence is divided into two types.

(i) Composition stoichiometry:This deals with the mass relationship among elements in a particular compound. For example, water(H

2O) contains

the elements H and O in the mass ratio 2:16 = 1:8 and methane(CH4) contains C and H in the ratio 12:4 =3:1 and

so on. The law of definite or fixed proportion is the basis of this stoichiometry i.e there is a fixed ratio of massesamong elements in a compound.PERCENT COMPOSITION, EMPIRICAL AND MOLECULAR FORMULA

PERCENTAGE COMPOSITIONThe percentage by mass of each element present in a compound can be found out and compared. Take thesimplest case of H

2O. What is the percentage of H and O present in H

2O? If in every 18gm(M.M) of H

2O,

there are 16gms of O and 2gm of H, For 100gms of H2O, there is (16/18)X100=88.89gms of O and

(2/18)X100=11.11gms of H. So % of O = 88.89% and that of H=11.11%. Mass of an element present in100gms of the compound gives the percentage composition of that element.SAQ 1: Calculate the percent by mass of O in Ca(ClO

3)

2.

SAQ 2: Find the percent of nitrogen in (i)NH4NO

3 and (ii)(NH

4)

2SO

4 and indicate which is a better

nitrogenous fertilizer?SAQ 3: Calculate the percent composition of K

2Cr

2O

7. (K=39, Cr=52, O=16)

SAQ 4: What is the percent by mass of Cu in CuO? From this, find mass of CuO that will be required toproduce 100kg of Cu? (Cu=63.5, O=16)SAQ 5: A sample impure Cu

2O contain 66.6% Copper. What is percentage of pure Cu

2O in the sample?

EMPIRICAL FORMULA AND MOLECULAR FORMULA:Empirical formula of a compound is the simplest whole number ratio of atoms which are present in amolecule. For example glucose has the molecular formula C

6H

12O

6 but its empirical formula will be obtained

by dividing all the coefficients by the highest common factor(HCF) 6. Hence its empirical formula is CH2O.

Molecular Formula, on the other hand, represents actual number of atoms of each kind present in a molecule.Empirical formula is related to the molecular formula of a compound as follows.

Molecular Formula = (Empirical Formula)n

(Where n is a whole number)

SAQ 6: The molecular formula of a compound is C7H

14, find its empirical formula.

SAQ 7: The molecular formula of a compound is C4H

8O

2. What is its empirical formula?

Empirical formula Mass:The mass(relative)obtained from the empirical formula is called empirical formula mass. For example, theempirical formula mass of CH

2O is (12+2+16)=30, but the molecular mass is mass obtained from the

molecular formula. The molecular mass of C6H

12O

6= 72+12+96=180. So if you know the molecular mass of

a compound and its empirical formula mass, you can easily find the molecular formula. Look to this example.Example: The empirical formula of a compound is CH

2O and its molecular mass is 180, find the

molecular formula.Solution: The empirical formula mass = 12+2+16=30

We know that Molecular Formula = (Empirical Formula)n:(Where n is a whole number)

So molecular mass = n × empirical formula massn = molecular mass/empirical formula mass = 180/30 =6So molecular formula = (CH

2O)

6 = C

6H

12O

6

Determination of Empirical Formula:The empirical formula of a compound can be calculated from the percent composition of the compound. Thisis just the opposite of finding percentage composition from the molecular formula of a compound discussedbefore. The difference is that here we cannot find the molecular formula. In stead we shall get the empiricalformula which may be same or different from the actual molecular formula. This is because both empiricalformula and molecular formula have the same percent composition of elements. For example, the empiricalformula CH

2 and the molecular formula C

3H

6 have the same % of C and H by mass.

2

Stoichiometry-I

Dr. S. S.Tripathy

Example: Find the empirical formula of a compound which contains 60% O and 40% S by mass.If its molecular mass is 80, what is its molecular formula.Solution: Let us take 100gms of the compound. This amount contains 60gms of O and 40gms of S.Let us calculate how many gm. atoms(mole of atoms) of each element present in these amounts and whatis their ratio. This is obtained by dividing the mass of the element by the atomic mass of the element. Theratio of gm atoms is same as ratio of the atoms present in the formula , hence it will give the empiricalformula. In this case,

the no. of gm. atoms of O= 60/16=3.75the no. of gm. atoms of S= 40/32= 1.25

So the ratio of gm. atoms and hence atoms of O and S present in the formula =3.75:1.25But you know that in the formula(empirical or molecular) there are whole number of atoms of the elements.So we have to convert it into simplest whole number ratio. This is done by dividing each value by the lowestof the values. In this case the lowest is 1.25.

So whole number ratio = 3.75/1.25 : 1.25/1.25 = 3 : 1Hence the empirical formula = O

3S or SO

3 : So its empirical formula mass=32+48=80

n= Molecular mass/empirical formula mass = 80/80=1So Molecular Formula = (Empirical Formula)

n,

So Molecular Formula =(SO3)

1=SO

3(same as its empirical formula).

STEPS FOR FINDING EMPIRICAL FORMULA:(i) First the percent composition data of the elements are divided by the respective atomic masses toget ratio of atoms present in the formula(ii) Make this ratio the simplest whole number ratio by dividing all the values by the lowest of the values.If this does not give the whole number ratio, multiply 2, 3 etc.to get the simplest whole number ratio.For example, if it comes 1:1.5:2.5 , you multiply throughout by 2 to make the ratio 2:3:5. If the ratio is1.33 : 2, then multiply by 3 to get the whole number ratio i.e 4:6 and similarly if the ratio is 1.25 :1, then multipyby 4 to get the whole number ratio i.e 5:4.(iii) Then empirical formula is written by placing these whole numbers as coefficients of the respectiveelements.(iv) Empirical formula mass is then found out. Then divide the molecular mass by the empirical formulamass to find the value of n. Usually the division does not give a whole number value to 'n'. Round it off toits nearest whole number to get the value of 'n'.(v) Multiply 'n' in all the coefficients of the empirical formula to get the molecular formula.To determine the empirical formula, it is better to calculate in tabular form. The previous example of SO

3

is given in the following table for better understanding. SO3 contains 60% S and 40% O by mass.

___________________________________________________________Element % At. Relative number Simple whole Empirical

Mass of atoms no. of atoms Formula___________________________________________________________

O 60 16 60/16=3.75 3.75/1.25=3SO

3

S 40 32 40/32=1.25 1.25/1.25=1___________________________________________________________

Example: A compound gave on analysis the following percent composition: K=26.57%,Cr=35.36%, O=38.07%.Derive the empirical formula.Solution: Let us take the help of the table for easy calculation.

___________________________________________________________Element % At. Mass Relative no. Simple Whole Empirical

of atoms No. of atoms Formula___________________________________________________________K 26.57 39 26.57/39=0.68 0.68/0.68=1X2= 2Cr 35.36 52 35.36/52=0.68 0.68/0.68=1X2= 2 K

2Cr

2O

7

O 38.07 16 38.07/16=2.379 2.379/0.68=3.5X2=7___________________________________________________________

Remember that fractional values like 3.95 or 3.99 which is close to a whole number is rounded off to thenearest whole number(4). But if it is 1.5, 1.33 or 2.25 etc. you cannot convert it to the nearest whole number.In that case you have to multiply a suitable minimum factor to make it a whole number.

3

Stoichiometry-I

Dr. S. S.Tripathy

SAQ 8: Find the empirical formula of a hydrocarbon that on analysis gave the following percent composition:C=85.63%, H=14.37%. If the molecular mass of the compound is 56, what is its molecular formula.SAQ 9: An oxide of nitrogen contains 30.4% nitrogen. What is its empirical formula?SAQ 10:Determine the simplest formula of a compound that has Cr=26.52%, S=24.52% and O=48.96%.Could you suggested the formula in proper order i.e in terms of acid and basic radicals with proper valences?

PRACTICE QUESTIONS1. A compound contains 21.6% sodium, 33.3% chlorine and 45.1% oxygen. Derive its empirical formula.2. Calculate the empirical formula of a compound formed when 7.3g of iron powder reacts completlywith 6.30gm of powder sulphur.3. A compound has the following percent composition: C=40%, H=6.66%, O=53.34%. Its molecularmass is 60. Derive its molecular formula.4. A compound consisting of 82.66% carbon and 17.34% hydrogen and has a vapour density 29.05.Determine its molecular formula.

(ii) Reaction StoichiometryThis deals with the quantitative relationship between the reactants and products in a chemical reaction. The lawof conservation of mass is the basis of this stoichiometry. Matter can neither be created nor destroyed i.e thetotal mass of the reactants should be equal to the total mass of the products in a chemical reaction.We know that chemical equations are used to describe chemical reactions.Balancing of an equation is doneonly to satisfy the law of conservation of mass.

2H2 + O

2 2 H

2O

2 ×2 32 2 × 18In the above reaction represented by a balanced equation, the total mass of the reactants is 36gm(2 g moles ofH

2 and one g. mole of O

2) and the total mass of the product is also 36g(2 g moles of H

2O)

Example: Calculate the mass of oxygen required to completely react with 24g of hydrogen gas.Solution: 4g of H

2 completely reacts with 32g of O

2.

So 24g of H2 will completely react with (32/4) × 24 = 192g of O

2

Example: Calculate the mass of water formed when 8g of oxygen completely reacts with excess of hydrogen.How much of H

2 is required for the reaction?

Solution: 32g of O2 on complete reaction produces 36g of water

So, 8g of O2 will produce (36/32) × 8 = 9 g of water(answer)

Again, 32g of oxygen requires 4g of hydrogen gasSo, 8g of oxygen will require (4/32) × 8 = 1 g of hydrogen gas(answer)That is why we have taken excess of hydrogen gas out of which 1g will be used to completely react with 8g ofoxygen gas and the rest of the hydrogen gas will remain unreacted. If we would have taken limited quantity ofhydrogen(say less than 1 g), then all the other reactant(oxygen in this case) would not have reacted completely.CONCLUSION: If the mass of any species i.e either any one of the reactants or any one of the products isgiven, the mass of any other species in the reactant or product side can be calculated from the balancedchemical equation.This is called stoichiometric calculations.So to study stoichiometric calculations two things are essential.(i)To predict the products correctly.(ii)To balance the equation correctly.If any one of the above two becomes wrong, the stoichiometric calculation also goes wrong.We shall study reaction stoichiometry in two parts.

(1)Stoichiometric calculations involving masses or gaseous volumes of the reactants and products

(2)Stoichiometric calculations involving the concentrations of solutions like molarity, normality etc. forthe reactants

4

Stoichiometry-I

Dr. S. S.Tripathy

STOICHIOMETRIC CALCULATIONS INVOLVING THE MASSES ANDGASEOUS VOLUMES OF THE REACTANTS AND PRODUCTSFor simplifying this study it is divided into three types.

(i)Mass-mass relationship(ii)Mass-volume relationship(iii)Gas analysis(Volume-volume relationship)

Mass-mass relationship is the original basis of stoichiometry. The other two are applicable for reactions involvinggases. Since mass of a gas is related to its volume at a particular temperature and pressure, we can directlystudy the relationship between mass of one species with the volume of the other species(gaseous) without usingthe mass of the latter. This is called mass-volume relatioship. Similary we can relate the volume of one gaseousspecies with the volume of another gaseous species instead of using their masses. This is called volume-volumerelationship.Mass-mass relationshipWorking steps:(i)Predict products and balance the equation correctly. For balancing redox reaction, follow ON method orother methods of your choice.(ii)Write the molecular masses for molecular substances and atomic masses for atomic substances below therespective reactants and products which are involved in stoichiometric calculation. Multiply the coefficientsfrom the balanced equation with the respective molecular masses or atomic masses.(iii)Then proceed with chemical arithmetic. Look to this example.Example: Find out the mass of KClO

3 that would produce 8 g of oxygen gas.

Solution: 2KClO3

2KCl + 3O2

2(39+35.5+48) 3 × 32g=2 × 122.5g

In this case, the species involved in the calculation are KClO3 and O

2. The mass of the latter is given while the

mass of the former is to be calclated . Since the mass of the product is given, it is to be written in the LHS whiledoing calculation by unitary method.3 ×32g of oxygen is produced by 2 ×122.5g of KClO

3

So, 8 g of oxygen is produced by (2×122.5)/(3×32) ×8 = 20.416g (answer)(Note that we had to use passive voice while constructing the sentence, since the product is written in the LHS).Example: Calculate the mass of chlorine gas obtained by the reaction of 8.7g of pure MnO

2 with excess of

dilute hydrochloric acid. Also calculate the mass of HCl consumed in the reaction.MnO

2 + 4HCl MnCl

2 + Cl

2 + 2H

2O

(55+32) 4×(1+35.5) 2×35.5

From the balanced equation, we know that one mole of MnO2 reacts with 4 moles of HCl to give one mole of

Cl2, one mole of MnCl

2 and 2 moles of H

2O.

(i)87g of MnO2(1 mole) produces 71 gm of chlorine

So, 8.7 g of MnO2 produces (71/87) × 8.7 = 7.1 gm of chlorine

(ii)Again, 87 gm of MnO2 reacts wit 4 × 36.5 gm of HCl

So 8.7 gm of MnO2 reacts with [(4×36.5)/87] × 8.7 = 14.6 gm of HCl

Alternative Method:We can use the mass of Cl

2 produced obtained in (i) to get mass of HCl required.

2 × 35.5 gm of Cl2 is produced by 4 × 36.5 gm of HCl

7.1 gm of Cl2 is produced by 14.6 gm of HCl

Hence we can use any data(either reactant or product) to find out the mass of any other reactant or product.SAQ 1: Calculate the mass of CO

2 obtained by completely burning 30gms of ethane with excess oxygen?

Also calculate the mass of H2O produced during the reaction.

SAQ 2: Calculate the number of moles and mass in gm of CaCl2 needed to react with excess of silver nitrate

to produce 6.6g of AgCl?(Ag=108, Ca=40)SAQ 3: Calculate the mass of CaO which will completely react with 6.92g of HCl?

5

Stoichiometry-I

Dr. S. S.Tripathy

SAQ 4: Calculate the mass of BaCO3 produced when excess CO

2 is bubbled through a solution containing

0.205mole of Ba(OH)2. (Ba = 137)

SAQ 5: What mass of sodium bicarbonate on strong heating will produce 1.5 mole of CO2?

Mass-Volume RelationshipIn this case we shall have to establish a relationship between the mass of one species(reactant or product)with the volume of another species(reactant or product). Here we shall make use of gram molar volumeof gases at NTP equal to 22.4 litres i.e one mole of any gas at NTP will have a volume equal to 22.4 litres.

Example: What volume of O2 gas be evolved at NTP when 2gms of KClO

3 is strongly heated?

3X22.42(39+35.5+48)KClO3 KCl O2+ 322

From the balanced equation, we know that 2 moles of KClO3 produces 3 moles of O

2 gas i.e 3×22.4 litres

of O2 gas at NTP. So can we not find the volume of O

2 that will be produced by 2gms of KClO

3 from these?

So for mass-volume relationship, you have to write the molecular mass with the coefficient multiplied withit below the species whose mass data is given or asked for and the molar volume (22.4litres) multiplied withits coefficient, below the species whose volume data is given or asked for. The rest of the procedure is sameas mass-mass relationship explained earlier. In the above example,

2(39+35.5+48)=2 ×122.5 gms of KClO3 produce 3×22.4 litres of O

2 gas at NTP

So 2 gms of KClO3 must produce 0.5485l =548.5ml of O

2at NTP.

Example : What mass of sodium nitrate on heating will produce 12 litres of O2 gas at NTP?

Solution: We know that NaNO3 on heating produces NaNO

2 and O

2. Let us first write the balanced equation

and the molecular mass and molar volume of the involved species i.e NaNO3 in LHS and O

2 in RHS.

22.4 l2(23+14+48)

22 +NaNO2 O2NaNO3

22.4 litres of O2 is produced by 2 × 85gms of NaNO

3

So 12 litres of O2 will be produced by = 91.07gms of NaNO

3.

SAQ 6: What volume of chlorine gas at NTP will be produced when 8.7gms of MnO2 reacts with excess

conc. HCl? (Mn=55, O=16, Cl=35.5)SAQ 7: What mass of sodium bicarbonate on reacting with excess dil. HCl will give 5.6 litres of CO

2 gas

at NTP?SAQ 8: What volume of NO

2 and what volume of O

2 be evolved at NTP when 10 gms of lead nitrate is

strongly heated?(Pb=207, N=14)SAQ 9: What mass of ammonium dichromate on strong heating will produce 11.2 litres of nitrogen gas atNTP?(N=14, Cr=52, O=16)

Procedure when volume of gas is given/asked at temperature and pressure other than NTP:(i) When volume data is given at temperature and pressure other than NTP, first you convert the givenvolume to NTP conditions by applying the combined gas equation(P

1V

1)/T

1 = (P

2V

2)/T

2.Then establish the

mass-volume relationship in the same way as done previously.(ii) When volume data is asked in the temperature and pressure other than NTP, then the conversion ofvolume from NTP conditions to the required conditions is done at the end.

Example : What mass of calcium carbonate on heating will produce 250ml of CO2 gas at 270C and

800mm of Hg pressure?Solution: Since volume of the gas has been given at a particular temperature and pressure, our first job isto convert it to NTP conditions.

273 A760mm X V2

(273+27)A =800mm X 250ml ⇒ V

2 = 239.47ml

Then we write the balanced chemical reaction. CaCO

3 CaO + CO

2

(40+12+48) 22400ml22400ml of CO

2 is evolved at NTP from 100gm of CaCO

3

So 239.47ml of CO2 is evolved at NTP from (100/22400) × 239.47 = 1.069gm(answer)

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Stoichiometry-I

Dr. S. S.Tripathy

Example: What volume of hydrogen gas will be evolved at 4500C and 700mm pressure by treating5gms of superheated iron with sufficient steam.Solution: Since in this case the volume of hydrogen is asked, we shall carry out the volume conversion atthe end. First write down the balanced equation.

3Fe + 4H2O Fe

3O

4 + 4H

2

3 ×56 gm 4 × 22400 ml

3×56gms of iron produces 4×22400ml of hydrogen gas at NPT.So 5gms of iron will produce (4×22400)/(3×56)×5= 2666.67ml of H

2 at NTP.

But we want to know the volume at some other conditions. So we shall convert now from the NTP conditionsto the given condition by using the combined gas equation.

760mm X 2666.67ml=273A

700mm X V 2(273+ 450) A ⇒ V

2 = 7667.6ml

So the volume of hydrogen collected at 4500C and 700mm pressure is 7667.7ml.

SUCCESSIVE REACTIONSIf the product of one reaction is used as a reactant in the second reaction and you are asked to establishthe relationship between the reactant of first reaction with product of second reaction, you can do so veryeasily by writing the balanced chemical reactions for both. The following example will make it more clear.Example: What mass of potassium chlorate on heating gives just sufficient oxygen gas to completelyburn 8gms of sulphur?Solution: Let us write the first reaction

2KClO3 2KCl + 3O

2

2(39+35.5+48) 3×32The oxygen evolved in the first reaction is used to burn sulphur in the second reaction.

S + O2 SO

2

32 32Since the reactant of first reaction has been asked for, we start to solve this problem from the second reactionand know how much of oxygen is necessary to completely burn 8gms of sulphur.2nd reaction:

32gms of sulphur needs 32gms of oxygen gas for complete burning.So 8gms of sulphur will need 8gms of oxygen gas

Now let us make use of the amount of oxygen (8gms) in the first reaction and know how much KClO3 on

heating will give 8gms of oxgyen.1st reaction:

3×32gms of O2 gas is produced by 2(39+35.5+48)=2×122.5gms of KClO

3

So 8gms of O2 gas will be produced by (2×122.5)/(3×32)×8=20.416gms of KClO

3.

Hence 20.416gms of KClO3 on heating will give enough oxygen gas(8gms) to burn 8gms of sulphur completely

to SO2.

SAQ 10: What mass of ammonium sulphate is required which on reacting with excess of alkali will produceenough ammonia which can reduce 2.5 gm of cupric oxide completely to metallic copper.(Cu=63.5)

(Hint: (NH4)

2SO

4 + NaOH → Na

2SO

4 + NH

3 + H

2O

CuO + NH3 → N

2 + Cu + H

2O

SAQ 11: 4 gm of CaCO3 was strongly heated and the residue obtained was mixed with excess of

ammonium chloride and heated. Calculate the volume of ammonia produced at 270C and 800mm pressure.SAQ 12: Calculate the mass of KMnO

4 required to produce enough chlorine(by reacting with conc. HCl)

which will completely react with 10gm of pure NaOH(hot and concentrated) to produce NaCl and NaClO3.

LIMITING REACTANT CONCEPTWhen the amounts of two reactants used in a reaction are known there can be two possibilities(i) The two reactants might be consumed wholly and there would be no excess reactant left after thereaction is over. This is the case when exact stoichiometric proportions of reactants are taken.(ii) If one reactant is completely exhausted and the other reactant remains in excess after the end thereaction.If in such a case, you are asked to calculate the amount of product formed, then how you proceed? Firstyou have to check on trial and error basis if both reactants are exhausted (type i) or one reactant is exhausted

7

Stoichiometry-I

Dr. S. S.Tripathy

(type ii). If it belongs to type (i), then you can take any one of the reactant data and calculate the amountof the product. However if the reaction belongs to type (ii), you have to use the limiting reactant i.e thereactant which has been exhausted completely to find the amount of the product. If you take the otherreactant which has been taken in excess for calculating the product, then you will get a wrong answer,because the excess amount of the reactant won't react as there is no equivalent amount of the other reactant.Example: 6 g of carbon was burned with 20 g of oxgyen gas, how much of CO

2 gas will be

formed? What volume of the gas will be formed at NTP.Solution:

C + O2 CO

2

12g 32g 44g ( or 22.4 l)Let us take the carbon amount:

12gm of carbon requires 32 gms of oxygen for complete burning.So 6gms of carbon must require 16gms of oxgyen for complete burning.

But 20gms of oxygen gas has been taken. So oxygen gas will remain in excess(4gms) after the end of thereaction. So carbon is the limiting reactant and we have to use the mass of carbon to find the amount ofproduct(not the mass of oxygen).

12gms of of carbon produce 44gms of CO2 gas,

So 6gms of carbon must produce (44/12)×6 =22gms of CO2 gas.

Again 12gms of carbon produces 22.4 litres of CO2 gas at NTP.

So 6gms of carbon must produce 11.2 litres of CO2 at NTP .(You can also convert mass of CO

2

directly into volume)Example : A mixture of 100 g H

2 and 100 g O

2 is ignited so that water is formed according to

the reaction, 2H2+O

2 2H

2O. How much water is formed?

Solution:Let us first take hydrogen data to check whether it is limiting reactant or excess reactant.

2H2 + O

2 2H

2O

2×2 32 2X184 gms of H

2 needs 32 gms of O

2 for complete reaction.

So 100gms of H2 must need (32/4)×100 =800gms of O

2.

But we have only 100gms of O2 in the reaction. So all H

2 cannot be used and therefore O

2 is the limiting

reactant and H2 will remains in excess after the reaction. Note that you can know which one is the limiting

reactant and which remains in excess by selecting any reactant data and then by finding the amount of theother reactant needed for it from the balanced equation. So now let us take the oxgyen data(limiting reactant)to find the amount of water formed.

32gms of O2 produce 36gms of H

2O

So 100gms of O2 must produce (36/32)×100= 112.5gms.

SAQ 13: Calculate the mass of NaCl produced by the reaction of 5.3 gm of Na2CO

3 with 5.3 gm of pure

HCl.SAQ 14: Find the mass of barium sulphate formed when 0.1 mole of barium chloride reacts with 0.05 moleof sodium phosphate. (Ba=137, P=31, O=16, Cl=35.5)

PROBLEMS BASED ON PURITY OF SAMPLE:In this case the mass of an impure sample which reacts with another reactant is given and the product amountis also given, you are asked to find the percentage of purity of the original impure sample. What you willdo in this case is to find the mass of the reactant that would produce the given quantity of the product fromthe balanced equation. In such case you will not use the mass data of the impure sample because thatamount contains impurity which does not react. After finding the mass of pure sample from stoichiometricanalysis, then find the percentage of purity.

percentage of purity = (mass of pure sample)/(mass of impure sample)×100.

Example: 4gms of an impure sample of calcium carbonate (containing sand as impurity) is treatedwith an excess of hydrochloric acid. 0.88g of CO

2 is produced. What is the percentage of pure CaCO

3

in the original sample?Solution:The impure sample taken may be lime stone, marble, chalk etc. which are the minerals containing CaCO

3.

The impure sample contains sand as impurity. Our interest is to find the mass of pure CaCO3 present in the

sample. Note that when the sample will react with acid, only CaCO3 reacts but not the impurity, according

8

Stoichiometry-I

Dr. S. S.Tripathy

to the equation.CaCO

3 + 2HCl CaCl

2 + CO

2 + H

2O

(40+12+48) 4444gms of CO

2 is produced by (40+12+48)100gms of CaCO

3

0.88gm of CO2 must be produced by (100/44)×0.88= 2gm of CaCO

3

So the amount of pure CaCO3 present in 4gm of impure sample is 2gms.

So percentage of purity = (2/4) ×100= 50SAQ 15: Calculate the amount of lime(CaO) that can be prepared by heating 200kg of limestone that is 95%pure CaCO

3?

SAQ 16: 1.0 gm sample of impure zinc on treatment with excess dil. H2SO

4 produced 250 mL of hydrogen

ver water at 200C and 780 mm pressure. Calculate the percentage of purity in the sample. (aqueous tensionat 200C = 15 mm of Hg) Zn = 65.5 (68.5)SAQ 17: A silver coin weighing 7.0 gm was completely dissolvoed in conc. HNO

3 and then excess of HCl

solution was added. The white precipitate formed was dried and weighed to be 9.0 gm. Calculate thepercentage of silver present in the coin.(Ag=108) (96.7)SAQ 18: 35.0 gm of an impure sample of potassium dichromate on complete reaction with excess of H

2S

deposited 9.6 gm of sulphur in the acidic medium. Calculate the % of purity in the sample. (Cr=52, K=39,S=32, O=16) (84)

BASED ON COMPOSITION OF A MIXTURE CONTAINING ONE REACTIVE SPECIES:This is similar to the problems based on purity discussed before. One of the components of the mixture isreactive while the other remains inert.Example: 10 gm of a mixture of KCl and KNO

3 was dissolved in water and treated with excess of AgNO

3

to produce a white precipitate weighing 14.35 g. Determine the composition of the mixture.(Ag=108, K=39)Solution:

Out of the two components present in the mixture, KCl takes part in the reaction with AgNO3 while

KNO3 remains as such without any change. The mass of KCl can be calculated by the help of the massof AgCl formed.

KCl + AgNO3 KNO3 + AgCl

(39 + 35.5) (108+ 35.5) 74.5 143.5143.5 gm of AgCl is produced by 74.5 gm of KClSo, 14.35 gm of AgCl is produced by (74.5/143.5) × 14.35 = 7.45 gm

Since the mass of the mixture is 10gm, mass of KNO3 = 10-7.45 = 2.55 gm

So percentage of KNO3 = (2.55/10) × 100 = 25.5% and percentage of KCl = 74.5%

SAQ 19: A 3.0 gm of sample of NaHCO3 and Na

2CO

3 lose 0.45 gm when heated strongly until constant

mas is attained. What is the percentage composition of the original mixture. (NaHCO3= 40.64%)

COMPOSITION OF A MIXTURE CONTAINING TWO REACTIVE SPECIES:In this case both the components in the mixture are reactive. In such case we have to solve by algebraicmethod taking the mass of one as x gm and the other as total mass minus x gm.Example: A mixture of NaCl and KCl weighing 3.0 gm was dissolved in water and treated with excess ofAgNO

3 solution. The white precipitate obtained was filtered and dried to give a constant mass of 6.3 gm.

Find the % of KCl in the mixture.Solution:In this case both NaCl and KCl react with AgNO

3 solution.

Let the mass of NaCl = x gm and hence mass of KCl = (3-x) gmNaCl + AgNO

3 NaNO

3 + AgCl

(23 + 35.5) (108+35.5)=58.5 =143.5

58.5 gm of NaCl produces 143.5 gm of AgClSo, x gm of NaCl will produce (143.5/58.5) × x gm of AgCl

KCl + AgNO3 KNO

3 + AgCl

(39+35.5) 143.5= 74.5

74.5 gm of KCl proudces 143.5 gm of AgCl

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Stoichiometry-I

Dr. S. S.Tripathy

So, (3-x) gm of KCl will produce [143.5/74.5) × (3-x) gm of AgClTotal mass of AgCl = 6.3 = (143.5/58.5) x + [143.5/74.5) × (3-x)On solving we get x = 1 gm(mass of NaCl)So, mass of KCl = 3-1 = 2 gm, hence its percentage= (2/3) X 100 = 66.67%

SAQ 20: 1.0 gm of an alloy of Al and Mg when heated with excess of HCl forms aluminium chloride andmagnesium chloride and hydrogen. The evolved hydrogen collected over mercury at 00C has a volume of 1.2litres at 0.92 atm. pressure. Calculate the composition of the alloy. (Al = 55%) (IIT 1978)

SAQ 21: A solid mixture(5.0 gm) consisting of lead nitrate and sodium nitrate was heated strongly to constantmass. If the loss in mass is 28%, find the amount of lead nitrate and sodium nitrate in the mixture.

[Pb(NO3)

2 = 3.33 gm] (IIT 1990)

GAS ANALYSIS: (GAY LUSSAC'S LAW)1. Volume-Volume Relationship:We know from Gay Lussac's Law of combining gaseous volumes that when gaseous reactants form gaseousproducts they do so in simple ratio with respect to their volumes, provided the reaction is carried out atconstant temperature and pressure. The ratio of their volumes is equal to the ratio of the stoichiometriccoefficients in the balanced equation. Look to this example.Example : 5 litre of hydrogen gas was allowed to react completely with 5 litres of chlorine gas atthe same temperature and pressure, how many litres of hydrogen chloride gas will be obtained at thattemperature and pressure?Solution: Write the balanced equation

H2(g) + Cl

2(g) 2 HCl(g)

1 mole of H2 reacts with 1 mole of Cl

2 to give 2 moles of HCl gas. Therefore

22.4litres of H2 gas reacts with 22.4 litres of Cl

2 to give 2×22.4 litres of HCl at NTP

1 litre of H2 reacts with 1 litre of Cl

2 to give 2 litres of HCl at the same temperature and pressure.

In general,1 volume of H

2 reacts with 1 volume of Cl

2 to produce 2 volumes of HCl gas at the same temperature

and pressure. The ratio of the volumes of reactants and products is simple i.e 1:1:2. This is Gay Lussac'sLaw. One can find this ratio by looking to the coefficients in the balanced equation. But one thing you mustnot forget that this law holds good for the gaseous reactants and gaseous products and not the product whichliquifies on cooling like H

2O. This will be made more clear later.

In the above example, 5 litres of H2 reacts with 5 litres of Cl

2 to give 10 litres of HCl gas.

Example : What maximum volume of HCl gas will be obtained if 5 litres of H2 is allowed to react

with 10litres of Cl2 gas at the same temperature and pressure? Which reactant gas remains unreacted

at the end and by what volume? What is the total volume of the mixture after the reaction?Solution: H

2 + Cl

2 2 HCl

From Gay Lussac's law we know that1 volume of H

2 will reacts with 1 volume of chlorine to produce 2 volumes of HCl gas (at the same

temperature and pressure conditions)Therefore 5 litres of H

2 needs only 5 litres of Cl

2. But you are given 10litres of Cl

2 out of which only 5 litres

will be used and remaining 5 litres of Cl2 will be left excess. So H

2 is the limiting reactant and we shall use

that to find the volume of HCl gas formed.5 litres of H

2 will react with 5 litres of Cl

2 to produce 10 litres of HCl gas.

The total volume of gaseous product along with excess gaseous reactant after the end of the reaction =volume of HCl gas + volume of excess Cl

2 not used in the reaction.

Total volume = 10 litre of HCl gas + 5 litres of unreacted Cl2 gas=15 litres.

Example : 10 litres of CO were allowed to burn with excess of oxygen gas to produce CO2 gas.

What volume of CO2 gas will be obtained at the same temperature and pressure? What volume of O

2

gas needed for the purpose at the same conditions?Solution: The balanced equation for the reaction is

2CO + O2

2CO2

2 litres 1 litre 2 litres (at same temperature and pressure)2 litres of CO will produce 2 litres of CO

2

So 10 litres of CO must produce 10 litres of CO2.

Again 2 litres of CO require 1 litre of O2

So 10 litres of CO must require 5 litres of O2.

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Stoichiometry-I

Dr. S. S.Tripathy

SAQ 22: 24 litres of N2 gas was allowed to react with excess of H

2 gas at a particular temperature

and pressure. What volume of NH3 gas at the same conditions would be formed if all the N

2 is made to

react? What volume of H2 gas is required for the purpose.(48L, 72L)

SAQ 23: 500ml of O2 gas was converted to O

3 gas at the same temperature and pressure and a 50ml

reduction in volume was observed at the end of the reaction. What is the volume of O3 gas formed? Find

the volume of the unreacted O2 gas and the volume of the mixture (ozonised oxygen). (100, 350, 450 mL)

SAQ 24: 20 litres of H2 gas was allowed to mix with 15 litres of O

2 gas in a vessel operated at

constant pressure (1atm) and temperature(1000C) conditions. An electric spark was formed inside the mixturewhere H

2O vapour is produced with a pop sound. Find out the volume of water vapour that would be formed

and which gas would remain in excess? What is the total volume of the resulting mixture? (20, 25 L)SAQ 25: 10 litres of O

2 gas was allowed to react with excess of sulphur. What volume of SO

2 gas

will be formed at the same temperature and pressure?SAQ 26: Calculate the volume of CO

2 gas and water vapour evolved by completely burning 3 litres

of CH4 gas at a particular temperature and pressure. (10 L)

PERCENTAGE OF GASEOUS MIXTURE:Example: 1 litre of a mixture of CO and CO

2 is passed through a tube containing red hot

charcoal. The volume now becomes 1.6 litres. The volumes are measured under the same conditionsof temperature and pressure. Find the composition of the original mixture.Solution: Out of the two gases present in the mixture, it is only CO

2 which will react with charcoal(C)

to be reduced to CO and the original CO present in the mixture will remain as such.CO

2(g) + C(s) 2CO(g)

1 litre 2 litresLet us assume that out of 1 litre of the mixture, the volume of CO is x litre and CO

2 is (1-x) litre.

From the balanced equation we know that1 litre of CO

2 produces 2 litres of CO.

So (1-x) litres of CO2 must produce 2(1-x) litres of CO.

So the total volume after the reaction = volume of original CO + the volume of CO formed from CO2 by

its reaction with C = x + 2(1-x)The initial volume before the reaction was 1 litre and the volume after the reaction is x + 2(1-x).In the question, it is given that the final volume is 1.6 litres.

x + 2(1-x) = 1.6 ⇒ x + 2 -2x =1.6, ⇒ x = 0.4litre(volume of CO).So the volume of CO

2 in the original mixture was 1-0.4= 0.6 litre.

So the percentage of CO = (0.4/1.0)X100= 40%So the percentage of CO

2 = 100-40= 60%

EUDIOMETRYThe composition of a mixture of gases is determined by this method. The mixture of gases along with O

2

gas is taken in a eudiometer tube which is a long and narrow glass tube closed at one end. The tubecontaining the mixture of gases is inverted over a mercury trough so as to keep always at same pressurecondition(1 atm. pressure). The reaction among the gases is allowed to take place by creating electric sparkinside the eudiometer tube. Reaction among gases takes place with an explosive sound (pop sound). Supposewe have a mixture of CH

4, CO and O

2 gases. CH

4 reacts with O

2 to give CO

2 and H

2O while CO reacts

with O2 to give CO

2. After the reaction the gaseous mixture is allowed to cool so that any H

2O formed in

the reaction is condensed to liquid. Thus the volume of the mixture is reduced because H2O vapour present

in the mixture is removed and liquid water formed has negligible volume compared to volume of gas and doesnot contribute to gaseous volume any more. Then the eudiometer tube is transferred carefully to a jarcontaining an alkali like caustic soda or caustic potash or lime water [Ca(OH)

2]. The tube is shaken well to

absorb all CO2 gas present in the mixture. The volume is further reduced. The volume reduced at this stage

is equal to the volume of CO2 present in the mixture. The mixture is then treated with alkaline pyrogallol

to absorb any excess (unreacted) O2 gas that might be present in the mixture. Again the volume decreases

and this decrease is equal to the volume of O2 present in the mixture. In this way the gases are selectively

removed with specific reagents one after the other and volume reduction is measured at each step. Thenusing the balanced equation and Gay Lussac's law, we can find the composition of the original mixture. Readthe following example mindfully.

11

Stoichiometry-I

Dr. S. S.Tripathy

(a) Composition of a gas mixture:Example: 20ml of a mixture of CO and C

2H

2(acetylene) are mixed with 30ml of O

2 and exploded

in a eudiometer tube. After cooling the residual gas occupied 34ml. After treatment with caustic potash,the residual oxygen gas occupied 8ml. Calculate the percentage composition of the original mixture.Solution: Let us write two separate combustion reactions, one for CO and the other for C

2H

2 reacting with

O2.

2CO(g) + O2(g) 2CO

2(g) (i)

2ml 1ml 2 ml2C

2H

2(g) + 5O

2(g) 4CO

2(g) + 2H

2O(l) (ii)

2 ml 5 ml 4ml (zero)Let us assume that out of 20ml of mixture, the volume of CO = x ml, and so the volume of C

2H

2=(20-x)ml

Equation (i)2ml of CO produces 2 ml of CO

2 gas

So x mol of CO will produce x ml of CO2 gas.

Again 2 ml of CO requires 1 ml of O2

So x ml of CO will require x/2 ml of O2

Equation (ii)2 ml of C

2H

2 produces 4 ml of CO

2 gas

(20-x)ml of C2H

2 will produce 2(20-x) ml of CO

2.

Again 2ml of C2H

2 requires 5 ml of O

2 gas

So (20-x) of C2H

2 will require (5/2) X (20-x)ml of O

2 gas.

Total CO2 formed = x + 2(20 -x)

The volume of oxygen consumed by both reactions=x/2 + (5/2) × (20-x)=(100-4x)/2

So the volume of oxygen left after the reaction = 30 - (100-4x)/2According to the question, the volume of unreacted oxgyen is 8ml.So 30-(100-4x)/2 = 8, ⇒ x = 14ml(volume of CO in the original mixture)So the volume of C

2H

2 = 20-14=6ml.

Alternative method: We can also solve this problem by taking the volume of CO2. Note that after the reaction,

the total volume was 34ml (which contain CO2 and unreacted O

2, not water which is in the liquid state)

and after absorbing the mixture by caustic postash the volume reduced to 8ml(for O2 only). So the

volume of CO2= 34-8 = 26ml

The total volume of CO2 is x +2(20-x)

So x + 2(20-x) = 26, ⇒ x =14ml(volume of CO) and hence volume of C2H

2 = 20-14 = 6ml.

Alternative method: We can also solve this problem by finding the total volume of the mixture after thereaction which is given to be 34ml.Total volume after the reaction = volume of CO

2 formed from both reactions + the volume of unreacted O

2

= x + 2(20-x) + [30-(100-4x)/2](since H

2O is converted to liquid state, its volume is negligible and hence has been taken as zero)

But it is given that the volume after the end of the reaction is 34ml.So x + 2(20-x) + [30-(100-4x)/2] = 34, ⇒ x = 14ml(volume of CO)

So the volume of C2H

2= 20-14= 6ml.

So you found that you can solve this problem in many ways and you can use the most economical way toget the result quickly. In the above example the first two ways are easier than the last method. Is'nt it?Example : 10ml of a mixture of CH

4 ,C

2H

4 and CO

2 were exploded with excess of air. After

explosion and cooling, there was a contraction of 17ml and after treatment with aq. KOH there wasfurther reduction of 14ml. Find the composition of the original mixture.Solution: CH

4 and C

2H

4 only react with O

2 to produce CO

2 and H

2O. The CO

2 present in the original mixture

remains as such.Let the volume of CH

4= x ml, volume of C

2H

4= y ml, So volume of CO

2 = 10-x-y

Now let us write two combustion reactions.CH

4(g) + 2O

2(g) CO

2(g) + 2H

2O(l) (i)

1 ml 2ml 1 ml (zero)C

2H

4(g) + 3O

2 2CO

2(g) + 2H

2O(l) (ii)

1ml 3ml 2ml (zero)Equation (i):

1 ml of CH4 produces 1 ml of CO

2

x ml of CH4 will produce x ml of CO

2

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Stoichiometry-I

Dr. S. S.Tripathy

1 ml of CH4 requires 2ml of O

2 gas,

x ml of CH4 will require 2x ml of O

2 gas.

Equation (ii):1 ml of C

2H

4 produces 2ml of CO

2

y ml of C2H

4 will produce 2y ml CO

2.

1 ml of C2H

4 requires 3ml of O

2

y ml of C2H

4 will require 3y ml of O

2.

So total CO2 formed during the reaction= (x + 2y)

The total CO2 present in the mixture = (x+2y) + (10-x-y)

Since the volume contraction by treatment with KOH is 14ml, the total volume of CO2 present after the

reaction is 14ml.So (x+2y) + (10-x-y) = 14ml, ⇒ y = 4ml(volume of C

2H

4)

Now let us find the total volume of mixture after the reaction= x+2y +(10-x-y)This is same as the volume of total CO

2 present since the volume of water formed in the liquid state is taken

to be zero.The volume of the mixture before the reaction = volume of mixture + O

2 consumed

= 10 + 2x + 3yNote that if there is any unreacted O

2 in the mixture, that would be cancelled while finding the difference.

Hence the difference between volumes of the reactant and product mixtures=volume contraction = 17ml(10+2x+3y) - (x+2y +10-x-y) = 17, ⇒ 2x+2y =17

Substituting the value of y, we get, 2x + 2×4=17, ⇒ x = 4.5ml(volume of CH4)

So the volume of CO2 present in the original mixture =10 - 4- 4.5=1.5ml

So volume of CH4

= 4.5ml, C2H

4 = 4ml and CO

2 = 1.5ml in the original mixture.

SAQ 27: What will be volume of the products in each case of the following.(i)When 40ml of H

2 gas is allowed to react with equal volume of Cl

2 gas (80 mL)

(ii)When 200ml of O2 is allowed to react with equal volume of CO gas. (300 mL)

SAQ 28: 10ml of a mixture of ethylene(C2H

4) and acetylene(C

2H

2) require 29ml of O

2 gas for complete reaction at

the same temperature and pressure. Determine the composition of the mixture. (8, 2 mL)(b)Determination of molecular formula of an unknown gasWe can find the molecular mass of an unknown gas by eudiometry. Go through the following example.Example: 100 mL of an unknown gas is mixed with 140 mL of oxygen in a gas burette and subjected toelectric spark. The residual volume was 190 mL and after absorbing the mixture by caustic potash, theresidual volume was 90 mL. Find the molecular mass of the unknown gas.Soultion: The voume of CO

2 formed = 190 - 90 = 100 mL

The volume of oxygen consumed = 140 - 90 = 50 mLHence 100 mL of the gas has reacted with 50 mL of O

2 gas to give 100 mL of CO

2

Applying Avogadro's law we have2 molecules of the unknown gas reacted with 1 molecule of O

2 to give 2 moclecules of CO

2

So, 1 molecule of the unknown gas reacted with ½ molecule of O2 to give 1 molecule of CO

2

Since ½ molecule of O2 contains 1 O atom and 1 molecule of CO

2 contains 1 C atom, the unknown gas

is CO.Alternative method:Let the formula of the unknown gas is C

xO

y (here we presumed that it is compound of C and O only)

CxO

y+ (2x - y)/2 O

2→ x CO

2

1 mL (2x-y)/2 mL x mL100 mL (2x-y)/2 X 100 mL 100x mL

Since the volume of CO2 is 100 mL (determined before)

100x = 100 ⇒ x =1Since the volume of O

2 consumed is 50 mL(determined before)

(2x-y)/2 × 100 = 50 ⇒ y = 1Hence the formula of the unknwon gas is CO.(Note that if you presume the gas to be a hydrocarbon having formula C

xH

y, then while solving for x you

shall get -ve value, which is absurd. Only presuming it be CxO

y you will get +ve value for both x and y.

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Stoichiometry-I

Dr. S. S.Tripathy

Example 2: 10 mL of a gaseous hydrocarbon is mixed with 100 mL of O2 in an eudiometer tube at 250C

and 1 atmosphere pressure. The mixture is exploded by an electric spark. The residual gas occupied 95 mL.On passing the mixture through KOH solution contraction of volume occurred by 20 mL. The residual gaswas completely absorbed by pyrogallol. Determine the molecular formula of the hydrocarbon.Solution: C

xH

y+ ½ ( 2x + y/2) O

2 x CO

2 + y/2 H

2O (l)

1 mL ½(2x + y/2) mL x mL zeroVolume of CO

2 formed = volume contraction by KOH = 20 mL

Volume of residual gas after explosion at room temp. = volume of CO2 + volume of excess O

2 = 95 mL

Hence volume of excess O2 = 95 - 20 = 75 mL (totally absorbed by pyrogallol)

Total volume of O2 used = 100 mL. Hence volume of O

2 consumed = 100 - 75 = 25 mL

From the above equation:1 mL of C

xH

y produces x mL of CO

2 gas

10 mL of CxH

y produces 10x mL of CO

2

Hence 10x = 20, ⇒ x = 2Again

1 mL of CxH

y requires 1/2(2x +y/2) mL of O

2 gas

So 10 mL of the gas requires 10 × 1/2(2x + y/2) mL of O2

Hence 10 × 1/2(2x +y/2) = 25Substituting the value of x in the above equation we get, y= 2Hence the molecular formula of the hydrocarbon is C

2H

2(acetylene)

SAQ 29: 0.9 g of a solid organic compound having molecular mass 90 containing C, H and O was heatedwith oxygen corresponding to volume of 224 mL at NTP. After combustion the total volume of the gas was560 mL at NTP. On treatment with KOH, the volume decreased to 112 mL. Determine the molecular formulaof the compound.

RESPONSE TO SAQs(Percent Composition, Empirical and Molecular Formula)

SAQ 1: M.M of Ca(ClO3)

2 = 40+ 2(35.5 + 48)= 207

207gms of Ca(ClO3)

2 contains 6×16=96gms of Oxygen

So 100 gms of Ca(ClO3)

2 contains (96/207) × 100=46.37gms

Hence the percent by mass of O = 46.37%SAQ 2: (i) M.M of NH

4NO

3 = 28 + 4+48= 80

80gms of NH4NO

3 contains 28gms of Nitrogen

100gms of NH4NO

3 contains (28/80) × 100 = 35gms.

So the percent of N in NH4NO

3 = 35%

(ii) M.M of (NH4)

2SO

4 = 2×18 + 32+ 64= 132

132gms of (NH4)

2SO

4 contains 28gms of Nitrogen

100gms of (NH4)

2SO

4 contains (28/132) × 100 = 21.21gms

So the percent of N in (NH4)

2SO

4 = 21.21%.

Comparing the % of N in both the cases, we find that NH4NO

3 is a better grade nitrogenous

fertilizer as it has greater percentage of nitrogen.SAQ 3: M.M of K

2Cr

2O

7 = 2×39+ 2×52 + 7×16= 294

If only percent composition is asked in the question, you are required to find the percent by mass of eachelement present in the compound.K: 294gms of K

2Cr

2O

7 contains 2×39gms of K

100gms of K2Cr

2O

7 contains (2×39)/294 × 100 = 26.53gms

Cr: 294gms of K2Cr

2O

7 contains 2×52gms of Cr

100gms of K2Cr

2O

7 contains (2×52)/294 × 100 = 35.37gms

O: 294gms of K2Cr

2O

7 contains 7×16=112 gms of O

100gms of K2Cr

2O

7 contains (112/294) X 100 = 38.1gms.

So percent of K=26.53%, Cr=35.37% and O=38.1%SAQ 4: The M.M of CuO= 63.5 + 16=79.5

79.5gms of CuO contains 63.5 gms of Cu100gms of CuO contains (63.5/79.5) × 100 = 79.87gms

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Stoichiometry-I

Dr. S. S.Tripathy

So the percent of Cu = 79.87%79.87gms of Cu can be produced by 100gms of CuO100kg i.e100000gms of Cu can be produced by 125203.455gm=125.203kg.

SAQ 5: Let us take 100gms of impure Cu2O. This contains 66.6 gms of Cu.

But let us see how much of Cu2O can contain 66.6gms of Cu from its molecular mass.

The M.M of Cu2O= 2×63.5 + 16 = 143;

127(=2×63.5)gms of Cu is present in 143gms of Cu2O

So 66.6gms of Cu must be present (143/127)×66.6=74.99gms of Cu2O.

So 100gm of impure Cu2O contains only 74.99gms of pure Cu

2O and the rest impurity.

So the percent of purity = 74.99%.SAQ 6: Dividing the coefficients by 7, we get the empirical formula CH

2.

SAQ 7: Dividing the coefficients by 2, we get the empirical formula C2H

4O.

SAQ 8:___________________________________________________________Element % At. Mass Relative no. Simple Whole Empirical

of atoms No. of atoms Formula___________________________________________________________C 85.63 12 85.63/12=7.135 7.135/7.135=1 CH

2

H 14.37 1 14.37/1=14.37 14.37/7.135=2

___________________________________________________________Empirical formula= CH

2 and empirical formula mass= 12+2=14

Molecular Mass = 56; But we know that Molecular Formula = (Empirical Formula)n

n= Molecular Mass/Empirical Formula mass = 56/14=4So Molecular Formula =(CH

2)

4= C

4H

8.

SAQ 9:___________________________________________________________Element % At. Mass Relative no. Simple Whole Empirical

of atoms No. of atoms Formula___________________________________________________________N 30.4 14 30.4/14= 2.17 2.17/2.17=1 NO

2

O 69.6 16 69.6/16=4.35 4.35/2.17=2___________________________________________________________

SAQ 10:___________________________________________________________Element % At. Mass Relative no. Simple Whole Empirical

of atoms No. of atoms Formula__________________________________________________________Cr 26.52 52 26.52/52=0.51 0.51/0.51=1×2=2S 24.52 32 24.52/32=0.766 0.766/0.51=1.5×2=3 Cr

2S

3O

12

O 48.96 16 48.96/16=3.06 3.06/0.51=6×2=12__________________________________________________________

It seems that the compound is a sulphate of chromium as 3(SO4) give S

3O

12. So arranging them in proper

order it is Cr2(SO

4)

3. From this formula it is evident that it is also the molecular formula. Anyway, we cannot

officially find the molecular formula without the molecular mass data.ANSWERS TO PRACTICE QUESTIONS1.

__________________________________________________________Element % At. Mass Relative no. Simple Whole Empirical

of atoms No. of atoms Formula__________________________________________________________Na 21.6 23 21.6/23=0.939 0.939/0.938=1Cl 33.3 35.5 33.3/35.5=0.938 0.938/0.938=1 NaClO

3

O 45.1 16 45.1/16=2.819 2.819/0.938=3

15

Stoichiometry-I

Dr. S. S.Tripathy

__________________________________________________________It appears that the empirical formula is also the molecular formula as it is the familiar compound sodiumchlorate.2. Total mass of the compound = 7.3+6.3 = 13.6gm

So % of Fe=(7.3/13.6) ×100 =53.68 , % of S= (6.3/13.6) ×100 =46.32After finding the % composition, we shall do in the same way as before.___________________________________________________________Element % At. Mass Relative no. Simple Whole Empirical

of atoms No. of atoms Formula___________________________________________________________Fe 53.68 56 53.68/56=0.958 0.958/0.958=1×2=2 Fe

2S

3

S 46.32 32 46.32/32=1.447 1.447/0.958=1.5×2=3___________________________________________________________Hence the empirical formula is Fe

2S

3.

3.___________________________________________________________Element % At. Mass Relative no. Simple Whole Empirical

of atoms No. of atoms Formula___________________________________________________________C 40 12 40/12=3.33 3.33/3.33=1H 6.66 1 6.66/1=6.66 6.66/3.33=2 CH

2O

O 53.34 16 53.34/16=3.33 3.33/3.33=1___________________________________________________________Empirical formula mass = 12+2+16=30; Molecular Mass(given)=60So n=60/30=2 and therefore Molecular Formula=(CH

2O)

2= C

2H

4O

2.

4.______________________________________________________El. % At. Mass Relative no. Simple Whole Empirical

of atoms No. of atoms Formula______________________________________________________C 82.66 12 82.66/12=6.88 6.88/6.88=1×2=2 C

2H

5

H 17.34 1 17.34/1=17.34 17.34/6.88=2.5×2=5______________________________________________________Since its vapour density(V.D)=29.05, its Molecular Mass=2×29.05=58.1Empirical formula mass= 24+5=29, So n = 58.1/29=2.0034 = 2So molecular formula = (C

2H

5)

2= C

4H

10.

RESPONSE TO SAQs(Stoichiometric Calculations)SAQ 1: The involved species are C

2H

6, O

2 and H

2O.

C2H6 O2 CO2 H2O+ +2 7 4 62(24+6) 4(12+32) 6(2+16)

. 2X30gms of C2H

6 on complete burning produces 4×44gms of CO

2 gas

So 30gms of C2H

6 produces (4×44)/(2×30) × 30= 88gms of CO

2

Again 2×30gms of ethane produces 6 × 18 gms of H2O

So 30gms of ethane will produce 54gms of H2O.

SAQ 2: Here the mass of AgCl is given and we are to find the mass of CaCl2.

CaCl2 AgNO3 AgCl Ca(NO3)2+ +2X(108+35.5)(40+ 2X35.5)

2 2

2×(108+35.5) gm of AgCl is produced by (40 + 2×35.5)gms of CaCl2.

6.6gms of AgCl must be produced by 2.55gms of CaCl2(do the calculation for yourself).

The number of moles of CaCl2 = 2.55/ 111 = 0.0229 mole.

SAQ 3: Here the mass of HCl is given and we have to find the mass of CaO.

CaO HCl CaCl2 H2O+ +2(40+16) 2X(1+35.5)

16

Stoichiometry-I

Dr. S. S.Tripathy

2×(1+35.5)gms of HCl requires (40+16) gms of CaOSo 6.92gms of HCl will require 5. 308gms of CaO.

SAQ 4: Here number of mole of Ba(OH)2 reacted is given.

Ba(OH)2 CO2 BaCO3 H2O+ +(137+12+48)1 mole

1 mole of Ba(OH)2 produces (137+12+48) gms of BaCO

3

0.205mole of Ba(OH)2 will produce 40.385 gms of BaCO

3.

SAQ 5: Here the moles of product formed is given.

NaHCO3 Na2CO3 CO2 H2O+ +22X(23+1+12+48) 1 mole

1 mole of CO2 is produced by 2X(23+1+12+48)gms of NaHCO

3

So 1.5moles of CO2 will be produced by 252gms of NaHCO

3.

SAQ 6: Here the mass of MnO2 is given and we are to find the volume of Cl

2.

2MnO2 HCl MnCl2 Cl2 H2O+ + +4(55+32) 22.4 l

(55+32)gms of MnO2 produces 22.4 litres of Cl

2 at NTP

So 8.7gms of MnO2 will produce 2.24 litres of Cl

2 at NTP.

SAQ 7: The volume of CO2 formed is given and we are to find the mass of NaHCO

3

NaHCO3 HCl NaCl CO2 H2O+ + +(23+1+12+48) 22.4 l

22.4 litres of CO2 at NTP is produced by (23+1+12+48)gms of NaHCO

3

5.6 litres of CO2 at NTP will be produced by 21 gms of NaHCO

3.

SAQ 8: Here you have to find both the volumes of NO2 and O

2.

Pb(NO3)2 PbO NO2 O2+ +2 2 42X[207 + 2X(14+48)] 4 X 22.4 l 22.4 l

2 X 331 gms of Pb(NO3)

2 produces 4 X 22.4 litres of NO

2 at NTP

10gms of Pb(NO3)

2 will produce 1.353 litres of NO

2 gas.

2 X 331gms of Pb(NO3)

2 produces 22.4 litres of O

2 gas at NTP.

So 10gms of Pb(NO3)

2 will produce 0.3383litre = 338.3 ml of O

2 gas.

SAQ 9: Here you have to find the mass of ammonium dichromate.

(NH4)2Cr2O7 N2 Cr2O3 H2O+ + 42X(14+4) + 2X52+112 22.4 l

22.4 litres of N2 at NTP are produced by 252 gms of (NH

4)

2Cr

2O

7

11.2 litres of N2 at NTP are produced by 126 gms of (NH

4)

2Cr

2O

7

SAQ 10: 2nd reaction:3 CuO + 2 NH

3 → 3 Cu + N

2 + 3 H

2O

3 ×(63.5+16)g 2 × 17 g

3× 79.5 g of CuO reacts with 2 × 17 g of NH3

So 2.5 g of CuO reacts with 0.356 g of NH3

1st reaction:(NH

4)

2SO

4+ 2NaOH → 2NH

3 + Na

2SO

4 + 2H

2O

[2(14+4) +32+64]g 2X17g=132 g2X 17g of NH

3 is produced by 132 g of NH

4)

2SO

4

So, 0.356 g of NH3 is produced by 1.38 g of ammonium sulphate

SAQ 11: CaCO3

→ CaO + CO2

(40+12+48)g (20+16)g100g of CaCO

3 produces 56g of CaO

4 g of CaCO3 produces 2.24 g of CaO

CaO + 2 NH4Cl → CaCl

2 + 2 NH

3+ H

2O

56g 2 X 22.4 L(NTP)

17

Stoichiometry-I

Dr. S. S.Tripathy

56 g of CaO produces 2 X 22.4 L of NH3 gas at NTP

2.24 g of CaO produces 1.792 L of NH3 at NTP

Using the combined gas equation, we can find the volume at the given conditions:The requied volume = 1.87 L

SAQ 12: 2nd reaction3 Cl

2+ 6 NaOH → NaClO

3 + 5 NaCl + 3 H

2O

3 × 71 g 6 × 40 g6×40 g of NaOH requires 3 × 71 g of Cl

2

10 g of NaOH requires 8.875 g of Cl2

1st reaction: Balance by partial equation method or NO method2KMnO

4 + 16 HCl → 2 KCl + 2 MnCl

2 + 5 Cl

2 + 8 H

2O

2(39+55+64)g 5×71g5× 71 g of Cl

2 is produced by 2 × 158 g of KMnO

4

So, 8.875g of Cl2 is produced by 7.9 g of KMnO

4

SAQ 13: Na2CO

3 + 2 HCl → 2 NaCl + CO

2+ H

2O

106g 2 × 36.5g 2 × 58.5 gFirst we have to find the limiting reactant.106 g of Na

2CO

3 reacts with 2 × 36.5 g of HCl

So, 5.3 g of Na2CO

3 reacts with 3.65 g of HCl

Since we have 5.3 g of HCl, the limting reactant is Na2CO

3

106g of Na2CO

3 produces 2 × 58.5 g of NaCl

So, 5.3 g of Na2CO

3 produces 5.85g of NaCl

SAQ 14: 3 BaCl2 + 2 Na

3PO

4 → Ba

3(PO

4)

2 + 6 NaCl

3 moles 2 moles 3×137+2(31+64)gLet us first find the limiting reactant.

3 moles of BaCl2 reacts with 2 moles of Na

3PO

4

so, 0.1 mole of BaCl2 reacts with 0.0667 mole of Na

3PO

4

But we have 0.05 mole of Na3PO

4, hence all BaCl

2 cannot be exhausted. So the limiting reactant is Na

3PO

4.

2 moles of Na3PO

4 produces 601 g of Ba

3(PO

4)

2

So, 0.05 mole of Na3PO

4 produces 15.025 g of Ba

3(PO

4)

2

SAQ 15: Mass of pure CaCO3 = 95% of 200kg = (95/100) × 200 = 190kg.

CaCO3

→ CaO + CO2

(40+12+48) (40+16)100gm of CaCO

3 produces 56gms of CaO(lime).

190,000gms of CaCO3 will produce (56/100) × 190,000 =106,400gm =106.4kg of CaO.

SAQ 16: Zn + H2SO

4→ ZnSO

4+ H

2

65.5g 22400mL at NTPBy using combined gas equation, the volume at the given conditions is converted to NTP.The NTP volume = 234.46 mL

22400 mL of H2 is produced by 65.5g Zn at NTP

So, 234.46 mL of H2 is produced by 0.685 g of ZnSo % of purity = (0.685/1)× 100 = 68.5%SAQ 17: Ag + 2 HNO

3→ AgNO

3 + NO

2 + H

2O

1 mole=108g 1 moleAgNO

3 + HCl → AgCl + HNO

3

1 mole 1mole =(108+35.5)g

In problems involving successive reactions, we can solve easily by mole method in stead of by mass methodby establishing mole relationship between the two involved molecules, one from the first reaction and the otherfrom the second reaction.1 mole of Ag(one gm. atomic mass) must produce 1 mole of AgCl as in one mole of AgCl there is one gm.atomic mass of Ag. So we can bypass the the product of the first reaction(AgNO

3) i,.e ignore the reaction

sequence through which AgCl has been prepared.143.5 g of AgCl is produced by 108 g of AgSo, 9 g of AgCl is produced by 6.77 g of AgSo % of purity = (6.77/7) × 100 = 96.7%

18

Stoichiometry-I

Dr. S. S.Tripathy

SAQ 18: Balance the equation by partial equation or ON method.K

2Cr

2O

7 + 4 H

2SO

4 + 3 H

2S → K

2SO

4 + Cr

2(SO

4)

3 + 3S + 7 H

2O

(2×39+2×52+112)g 3×32g96g of S is deposited by 294 g of K

2Cr

2O

7

9.6 g of S is deposited by 29.4 g of K2Cr

2O

7

So % of purity = (29.4/35) × 100 = 84%SAQ 19: 2 NaHCO

3 → Na

2CO

3 + H

2O + CO

2

2×(23+1+12+48)g 18g 44g(18+44)g of volatiles are lost by 2 × 84 g of NaHCO

3

0.45 g of volatiles are lost by 1.219 g of NaHCO3 (Note that Na

2CO

3 is stable to heat)

So, % of NaHCO3 = (1.219/3)×100 = 40.64%, hence % of Na

2CO

3= 59.36%

SAQ 20: 2Al + 6 HCl → 2 AlCl3 + 3 H

2

2×27g 3×22.4 L(at NTP)Mg + 2 HCl → MgCl

2 + H

2

24g 22.4 L (at NTP)First the volume at the given conditions is converted to volume at NTP by using combined gas equation.

NTP volume = 1.1 LLet the mass of Al = x g and hence mass of Mg= (1-x)g

54g of Al produces 3× 22.4 L of H2 at NTP

So, x g of Al produces 1.244 x L of H2 at NTP

24g of Mg produces 22.4 L of H2 at NTP

So, (1-x) g of Mg produces 0.933(1-x) L of H2 at NTP

Hence, 1.244x + 0.933(1-x) = 1.1 (total volume of H2 at NTP)

Solving the above equation, we get, x=0.55 g. So % of Al= 55% hence % of Mg= 45%SAQ 21: Pb(NO

3)

2 → PbO + 2 NO

2 + ½ O

2

[207+2(14+48)]g 2×46g 16gNaNO

3→ NaNO

2 + ½ O

2

(23+14+48)g 16gLet the mass of Pb(NO

3)

2 = x g and NaNO

3 = (5-x) g

331 g of Pb(NO3)

2 produces (92+16)g of volatiles(gases)

x g of Pb(NO3)

2 produces 0.326 x g of volatiles

85 g of NaNO3 produces 16g of volatiles

(5-x)g of NaNO3 produces 0.188(5-x) g of volatiles

Hence, 0.326x + 0.188(5-x) = (28/100) × 5On solving the above equation we get, x=3.33 g and NaNO

3 = 1.67g

SAQ 22: N2

+ 3H2

→ 2NH3

1vol. 3vol. 2vol.1litre of N

2 produces 2 litres of NH

3( at the same temperature and pressure)

24litres of N2 must produce 48litres of NH

3.

Again 1 litre of N2 requires 3 litres of H

2 gas

So 24 litres of N2 will require 72 litres of H

2 gas.

SAQ 23: 3O2

→ 2O3

3 vol. 2 vol.3 ml of O

2 reacts to form 2 ml of O

3 at the same temperature and pressure. So there

is a decrease of volume of 1 ml.For the decrease in volume by 1 ml, the volume of ozone(O

3) formed = 2ml

Hence for the decrease in volume by 50ml, the volume of O3 formed = 100ml.

If 2 volumes of O3 = 100ml, then what is 3 volumes? It is 150ml.

So 150ml of O2 reacted to form 100ml of O

3 in the volume ratio 3:2.

The volume of O2(unreacted) remaining= 500 - 150 = 350ml

So the total volume after the reaction = 350(O2) + 100ml(O

3) = 450 ml.

Alternative Method If 3 mL of O2 produces by 2 mL of O

3, then x mL of O

2 will produce 2x/3 mL of O

3.

So volume after reaction = 500 - x +2x/3; Then volume reduction = [500 - (500 -x + 2x/3)] =50, ⇒ x=150 mL(volume O

2 used up). So volume of unreacted O

2 = 500-150=350 and volume of O

3=2x/3=100mL,

So the total volume = 350 + 100 = 450 mL

19

Stoichiometry-I

Dr. S. S.Tripathy

SAQ 24: 2H2(g) + O

2(g) → 2H

2O(g)

2 vol. 1 vol. 2 vol.2 litres of H

2 reacts with 1 litre of O

2 to form 2 litres of water vapour

So 20 litres of H2 will react with 10 litre of O

2 to form 20 litres of water vapour

Since 15 litres of O2 gas has been taken, 5 litres of it will remain in excess.

Total volume =20 litres(water vapour) + 5 litres (excess O2) = 25 litres.

SAQ 25: S(s) + O2(g) → SO

2(g)

1 vol. 1 vol.10 litres of O

2 will produce 10 litres of SO

2. Note that S is in the solid state and therefore its volume

has not been written.SAQ 26: CH

4(g) + 2O

2(g) → CO

2(g) + 2H

2O(g)

1 vol. 2 vol. 1 vol. 2 vol.3 litre of CH

4 produces 3 litre of CO

2 and 6 litres of water vapour

SAQ 27:(i) H2(g) + Cl

2(g) → 2HCl(ga)

40 ml of H2 will react with 40ml of Cl

2 to form 80ml of HCl gas

(ii) 2CO(g) + O2(g) → 2CO

2(g)

200ml of CO will react with 100ml of O2 to produce 200ml of CO

2 gas. Here 100ml(200-

100) of O2 gas will remain unreacted.

SAQ 28: C2H

4 + 3O

2 → 2CO

2 + 2H

2O (i)

2C2H

2 + 5O

2 → 4CO

2 + 2H

2O (ii)

Let the volume of C2H

4 = x ml; Hence the volume of C

2H

2 = (10-x) ml

Eqn.(i) 1 ml of C2H

4 requires 3mls of O

2.

So x ml of C2H

4 must require 3x ml of O

2.

Eqn.(ii) 2ml of C2H

2 requires 5 ml of O

2.

So (10-x) ml of C2H

2 must require (5/2)×(10-x) ml of O

2.

Total volume of O2 needed = 3x + (5/2) × (10-x) = 29 (according to question)

⇒ x = 8 ml (volume of C2H

4); Hence the volume of C

2H

2 = 10-8 = 2ml.

SAQ 29 : CxH

yO

z + 1/2(2x+y/2-z) O

2 → x CO

2 + y/2H

2O(l)

1 mole 1/2(2x+y/2-z)×22400mL x × 22400mL zero

After combustion, the mixture contains CO2 + unreacted O

2 = 560 mL

Volume after absorption by KOH = 112 mL = volume of unreacted O2

So voume of CO2 = 560 - 112 = 448 mL

So volume of O2 consumed = 224 - 112 = 112 mL

90g of the hydrocarbon produces 22400 x mL of CO2 at NTP

So, 0.9 g of the hydrocarbon produces 224x mL of CO2 at NTP

Hence 224x = 448 ⇒ x = 2 (1)Again 90g of the hydrocarbon consumes 22400/2 ( 2x+y/2-z) mL of O

2

So, 0.9g of hydrocarbon consumes 112(2x+y/2-z) mL of O2

112(2x+y/2-z) = 112 (2)From the molecular mass of C

xH

yO

z, we get the equation

12x + y + 16z = 90 (3)Solving simultaneously equations 1,2 and 3, we get, z=4 and y =2Hence the molecular formula of the compound is C

2H

2O

4 (oxalic acid)

20

Stoichiometry-I

Dr. S. S.Tripathy

PRACTICE QUESTION

1. 200Kg of rock salt(NaCl) which contains 95% NaCl is allowed to react with excess of H2SO

4. How

much sodium sulphate will be formed? (230.5Kg)2. What volume of H

2 gas be evolved at NTP when 2g of superheated iron is allowed to completely

react with excess of superheated steam? Also find the the mass of the product residue? (Fe = 55.8)(1.070L, 2.764g of

Fe3O

4)

3. What volume of oxygen be evolved at NTP when 2.94g of K2Cr

2O

7 completely reacts with with

excess of hydrogen peroxide in presence of dil. H2SO

4?(K=39, Cr=52) (672mL)

4. How many grams and litres of CO2 formed at NTP when 1g of sucrose(C

12H

22O

11)is completely

burnt in excess of air? (1.54g and 785.96mL)5. Calculate the mass of CO

2 and water that will be produced by completely burning 1.5g of an organic

compound having molecular formula C3H

6O? (3.4g and 1.3965g)

6. When a mixture of ammonia and oxygen is passed over platinum maintained at 8000C, how muchof water will be formed from 0.5g of ammonia? Assume complete reaction.

(Hint: NH3 + O

2 ----Pt----> NO + H

2O) (0.794g)

7. 50mg of an impure sample of aluminium gave on treatment with excess dil. HCl, 50 mL of moisthydrogen at 270C and 760mm presure. Calculate the percentage of purity in the sample. (73.3)8. 1.1g of a mixture of KClO

3 and KCl gave on ignition 0.85g of residue. Find the composition of the

mixture. (KClO3=58%, KCl= 42%)

9. 1.5g of a mixture of BaCO3 and CaCO

3 was heated in a platinum crucible to a constant mass. The

mass of the residue was 1.05g. Calclulate the percentage of each metal carbonate present in themixture.(Ba=137, Ca=40) (BaCO

3=64.6%, CaCO

3=35.4%)

10. What mass of potassium chloride will be formed by passing 10L of Cl2 gas at 200C and 780mm

pressure into hot conc. KOH solution. (53g)(Hint: Cl

2 + KOH KCl + KClO

3 + H

2O)

11. 2.4g of a sample of sodium bicarbonate when strongly heated gives 300 mL of CO2 measured at 230C

and 780 mm pressure. Calculate percentage of purity of the sample. (88.7)12. What volume of oxygen gas at NTP is necessary for complete combustion of 10 litres of butane(C

4H

10)

measured at 270C and 760mm pressure? (59.15L)13. Calculate the volume of air containing 21% by volume of oxygen at NTP required to convert 400mLof SO

2 into SO

3 under same conditions. (952.38mL)

14. A portable hydrogen generator utilises the reaction of hydrolysis of CaH2. How many litres of H

2 gas

at NTP can be produced by 50g of CaH2? (53.33L)

15. 1.2g of a mixture of CaCO3 and MgCO

3 on complete ignition produced 42mL of CO

2 measured at

270C and 780mm pressure. Calculate the percentage composition of the mixture.(CaCO3=92%, MgCO

3=8%)

16. 50mL of solution containing 0.25g of an impure sample of H2O

2 reacts completely with 0.3g of

KMnO4 in presence of sulphuric acid. (i)Calculate the volume of dry oxygen gas evolved at 270C and 810mm

pressure. (ii)Find the percentage of purity of H2O

2 solution. (109.6mL, 64.4%)

17. How many grams of CaCO3 should be heated to get enough CO

2 which can convert 0.2 mole of

Na2CO

3 into NaHCO

3? (20g)

18. What mass each of magnesium and dilute H2SO

4 are required to give hydrogen just sufficient to

reduce 6.5g of zinc oxide?(Zn=65, Mg=24) (Mg=1.92g, H2SO

4=7.84g)

19. How much KCl is produced by the reaction of 1.5Kg of potassium and 2Kg of chlorine gas?(Hint: First find out which is the limiting reactant) (2.86Kg)

20. 32g of sulphur was burnt in the presence of 4g of oxygen in a closed vessel at certain temperatureand pressure. What mass of SO

2 will be formed? What is the volume of the resulting gas at NTP?

(Hint: First find out which is the limiting reactant) (8g, 2.8L)21. 5g of a natural gas consisting of methane(CH

4) and ethylene(C

2H

4) was burnt in excess of O

2producing 14.5g of CO

2 and some water as products. Find the mass percentage of ethylene in the natural

gas. (38.4%)22. A sample of gaseous hydrocarbon occupying 1.12L at NTP when completely burnt in oxygen produced1120mL of CO

2 and 1.8g of water. Calculate the mass of the hydrocarbon and the volume of O

2(at NTP)

required for its burning. Also find the molecular formula of the hydrocarbon. (0.8g, 2.24L, CH4)

23. A gaseous hydrocarbon is exploded with oxygen. The volume of O2 needed for complete combustion

and CO2 formed is in the ratio 7:4. Find the molecular formula of the hydrocarbon. (C

2H

6)

21

Stoichiometry-I

Dr. S. S.Tripathy

24. 50mL of a mixture of nitrous oxide(N2O) and nitric oxide(NO) was exploded with excess of hydrogen.

If 30mL of N2 was formed, calculate the percentage composition(mole %) of the mixture.

(N2O=20%, NO=80%)

25. When 500mL of hydrocarbon were exploded with excess of oxygen, 2500 mL of CO2 and 3L of

water vapour formed at NTP. Determine the molecular formula of the hydrocarbon. (IIT 74) (C5H

12)

LEVEL-II1. How many Kg of H

2SO

4 can by prepared from 3 Kg of cuprite(Cu

2S)? (Cu=63.5, S=32)(1.84Kg)

2. A certain mass of iron pyrite(FeS2) was roasted with excess of oxygen to produce SO

2 gas. SO

2

reacted with excess of O2 and presume that all SO

2 is completely converted to SO

3 gas at 5000C and 200

atm. pressure in presence of V2O

5 catalyst. From SO

3 4.4 g of pure H

2SO

4 was produced by an indirect

method. Find the mass of iron pyrite originally taken.(Fe=56, S=32) (2.69g)3. 3.4g of pure H

2O

2 on decomposition gives enough oxygen at 1000C and 760mm pressure which

oxidise excess of NH3 at that temperature. to N

2 and H

2O. Calculate the volume of N

2 produced.(1.02L)

4. 20 mL of a solution of KCl containing some NaCl as impurity on evaporation gave 2g of the residue.It was again made soluble in water and treated with excess AgNO

3 solution. The white precipitate of AgCl

was filtered and dried to give 4.27g. Find how much of NaCl was present in the 20 mL of the above solution.(Ag=108, K=39, Na=23, Cl=35.5) (0.8g)

5. 250 mg of a commercial sample of mercuric oxide on decomposition liberates enough oxygen forcomplete combstion of 40 mL of CO at 2000C and 760mm pressure. Caculate the percentage of purity ofthe mercuric oxide sample (Hg=200, O=16) (88.4%)6. An alloy of zinc and copper weighing 1g when treated with excess of dil. H

2SO

4 gave 224 mL of

dry hydrogen at 270C and 720 mm pressure. What is the % by mass of copper in the alloy? (44%)7. 10g of impure sample of Na

2CO

3 is dissolved in water and reacted with a solution of excess CaCl

2.

The resulting precipipite after filtration and drying was found to react completely with 5.6g of pure HCl.Calculate % of purity of Na

2CO

3. Assume that CaCl

2 bring about precipitation of Na

2CO

3 only. (81.3%)

8. Calculate the volume of CO2 evolved at 1000C and 700 mm pressure that can be obtained by heating

2 kg of limestone which is 90% pure. (598.1 L)9. 2.1 g of miixture of KNO

3 and NaNO

3 was heated to constant weight and found to have lost 0.373g.

What is the % of KNO3 in the mixture. (35.71)

10. What would be the volume at STP of each of the products formed by burning 10g of carbon disulfide.The gaseous mixture produces a burning sulfur smell. What amount caustic soda would be necessary toproduce the acidic salt of the whole of the above products?)

(CO2=2.947L, SO

2=5.894 L, acidic salt = 15.78g)

11. 1.1g of solder(a lead-tin alloy) was heated in a current of dry chlorine till the metals were completelyconverted to their chlorides PbCl

2 nd SnCl

4. If the mass increased by 52%, find the percentage composition

of solder. (Pb=207, Sn=119) (Pb=79%, Sn=20.9%) (13)12. 98 L of air at 270C and 760mm pressure were bubbled through baryta solution and yield 0.35 g ofBaCO

3 precipitate. What is the partial pressure of CO

2 in the air.(baryta=Ba(OH)

2) Ba=137 (0.339 mm)

13. A mixture of NaHCO3 and Na

2CO

3 weighing 1g was treated with excess of baryta solution to form

2.1 g of BaCO3. Find the % of NaHCO

3 in the original mixture. (50%)

14. Ignition of MnO2 converts quantitatively to Mn

3O

4. A sample of pyrolusite is of the following composition.

MnO2 80%, SiO

2 and other inert constituents 15% and rest being water. The sample is ignited in air to

constant weight. What is the percentage of Mn in the ignited sample?(Mn=55) (59.36)15. One litre of an acidified solution of KMnO

4 containing 15.8 g of KMnO

4 is decolorized by passing

sufficient amount of SO2. If the SO

2 is produced by roasting of iron pyrite(FeS

2) what will the amount of

pyrite required to produce necessary amount of SO2? (K=39, Mn=55, Fe=56, S=32) (15g)

16. 2.5 g of a mixture of CaCO3, MgCO

3 and NaHCO

3 suffered a loss of 1.16g on heating. The residue

on treatment with excess HCl gave 117.3 mL of CO2 at NTP. Calculate teh mass of each component.(CaCO

3=0.12g, MgCO

3=1.5g, NaHCO

3=0.88g)

17. In the analysis of a 0.5g of feldspar, a mixture of the chlorides of sodium and potassium is obtainedwhich is 0.118g. Subsequent treatment of the mixed chloride with AgNO

3 gives 0.2451g of AgCl. What the

% of Na2O and K

2O in feldspar? (Ag=108, Cl=35.5, K=39) (Na

2O =3.6%, K

2O=10.59%)

18. A flash bulb used for taking photograph in bar light contains 30mL of O2 at pressure of 780mm and

270C. Suppose the metal wire flashed is pure aluminium which is oxidised to aluminium oxide in the processof flashing. Calculate the minimum mass of aluminium wire that is to be used for the maximum efficiency.

(0.045g)

22

Stoichiometry-I

Dr. S. S.Tripathy

19. 500mL of O2 at NTP were passed through an ozoniser when the resulting volume was 444 mL at

NTP. This quantity of ozonised oxygen is passed through excess of KI solution. Calculate the mass of I2

liberated. (I=127) (1.27g)20. Caclulate the mass of phosphorous obtained by strongly heating 1g of calcium phosphate with excessof silica and coke in an electric furnace in the absence of air. Also find out the volume of the gas evolvedat NTP? (Ca=40, P=31) (0.2g, 0.3612L)21. What mass of P

4O

6 and P

4O

10 will be produced by the combustion of 2g of P

4 in 2g of oxygen gas

leaving no unfinished reactant(neither P4 nor O

2)? (P

4O

6=1.995g, P

4O

10 = 2.005g)

22. 0.5 g of a mixture of K2CO

3 and Li

2CO

3 produced 32.5 mL of CO

2 at NTP. Find out the composition

of the mixture. (Li=7) (Li2CO

3=0.107g, K

2CO

3=0.393g)

23. A mixture contains NaCl and an unknown chloride MCl. (i)1g of this is dissolved in water. Excessof acidified AgNO

3 solution is added to it. 2.567g of a white dry precipitate is obtained. (ii) 1g of the original

mixture is heated to 3000C. Some vapours came out which are absorbed in acidifed AgNO3, 1.341g of a white

precipitate is obtained. Find the molecular mass of the unknown chloride. (IIT 1980) (53.5)24. 10mL of a gaseous organic compound containing C, H and O only was mixed with 100 mL of O

2

and exploded under conditions which allowed the water formed to condense. The volume of the gas afterexplosion was 90 mL. On treatment with potash solution a further contraction in volume of 20mL wasobserved. Given that the vapour density of the compound is 23, deduce the molecular formula. All volumemeasurements were done under same conditions) (IIT 1977) (C

2H

6O)

Solution to Practice Questions(Level II)1. Cu

2S → → → → H

2SO

4

1 mole 1 moleCu

2S is converted to H

2SO

4 in a series of reactions. But for establishing a relationship between the starting

material and final product, we should look to the number of atoms(or moles)of one element undergoingchange. In this case, Cu

2S contains one S atom and so also H

2SO

4 contains one S atom.

So 1 mole of Cu2S must produce 1 mole of H

2SO

4

2. FeS2

→ → → → 2H2SO

4

1mole 2 molesIn FeS

2, there are two S atoms and in H

2SO

4 there is one S atom. On balancing for S atoms, we find that

one mole of FeS2 will produce 2 moles of H

2SO

4. We are least concerned about the sequence of reactions

by which the final product is obtained.3. Repeated calculation method:

H2O

2 → H

2O + ½ O

2

34g 11.2L(at NTP)3.4g 1.12L4NH

3 + 3O

2 → 2 N

2+ 6 H

2O

4×22.4L 2×22.4L(NTP)1.12L 0.746L

Converting this volume from the NTP conditions to the given condition, we get V=1.02LAlternative method(One step calculation)In stead of calculating for each step, we can add up all the step equations and establish a relationship betweenthe starting reactants with the final products. Although the overall reaction resulted from the addition of thesteps, may not really occur, it is used for stoichiometric calculation.

H2O2 H2O O 22 2

NH3 O 2 N2 H 2O4 63 2

+

+ +

X 3

H2O2 N H3 N2 H 2O6 4 2 12+ +

H2O2 N H3 N2 H 2O3 2 6+ +

3×34g 22.4L 3.4g 0.746L

Converting this volume at NTP to the given conditions we get the volume=1.02L4. We have 2g of (KCl+NaCl) mixture. Let KCl = xg and NaCl = (2-x)g

KCl + AgNO3

→ AgCl + KNO3

23

Stoichiometry-I

Dr. S. S.Tripathy

(39+35.5)g (108+35.5)gNaCl + AgNO

3→ AgCl + NaNO

3

58.5g (108+35.5)gFind the mass of AgCl produced by x g of KCl and (2-x)g of NaCl by using the above two equations. Thenadd the two to be equal to 4.27g. Then solve for x.5.

HgO Hg O 2

CO O 2 CO2

+ 12

2

1+

HgO CO Hg CO2+ +(200+16) 22400mL

First convert the volume from the given conditions to NTP conditions.NTP volume= 23 mL22400mL of CO reacts with 216g of HgO at NTP23 mL of CO reacts with 0.221g of HgO

So percentage of purity = (0.221/0.25)×100 = 88.4%Note that we solved this problem by additon method.You could have as well solved by repeated calculationmethod, i.e first finding the volume of O

2 required from the 2nd rection and then the mass of HgO required

from the 1st reaction.6. Zn + H

2SO

4(dil) → ZnSO

4 + H

2

Cu cannot displace H2 from dil. H

2SO

4.First find volume at NTP and solve from the above equation.what

mass of zince is required.7.

Na2CO3 CaCl2 CaCO3 N aCl

CaCO3 HCl CaCl2 CO2 H2O

+ +

+ + +

2

2

Na2CO3 HCl NaCl CO2 H2O2 2+ + +106g 2X36.5g

1×36.5g of HCl reacts with 106g of Na2CO

3

5.6g of HCl reacts with 8.13g of Na2CO

3.

% of purity = 81.3%8. CaCO3 → CaO + CO2

100g 22.4L(NTP)90% of 2Kg = 1.8Kg = 1800g100g of CaCO3 gives 22.4L at NTP1800g of CaCO3 gives 403.2L at NTP

Then covert this volume from NTP to the given conditions(598.1L).9. KNO

3 → KNO

2 + ½ O

2

NaNO3 → NaNO

2 + ½ O

2

Let KNO3 = x g, NaNO

3 = (2.1 - x)g

Then find the total mass of O2 lost in two separate equations. Add them to 0.373g and solve for x.

10. CS2

+ 3 O2 → CO

2 + 2 SO

2

(12+64)g 22.4L 2×22.4L(NTP)NaOH + CO

2 → NaHCO

3(acidic salt)

40g 22.4LNaOH + SO

2 → NaHSO

3 (acidic salt)

40g 22.4LFind the total mass of NaOH required.11. Pb + Cl

2 → PbCl

2

207g (207+71)gSn + 2Cl

2 → SnCl

4

119g (119+4×35.5)gTotal mass after the reaction = 1.1 + 0.52X1.1 = 1.672gLet Pb = x g and Sn = (1.1 - x)gFind the mass of PbCl

2 and SnCl

4 from the above reactions, and add them to 1.672. Then solve for x.

24

Stoichiometry-I

Dr. S. S.Tripathy

12. Ba(OH)2 + CO

2 → BaCO

3 + H

2O

22.4L (137+12+48)g197g of BaCO

3 is produced by 22.4L of CO

2 at NTP

0.35g of BaCO3 is produced by 0.03979L of CO

2 at NTP

Then convert this volume from NTP conditions to the required conditions.(V=0.0437L)So volume fraction = mole fraction = (0.0437/98) = 0.00044So partial pressure of CO

2 = 0.000446 × 760 = 0.339 mm of Hg

13. NaHCO3 + Ba(OH)

2 → BaCO

3 + NaOH + H

2O

84g 197gNa

2CO

3 + Ba(OH)

2 → BaCO

3 + 2NaOH

106g 197gLet NaHCO3 = x g, Na

2CO

3 = (1-x)g. Find the mass of BaCO

3 formed from both the reactions separately

and add them to 2.1. Then solve for x.14. 3MnO

2

→ Mn

3O

4 + O

2

3×87g 229gLet us take 100g of pyrolusite. MnO

2=80g, SiO

2=15g and H

2O = 5g

On ignition MnO2 changes to Mn

3O

4 and H

2O evaporates.

3×87g of MnO2 produces 229g of Mn

3O

4

80g of MnO2 produces 70.19g of Mn

3O

4

229g of Mn3O

4 contains 3×55g of Mn

70.19g of Mn3O

4 contains 50.57g of Mn

Total residue after ignition = 70.19(Mn3O

4) + 15(SiO

2) = 85.19g

So % of Mn in the ignited sample = (50.57/85.19) × 100 = 59.36%15. Balance the two equations independently by partial equation method or ON method and then add thetwo to get a composite equation.

FeS2 O 2 Fe2O3 S O24 11 2 8+ +

KMnO4 H2O SO2 K2SO4 MnSO4 H2SO42 2 5 2 2 + + + +

X 5

FeS2 O 2 K MnO4 H2O Fe2O3 K2SO4 MnSO4 H2SO420 55 16 16 10 8 16 16+ + + + + +

20×(56+64)g 16×158g16×158g of KMnO

4 requires 20×120g of FeS

2

15.8g of KMnO4 requires 15g of FeS

2

16. CaCO3

→ CaO + CO2

100g 56g 44gMgCO

3 → MgO + CO

2

84g 40g 44g

NaHCO3 Na2CO3 CO2 H2O22 X 84g 106g 44g 18g

+ +

Let CaCO3 = x g, MgCO

3= yg and NaHCO

3 = (2.5-x-y)g

(44/100)x + (40/84)y + (62/168)(2.5-x-y) = 1.16 (1)In the second reaction, only Na

2CO

3 from the residue reacts with HCl to produce CO

2 gas.

Na2CO

3 + 2HCl → 2NaCl + CO

2 + H

2O

106g 22400mL(NTP)(22400/106)×(2.5-x-y) =117.3 (2)Solving equations (1) and (2) simultaneously we get the results.

17. NaCl + AgNO3 → AgCl + NaNO

3

58.5g 143.5gKCl + AgNO

3 → AgCl + KNO

3

74.5g 143.5gLet NaCl = x g and KCl = (0.118 -x)g

(143.5/58.5) x + (143.5/74.5)× (0.118-x) = 0.2451,

25

Stoichiometry-I

Dr. S. S.Tripathy

Hence x= 0.034g(NaCl) and KCl = 0.084gNa

2O → 2NaCl

62g 2×58.5g2×58.5g of NaCl is produced by 62g of Na

2O

0.034g of NaCl is produced by 0.018g of Na2O

K2O → 2KCl

94g 2×74.52×74.5g of KCl is produced by 94g of K

2O

0.084g of KCl is produced by 0.053g of K2O

So % of Na2O in the original feldspar = (0.018/0.5)×100 = 3.6%

% of K2O = (0.053/0.5)×100 = 10.59%

18. 2Al + 3/2 O2 → Al

2O

3

2×27g 3/2×22400mLFirst the given volume is converted to NTP condition. NTP volume=28.018mL

3/2 × 22400 mL of O2 is produced by 54g of Al

28.018mL of O2 is produced by 0.045g of Al

19. 3 O2 → 2 O

3

3mL 2mLThe volume reduction = 500 - 444 = 56 mL

For a volume reduction of 1 mL the volume of O3 formed = 2mL

Hence for a volume reduction of 56mL the volume of O3 formed = 112mL

2KI + O3 + H

2O → 2KOH + O

2 + I

2

22400mL 2×127g22400mL of ozone produces 2×127g of iodine112mL of ozone produces 1.27g of iodine

Note that O2 does not react with KI.

Alternative method:3 O

2 → 2 O

3

3mL 2mLLet x mL of O

2 is converted to O

3

3mL of O2 forms 2 mL of O

3,

So x mL of O2 will form (2/3)x mL of O

3

Total volume after the reaction = (2/3)x + (500-x) = 444 ⇒ x=168mLSo the volume of O

3 = (2/3)x = 112mL

The subsequent calculation is same as the previous method.20. 2 Ca

3(PO

4)

2 + 6 SiO

2 + 10C → P

4 + 6 CaSiO

3 + 10CO

2×[3×40 +2(31+64)]g 4×31g 10×22.4LFrom this we can find out the mass of P

4 and volume of CO formed at NTP.

21. In this case no reactant remains in excess after the reaction. So this is not a problem involving limitingreactant that we have solved before.

P4

+ 3O2

→ P4O

6

4×31g 3×32g (3×31+96)gP

4+ 5O

2→ P

4O

10

4×31g 5×32g (4×31 + 160)gLet P

4 used in the first reaction = x g and so P

4 used in the 2nd reaction = (2-x)g

1st reaction: 124g of P4 consumes 96g of O

2

x g of P4 consumes (96/124) x g of O

2

2nd reaction: 124g of P4 consumes 160g of O

2

(2-x)g of P4 consumes (160/124)(2-x)g of O

2

According to the data(96/124) x + (160/124)(2-x) = 2

⇒ x = 1.125g (P4 used in 1st reaction), (2-x) = 0.875g(P

4 used in the 2nd reaction)

1st eaction: 124g of P4 produces 220g of P

4O

6

So 1.125g of P4 produces 1.995g of P

4O

6

2nd reaction: 124g of P4 produces 284g of P

4O

10

0.875g of P4 produces 2.005g of P

4O

10

26

Stoichiometry-I

Dr. S. S.Tripathy

Note that the total mass of products = 1.995 + 2.005 = 4g (law of conservation of mass)Alternative method:Let all 2g of P

4 is converted to P

4O

6. Let us calculate the amount of O

2 needed and amount of P

4O

6 formed

in the first step.124g of P

4 requires 96g of O

2

2g of P4 requires 1.548g of O

2

Hence the amount of O2 left out in this reaction = 2-1.548 = 0.452g

This amount will be used to convert a part of P4O

6 to P

4O

10

124g of P4 forms 220g of P

4O

6

2g of P4 forms 3.548g of P

4O

6

P4O

6 + 2O

2 → P

4O

10

2×32 g of O2 requires 220g of P

4O

6

0.452g of O2 requires 1.553g of P

4O

6

So the amount of P4O

6 left = 3.548-1.553=1.995g

64g of O2 forms 284g of P

4O

10

0.452g of O2 will form 2.005g of P

4O

10

22. Li2CO

3 → Li

2O + CO

2

74g 22400mL(NTP)Note that K

2CO

3 in the mixture does not decompose.So from the above reaction we can find the mass of

Li2CO

3 present in the mixture. The mass of K

2CO

3 can be found out.

23. NaCl + AgNO3 → AgCl + NaNO

3

58.5g 143.5gMCl + AgNO

3 → AgCl + MNO3

yg 143.5g1st experiment: Let NaCl = x g, and so MCl = (1-x)g

(143.5/58.5) x + (143.5/y) (1-x) = 2.567 (1)

2nd experiment: MCl gasAgNO3

AgCl1mole(yg) 1mole

y g of MCl produces 143.5g of AgCl(1-x)g of MCl produces (143.5/y)(1-x) g of AgClAccording to the data; (143.5/y)(1-x) = 1.341 (2)Solving equations (1) and (2), we get x and y. y= 53.5. So the MCl is NH

4Cl

which on heating gives NH3 and HCl.

24. CxH

yO

z + ½ (2x+y/2 -z) O

2 → x CO

2 + y/2 H

2O(l)

1mL 1/2(2x+y/2-z)mL x mL zeroVolume of CO

2 = 20mL

1mL of hydrocarbon gives x mL of CO2

10mL of hydrocarbon gives 10x mL of CO2.

10x =20 ⇒ x = 2 (1)1mL of hydrocarbon comsumes (x + y/4 -z/2) mL of O

2

10mL of hydrocarbon consumes 10(x+y/4-z/2)mL of O2

Volume after explosion = CO2 + excess O

2

10x + [ 100 - 10(x +y/4 -z/2)] = 90 (2)Molecular mass = 2×23 =4612x + y + 16z = 46 (3)

Solving these three equations simultaneously we get, y=6 and z =1So the molecular formula = C

2H

6O

27

Stoichiometry-I

Dr. S. S.Tripathy

SUPPLIMENTARY PRACTICE QUESTIONS(UNSOLVED)1. Chlorophyl, the green colouring matter of plants responsible for photosynthesis, contains 2.68% ofmagnesium by mass. Calculate the number of magnesium atoms in 2.0 g of chlorophyll.(1.33 X 1021)2. When 10.0g of marble chips(CaCO

3) are treated with 50 mL of HCl(d = 1.069 g/mL). the marble

dissolves giving a solution and releasing CO2 gas. The solution weighs 60.4g. How many litres of CO

2 gas

are released ? The density of the gas is 1.798 g/L. (2.4 L)3. 105 mL water at 40C is saturated with NH

3 gas, producing a solution of d = 0.9 g/mL. If the solution

contains 30% NH3 by mass, calculate its volume. (166.6 mL)

4. The atomic mass of A and B are 20 and 40. If x g of A contains Y atoms, how many atoms arepresent in 2x g of B ? ( y atoms)5. Density of a gas relative to air is 1.17. Find the molecualr mass of the gas. [ M

air = 29 g/mol]. (33.9)

6. Equal masses of mercury and iodine were allowed to react completely to form a mixture of mercurousand mercuric iodide leaving none of the reactants. Calculate the ratio of masses of Hg

2I

2 and HgI

2 formed.

( 0.532 : 1)7. Titanium which is used to make air plane engines and frames, can be obtained from titanium tetrachloridewhich in turn is obtained from titanium oxide by the following process :

3 TiO2(s) + 4C(s) + 6Cl

2(g) --------> 3 TiCl

4(g) + 2CO

2(g) + 2CO(g)

A vessel contains 4.15 g of TiO2, 5.67 g of C and 6.78 g of Cl

2, suppose the reaction goes to completion

as written, how many grams of TiCl4 can be formed ? (Ti = 48) (9.063g)

8. When you see the tip of a match fire, the chemical reaction is likely to beP

4S

3 + 8O

2 -----> P

4O

10 + 3SO

2What is the minimum amount of P

4S

3 that would have to be burned to produce at least 1.0 g of P

4O

10 and

at least 1.0 g of SO2. (1.14 g)

9. The action of bacteria on meat and fish produces a poisonous compound called cadaverine. It is58.77% C, 13.81%H and 27.42% N. Its molar mass is 102 g/mol. Determine the molecular formula ofcadaverine. (C

5H

14N

2)

10. On combustion analysis, a 0.45 g sample of caproic acid ( contained only C, H and O) gives 0.418g of H

2O and 1.023 g of CO

2. What is the empirical formula of caproic acid? If the MM of caproic acid

is 116 amu, what is its MF ? ( C3H

6O, C

6H

12O

2)

11. A sample of pure metal M weighing 1.35g was quantitatively converted to 1.88 g of pure MO.Calculate atomic mass of M. (41)12. Cu

2S and M

2S are isomorphous. The % of S in former is 20.14 and in the latter 12.94. Atomic mass

of Cu = 63.57. Calculate the atomic mass of M. (108.16)13. A compound which contains one atom of X and two atoms of Y for each three atoms of Z is madeby mixing 5.00 g of X, 1.15 X 1023 atoms of Y and 0.03 mole of Z atoms. Given that only 4.40 g of compoundresults. Calculate atomic mass of Y if the atomic mass of X and Z are 60 and 80 respectively. (70)14. A 0.6025 g sample of a chloride salt was dissolved in water and the chloride precipitated by addingexcess of silver nitrate. the precipitate of AgCl was filtered, washed, dried and found to weigh 0.7134 g.Calcualte the pecentage of chlorine in the sample. (29.29%)15. A 0.4852 g sample of iron ore is dissolved in acid, the iron oxidised to +3 state, and then precipitatedas hydrated oxide, Fe

2O

3.xH

2O. The precipitate is filtered, washed and ignited to Fe

2O

3 which is found to

weigh 0.2481 g. Calculate the percentage of iron in the sample. (35.77%)16. A mineral consists of an equimolar mixture of the carbonate of two bivalent metals. One metal ispresent to the extent of 13.2% by mass. 2.58 g of the mineral on heating lost 1.233 g of CO

2. Calculate the

% by mass of the other metal. (21.68%)17. A 10 g sample of a mixture of CaCl

2 and NaCl is treated with Na

2CO

3 to precipitate calcium as

calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is

1.62 g. Calculate the % by mass of NaCl in the original mixture. (67.9%)18. 10 mL of a mixture of CO, CH

4 and N

2 exploded with excess of oxygen gave a contraction of 6.5

mL. There was a further contraction of 7 mL, when the residual gas treated with KOH. What is thecompostion of the original mixture. (CO = 5 mL, CH

4 = 2 mL)

19. When 100 mL of a O2 - O

3 mixture was passed through turpentine, there as reduction of volume by

20 mL. If 100 mL of such a mixture is heated , what will be the increase in volume ? (10 mL)20. 40 mL of a mixture of H

2, CH

4 and N

2 was exploded with 10 mL of O

2. On cooling, the gases

occupied 36.5 mL. After treatment with KOH the volume reduced by 3 mL and again on treatment withalkaline pyrogallol, the volume further decreased by 1.5 mL. Determine the composition of the mixture.

( H2 = 12.5%, CH

4 = 7.5%)

21. 5 mL of a gaseous hydrocarbon was exploded to 30 mL of O2. The resultant gas, on cooling is found

to measure 25 mL of which 10 mL are absorbed by KOH and the remainder by pyrogallol. Determine themolecular formula of the hydrocarbon. (C

2H

4)

28

Stoichiometry-I

Dr. S. S.Tripathy

SOLUTION STOICHIOMETRY(VOLUMETRIC ANALYSIS)

MOLARITY AND PERCENT STRENGTH

Concentration or strength of a solution:When a solution is prepared by dissolving a solute in a solvent, it is always necessary to know how much

of solute is present in how much of the solvent i.e their relative proportions in the mixture(solution). If moreamount of solute is present in a given solution, it is called a concentrated solution and if less amount of solute ispresent, it is called the dilute solution. In order to know the concentration or strength of the solution i.e to knowhow concentrated or how diluted a solution is, different terms are used. These terms are Molarity, Molality,Normality and % strength. Note that the volume of a solution should not be confused with gaseous volume. Boththe types of volumes are measured in the same unit i.e mL or cc and L. Gas laws are applicable to gaseousvolume but not to volume of solutions.Preparation of a solution of desired concentration

An accurately weighed amount of solute(solid or liquid) is taken in a measuring flask of a fixedvolume(capacity). Measuring flasks of various capacities such as 2 litres, 1 litre, 500 ml, 250 ml, 100 ml etc. areavailabe in the laboratory. Each flask has a mark at the upper part of the stem of the flask. The solute is taken ina measuring flask first and slowly distilled water is added from a wash bottle and the solute is allowed to mix withthe solvent by gentle shaking of the flask. Water is added in small amounts slowly with constant shaking upto themark located in the upper part of the flask. Finally the flask is stopperd and is shaken thoroughly. The volume ofsolution becomes exactly equal to the capcity of the flask which is written on the flask. Don't add more water soas to cross the mark on the glass. If you do so then the total volume will be more and cannot be properly recorded.Note that the volume of the solution is the total volume of solute and solvent. In case of solid solutes, particularlyionic solutes like NaCl, Na

2CO

3 etc. the dissolution involves the formation of free ions which occupy the intermolecular

space of the solvent molecules. Hence the volume change that occurs when such a solute dissolves in a solventis very small. Hence in such case the volume of the solution is approximately equal to the volume of the solvent.Of course this is true for dilute solutions prepared from solid solutes. However when a liquid solute such asH

2SO

4 is used, the addition of water brings about change in the volume of solution compared to the volume of

solvent. This is because H2SO

4 exists in the liquid state and itself possesses a lot of free intermolecular volume

like the solvent molecules. So the volumes of solute and solvent approximately add to give the volume of thesolution.

MOLARITY:Molar Solution:(1M or M solution)

If one mole of solute is present in 1 litre(1000 mL) of solution, it is called a Molar(1M or M)solution.One mole of H

2SO

4(98 gms) present in 1 litre solution is called 1M H

2SO

4.Similarly when 40gms(1mole) of

NaOH is used to prepare 1000 cc of solution, it is Molar(M) or 1M solution of NaOH.____________________________________________________________________________________SAQ 1: How much of solute in gm. will be required per litre of solution to prepare 1M solution in the followingcases.

(i)Na2CO

3(ii)HCl (iii)KMnO

4(iv)FeSO

4

SAQ 2: The solutions have the following amounts of the solutes per litre of the solution. What are their molarities.(i)56gms of KOH (ii)106gms of Na

2CO

3(iii)98gms of H

2SO

4

____________________________________________________________________________________Molarity:

The number of moles of solute present in 1000 mL(1 litre) of the solution is called the molarity of asolution.

4gms(0.1 mole) of NaOH present in 1000 cc of the solution is called 0.1M solution or the molarity of thesolution is 0.1M.

29

Stoichiometry-I

Dr. S. S.Tripathy

2gms(0.05mole) of NaOH present in 1000cc of the solution is called 0.05M solution.80gms(2 moles) of NaOH present in 1000cc of the solution is called 2M solution and so on.

SAQ 3: Find out the molarity of the solutions if the following amounts are dissolved to make 1 litre solution ineach case.

(i)49gm of H2SO

4(ii)3.65gms of HCl (iii)0.7kg of H

2SO

4

SAQ 4: A certain bottle of H2SO

4 is marked as 2M H

2SO

4 and another bottle of H

2SO

4 is marked 0.5M H

2SO

4.

What do they mean? Which is a diluted solution and which concentrated? How many times the one is moreconcentrated than the other.

Let us consider another situation. Supposing you do not have a 1 litre measuring flask with you, in steadyou have a 500 mL flask with you. How can you prepare a 1M solution of H

2SO

4? This is very simple. If 1

litre(1000cc) of solution requires 98gms of H2SO

4 to make 1M solution, 500ml of solution will need half of it i.e 49

gms. So 49gms of H2SO

4 taken to make 500 cc solution has a molarity 1M.

SAQ 5: (i)How much of NaOH will be needed to produce a 1M solution by using 250ml measuringflask?

(ii)How much of H2SO

4 will be needed to produce a 1M solution by using 100ml measuring

flask?

Now let us consider still a different situation. Supposing you want to prepare 0.2M solution of NaOH and youhave 500ml measuring flask, how shall you do? This is again very simple.

If 1000ml solution needs 0.2 mole i.e 0.2×40=8gms of NaOH for making 0.2M solution.500ml solution will need 8/2 = 4gms of NaOH to make the same 0.2M solution.

So in all these cases you have to find out in an unitary method how much of the solute is required for the givenvolume of the solution.SAQ 6: Calculate the amounts of solute present in each case of the following solutions.

(i)250ml of 0.5M H2SO

4(ii)2litres of 0.1M NaOH

(iii)100ml of 4M Na2CO

3(iv)500ml of 12M HCl

FORMULA METHOD:

Molarity =number of moles of solute

volume of solution in litre

⇒ Number of moles of solute = Volume of solution in litres × Molarity

Example: How many moles and what mass of each of the following solutes are required to prepare thefollowing solutions.

(i)500 ml of 0.8M H2SO

4(ii)100ml of 2M Na

2CO

3

(iii)2 litres of 0.1 M HCl (iv)250ml of 10M NaOHSolution: Unitary Method

(i) 1litre of 0.8M solution of H2SO

4 contains 0.8 mole of H

2SO

4

So 500ml or 0.8l of this solution must contain 0.5 ×0.8= 0.4 moleFormula method:

volume of the solution =0.5 L and molarity is 0.8M.So the number of moles of H

2SO

4= 0.5×0.8=0.4

Once you find the number of moles of the solute, it is an easy task to find the mass in gm by multiplying thenumber of moles with molecular mass.

mass of 0.4 mole of H2SO

4 = 0.4×98= 39.2gms.

(ii) Moles of Na2CO

3= volume in litre × molarity= 0.1 litre× 2= 0.2.

Mass= 0.2×106 (M.M)= 21.2gms.(iii) Moles of HCl = 2× 0.1 =0.2, so the mass of HCl= 0.2×36.5=7.3gms.(iv) Volume in litre= 0.25 l , Molarity=10, So number of moles of NaOH= 0.25×10=2.5

So the mass of NaOH=2.5×40=100gmsSAQ 7: Find out the mass of the following solutes required to prepare 2 litres of M/20 solution of

(i)H2SO

4(ii)KBr

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Stoichiometry-I

Dr. S. S.Tripathy

____________________________________________________________________________________DETERMINATION OF MOLARITY OF A SOLUTION:Example: Find the molarity of a H

2SO

4 solution, 7gms of which is present in 250 cc solution.

Solution: Unitary method:7gms of H

2SO

4 = 7/98=1/14 mole

250cc of the solution contains 1/14 mole1000cc of the solution contain [(1/14)1000]/250 = 4/14= 0.286 mole

So the molarity of the solution= 0.286MFormula MethodWe know the formula that

volume in litre= Number of moles Molarity

In the above example , we first have to find the number of moles of solute.7gms of H

2SO

4 = 7/98=1/14 mole, volume of solution in litre= 0.25 l

So Molarity = 1/14/0.25= 0.286M.(same as obtained by unitary method)SAQ 8: Find out the molarity of the following solutions.

(i)2gms of NaOH present in 100ml solution(ii)5.3gms of Na

2CO

3 present in 2liters of solution

(iii)0.365gm of HCl present in 250ml of solution

MILLIMOLE(MMOL) METHOD:Note that in stead of solving these problems in terms of moles which often appears as a fraction, it is

more convenient to work in terms of millimoles(mmols). You know that1 mole=103 mmoles=1000 mmoles

volume in litre= Number of moles Molarity

Molarity Number of mmoles = volume in ml or ccVolume in litre × molarity = number of molesVolume in ml(cc) × molarity = number of mmoles

If we express volume of the solution in ml(cc) in stead of litre, we get millimoles(mmoles) in stead of moles.

Example: Find out the molarity of 200 ml solution containing 0.7 gm of H2SO

4

Solution: Unitary Method: 200 ml of solution contains 0.7/98 mole of H2SO

4

1000ml of solution contains 0.7/98 X1000/200 =5/140=1/28moleSo the molartiy of the solution is 1/28M

Formula method(1): volume in litre= Number of moles Molarity

The number of moles= 0.7/98= 1/140, volume of the solution=200ml=0.2 litreHence Molarity= 1/140/0.2= 1/28M

Formula method(2): Molarity Number of mmoles = volume in ml or ccThe number of moles= 0.7/98, Hence number of mmoles = 103 × no. of moles= 1000 ×0.7/98=100/14;

volume in ml= 200ml molarity =100

14 X200= 1/28 M

(You can adopt any method you like)SAQ 9: Find out the number of mmoles of the solute present in the following solutions.

(i)100ml of 0.5M NaOH solution (ii)3litres of 0.04M H2SO

4 solution

(iii)500ml of 3M Na2CO

3 solution

SAQ 10: Find out the masses of solute in each case of SAQ 9.SAQ 11: Find out the number of Na+ ions present in 100ml of M/10 Na

2CO

3 solution.

SAQ 12: You are given 500ml of 2M solution of H2SO

4. You are asked to divide this solution in two bottles, one

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Stoichiometry-I

Dr. S. S.Tripathy

containing 200ml and the other containing 300ml. What will be molarity of H2SO

4 in the first and second bottles?

PERCENT STRENGTHLike molarity, there is another way to express the concentration of a solution. This is called percent

strength. There are two types of percentages used to find the concentration of a solution. These are(i)Volume percent(w/v) and (ii) weight percent(w/w)

(i) Volume Percent(w/v):The mass of the solute( in gram) present in 100ml of the solution is called percent strength of the

solution by volume. This pecentage is denoted by the symbol w/v i.e weight of the solute(w) and volume of thesolution(v) are considered. Say for example, 2gms of sugar used to prepare 100ml solution is a 2% solution byvolume(w/v). In molarity, we have to find out the number of moles in stead of the mass in grams and moreover thevolume of the solution is taken to be 1000ml to find molarity whereas it is 100ml in percent strength by volume.Example: Find the percent strength of a solution containing 49gms of H

2SO

4 in 10 litres of solution.

Solution: In 10 litres i.e 10,000 ml of solution, the mass of solute=49gmsSo in 100ml of solution, the mass of solute would be (49/10,000) ×100=0.49So percent strength by volume is 0.49% (w/v)

SAQ 13:Find the percent strength of the following solutions by volume.(i)NaOH solution containing 25gms of NaOH in 500ml solution(ii)Na

2CO

3 solution containing 0.005moles of Na

2CO

3 in 1 litre solution

(iv)H2SO

4 solution which has a molarity equal to 0.05M

FINDING PERCENT STRENGTH FROM MOLARITY OF SOLUTION:Example: Find the percent strength by volume of a NaOH solution whose molarity is 0.01M.

A 0.01M solution means, 1000ml of the solution contains 0.01mole of NaOH i.e 0.01×40=0.4gm.If 1000ml of the solution contains 0.4gm of NaOHThen 100 ml solution must contain (0.4/1000)×100= 0.04gm of NaOH.So the percent strength of the solution by volume= 0.04%

SAQ 14: Find the percent strength by volume in respect of the following solutions.(i)3M solution of Na

2CO

3(ii)0.5M solution H

2SO

4(iii)500ml of 2M KOH

solution.

(ii) Weight(Mass) Percent (w/w)The mass of the solute in grams that is present in 100gms of the solution is called the weight(mass)

percent of the solution. Note that in this case, for both solute and solvent we consider the mass. Therefore it isdesignated as w/w i.e weight/weight method. Let us take an examle. Supposing we took 2gms of sugar andadded with it 98gms of water, so as the make the total mass of the solution 100gms. Note that here we will notuse any measuring flask which measures exact volume of the solution. In this case the percent strength byweight is 2%(w/w).Example: 500gms of H

2SO

4 is mixed with 1kg of water to prepare a solution of H

2SO

4. Find the percent

strength by weight.Solution: Mass of the solute=500gms, Mass of the solvent= 1kg=1000gm.

So the mass of the solution= 500+1000=1500gms.If 1500gms of the solution contains 500gms of H

2SO

4

Then 100gms of the solution must contain (500/1500)×100=33.33gmsSo the percent strength by weight is 33.33%.

SAQ 15: 15gms of pure HNO3 is mixed with 150gms of water to prepare a solution. What is its percent strength

by weight?

FINDING THE WEIGHT PERCENTAGE IF VOLUME OF THE SOLUTION IS KNOWN INSTEAD OF MASS BY USIND DENISTY OF THE SOLUTION

For finding the weight percentage from the volume of the solution, we need to know the density of thesolution.

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Stoichiometry-I

Dr. S. S.Tripathy

Example: Find out the percent strength by weight of a H2SO

4 which contains 49gms of H

2SO

4 in

500ml solution. The density of the solution is 1.32gm/ccSolution: We know that density of the solution= (mass of the solution)/(volume of the solution)d=m/v; Volume of the solution=500ml(data);

so mass= d ×V =1.32gm/ml × 500ml= 660gm.Mass of the solution is 660gms and out of which 49gms is the mass of solute.660gms of the solution contains 49gms of H

2SO

4

So 100gms of the solution must contain (49/660) ×100= 7.42gmsSo the percent strength by weight is 7.42%(w/w).

SAQ 16: Find the percent strength by volume and by weight of 18M molar solution of H2SO

4. The specific

gravity of the solution is 1.87.SAQ 17: A bottle of commercial sulphuric acid has a density 1.787gm/ml. It is levlled as 86% by weight.Find the molarity of the acid and also the percent strength by volume.

STOICHIOMETRIC CALCULATION BASED ON MOLARITY AND STRENGTH OFSOLUTION

In the previous chapter we have discussed molarity and percentage strength of the solutions. Nowwe shall use it in solving stoichiometric problems. Read this example.Example 1: How much zinc will completely react with 200ml of 2M H

2SO

4 solution? Also calculate

the volume of hydrogen gas evolved at 270C and 900mm pressure.(Zn=65)Solution: M.M of H

2SO

4 is 98.

1000ml of a 2M solution contains 2 moles i.e 2×98gms of H2SO

4

So 200ml of such solution must contain (2×98)/(1000)×200= 39.2gms.Alternative (mmole) method:

Number of mmole of H2SO

4 = volume in ml × molarity= 200×2=400

Mass of H2SO

4 = 400 × 10-3 × 98 = 39.2gms.

( Note that when mmole is converted to mol, a factor of 10-3 is multiplied)Now let us write the balanced equation.Zn + H

2SO

4 ZnSO

4 + H

2

65 g 98 g 22.4L98gms of H

2SO

4 completely reacts with 65gms of zinc

39.2gms of H2SO

4 must react with (65/98)×39.2= 26gms of Zn.

98 gms of H2SO

4 produces 22.4 litres of H

2 gas at NTP

So 39.2 gms of H2SO

4 must produce (22.4/98)×39.2= 8.96 litres of O

2 at NTP.

(Note that you can also take the data of zinc found above to calculate the volume of H2)

Now we have to convert the volume from NTP to the given conditions.

760 mm X 8.96 l273A

=900mm X V2

(273+27)A ⇒ V2 = 8.31 litres.

Example 2: What volume of 0.3M H2SO

4 is required to exactly neutralize 200ml of 0.5M NaOH

solution?Solution:

Number of mmoles of NaOH= 200 × 0.5 = 100Let us write the balanced chemical equation.2NaOH + H

2SO

4 Na

2SO

4+ H

2O

2 moles 1 mole2 mmoles 1 mmole2 mmoles of NaOH reacts with 1 mmole of H

2SO

4.

So 100 mmoles of NaOH must react with 50 mmoles of H2SO

4.

mmoles = volume in ml × Molarity ⇒ volume = 50/0.3= 166.67 ml.(You are advised also to solve this problem in terms of mass and check your answer. For that you have toconvert the mmoles of NaOH to its mass and then find out the mass of H

2SO

4 required from the equation.

Finally the volume of the 0.3M H2SO

4 solution required is found out as usual by the unitary method).

SAQ 18: How many ml of 3M HCl solution should be needed to react completely with 16.8gms of NaHCO3?

SAQ 19: What volume of 0.025M HBr solution is required to neutralize 25.0 ml of 0.02M Ba(OH)2 solution?

SAQ 20: Calculate the molarity of a HCl solution if 2.5ml of the solution took 4.5ml of 3M NaOH solution

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Stoichiometry-I

Dr. S. S.Tripathy

for complete neutralization.SAQ 21: What volume of M/10 solution FeSO

4 solution will be required to completely react with

1.58gms of KMnO4 in dilute H

2SO

4 medium?

PROBLEMS BASED ON VOLUME STRENGTH OF HYDROGEN PEROXIDE SOLUTIONWhen we buy hydrogen peroxide from the market, we get them usually in dark coloured bottles which

are labelled as “10 Volumes” or “20 Volumes” or in general “x volumes”. In fact these are not pure H2O

2,

rather are aqueous solutions.“x Volumes” H

2O

2 solution means if 1 volume of such H

2O

2 solution(liquid) at NTP is decomposed,

‘x’ volumes of O2 gas are evolved. Note that here the two volumes are of different types. One is volume

of the solution(1 volume i.e 1L or 1 mL) and the other is gaseous volumes(i.e ‘x’ L or ‘x’ mL).“10 Volumes” of H

2O

2 means, 1 mL of such a solution on decomposition will produce 10 mL of O

2

gas at NTP. This is called volume strength of H2O

2. Volume strength can be converted to percent strength

by volume or vice versa. Look to the following examples.Example 1: Calculate the percent strength by volume of a H

2O

2 solution if its volume strength is “22.4

Volumes”.Solution: 2 H

2O

2 → 2 H

2O + O

2

2(2+32)g 22.4L(NTP)“22.4 Volumes” solution means 1 mL of such a solution liberates 22.4mL of O

2 at NTP.

1mL of the solution liberates 22.4mL of O2 at NTP

So, 100mL of the solution will liberated 2240mL of O2 at NTP

From the balanced equation,22400mL of O

2 is liberated by 2×34g of H

2O

2

So, 2240mL of O2 is liberated by 6.8g of H

2O

2

Hence 100mL of the solution contains 6.8g of H2O

2. So the percent strength is 6.8%(w/v).

Example 2: 20mL of a solution of H2O

2 labelled “15 volumes” just decolorized 100mL of KMnO

4 acidified

with dil. H2SO

4. Calculate the mass of KMnO

4 in the given solution.(K=39, Mn=55)

Solution: “15 volumes” means 1mL of the solution liberates 15 mL of O2 gas at NTP.

So 20mL of this solution will liberate 20×15 = 300mL of O2 at NTP.

2KMnO4 + 3H

2SO

4 + 5 H

2O

2 → K

2SO

4 + 2MnSO

4 + 5O

2 + 8H

2O

2×158 5×22400mL(NTP)5×22400mL of O

2 is produced by 2×158g of KMnO

4

So, 300mL of O2 is produced by 0.846g of KMnO

4

Hence the mass of KMnO4 required is 0.846g.

SAQ 22: Which is more concentrated H2O

2 solution, “10 volumes” or “20 volumes” and why?

SAQ 23: Calculate the % strength by volume of a H2O

2 solution labelled as “20 volumes”

SAQ 24: Find the volume strength on the basis of available oxygen of a H2O

2 solution which is 12% by

volume.SAQ 25: 200 mL of 0.1M BaCl

2 was mixed with 100 mL of 0.05M Na

3PO

4. What mass of barium phosphate

will be formed ? (Ba = 137)

PRACTICE QUESTIONS1. Find the number of moles of solute present in each case

(i)1 litre of 2M H2SO

4(ii)200ml of 0.02M HNO

3(iii)3.5litres of 17M HCl

(iv)500ml of 0.1M Na2CO

3

2. Find the mass of the solute required to prepare the following solutions.(i)750ml of 5M KOH (ii)2 litres of 1M H

2SO

4(iii)200ml of 0.002M HCl

(iv)20ml of 18M H2SO

4

3. Find the Molarity of the following solutions.(i)1gm of NaOH present in 50ml solution. (ii)2.12gms of Na

2CO

3 present in 250ml

solution (iii)196gms of H2SO

4 present in 2.5litres solution

4. What is the molarity of NaOH in a solution which contains 24.0gms of NaOH dissolved in 300ml solution?5. What volume of 3M NaOH solution can be prepared with 84.0g NaOH?6. What volume of 1.71M NaCl solution contains 0.2 mol NaCl?7. 0.585gm of NaCl is present in 50ml solution. What is the strength of the solution in percent by volume?8. What is the molarity of a 4% NaOH solution by volume?

34

Stoichiometry-I

Dr. S. S.Tripathy

9. 9.8gms of H2SO

4 is present in 25ml solution. The density of the solution is 1.7gm/ml. What is the percent

strength by weight.10. How many grams of a 5% NaCl solution by weight are necessary to yield 3.2g of NaCl?11. To prepare 100gm of 19.7% by weight solution of NaOH, how many g of each of NaOH and H

2O are

needed?12. What volume of dilute nitric acid of density 1.11g/mL which is 19% HNO

3 by weight contains 10g

HNO3?

13. How many gm. of conc. hydrochloric acid which is 37.9 % by weight will contain 5.0gms of HCl?14. An aqueous solution of HCl contains 28% HCl by weight and has a density of 1.2gm/ml .Find the molarityof the solution.15. What volume of 96% H

2SO

4 solution(density 1.83gm/ml)is required to prepare 2.0L of 3M H

2SO

4 solution?

16. Calculate the mass of 80% sulphuric acid by weight required to completely react with 25g of CaCO3.(30.62g)

17. Calculate the (i)mass of MnO2 and (ii)the volume of hydrochloric acid having density 1.2g/mL and

containing 50% HCl by weight needed to produce 2L of chlorine gas at NTP? ( 7 . 7 6 g ,21.7mL)18. 3.5g of MgCO

3 were added to double its mass of H

2SO

4 solution. After complete reaction, 0.5g of

MgCO3

remained unreacted. Calculate percent strength of H2SO

4 by weight and volume of CO

2 formed

at 270C and 760mm pressuire.(50%)

19. Calculate the mass of NaCl formed when 50g of Na2CO

3 reacted with 100g of 50% HCl by weight. Also

calculate the mass of excess reagent which remained unreacted. (55.18g, 65.57g ofHCl)20. A solution of nitric acid contains 65% HNO

3 by weight. What mass of this acid will be necesary to

dissolve 2g of zinc oxide. (Zn =65)(4.78g)21. Most commercial HCl is prepared by heating NaCl with conc. H

2SO

4. How much of sulphuric acid

containing 90% H2SO

4 by weight is needed for the production of 100Kg of conc. HCl containing

42% HCl by weight.(62.64Kg)

22. 10mL of a solution of H2O

2 labelled “20 volumes” just decolorizes 50mL of KMnO

4 solution acidified

with dil. H2SO

4. Calculate the amount of KMnO

4 present in the solution.

(0.564g)23. How many mL of water must be added to 250mL of 0.85M HCl to dilute the solution to 0.25M?(600mL)24. Calculate the molarity of H

3PO

4 solution if 25mL of this solution is completely neutralised by 45mL of

0.85 M/10 Ba(OH)2 solution. (0.022M)

25. Calculate the molar concentration of each of the ionic species in solution.(a)250mL of 2M BaCl

2 is diluted to 750mL

(b)200mL of 2M H2SO

4 is mixed with 100mL of 0.5M NaOH and the mixture diluted to 500mL

(c)50mL of 0.12M Fe(NO3)

2 + 100mL of 0.1M FeCl

2 + 50mL of 0.26M Mg(NO

3)

2(BITS-1990)

26. What volume of 0.5M K2Cr

2O

7 solution will react completely with 100mL of M/10 (NH

4)

2C

2O

4 solution

in acidic medium. (Cr =52, K=39)(6.66mL)27. Calculate the pecent of BaO in 35g of a mixutre of CaO and BaO which just reacts with 150mL of 6M

HCl.(Ba=137) (43.91%)28. A solution of density 1.6g/mL is 67% by weight. What will be % strength by weight of solution if it isdiluted to have density 1.1 g/mL?(16.24%)

RESPONSE TO SAQs(Solutions: Molarity)

SAQ 1: In each case we require 1 mole or gm molecular mass of the solute(i) M.M of Na

2CO

3 =106, so we need 106gms per litre

(ii) M.M of HCl = 36.5, and so we need 36.5 gms per litre(iii) M.M of FeSO

4 = 56+32+64=152, so we need 152gms per litre

SAQ 2: (i) 1M(one Molar or Molar) since 1 mole of KOH=39+16+1=56(M.M)

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Stoichiometry-I

Dr. S. S.Tripathy

(ii) 1M since M.M of Na2CO

3 = 46+12+48=106(1 mole)

(iii) 1M since M.M of H2SO

4 =98(1 mole)

SAQ 3: (i) 98 gms of H2SO

4=1 mole, so 49 gms of H

2SO

4 = 0.5 mole.

Since 0.5mole of it is present in 1 litre solution, the molarity = 0.5M(ii) 36.5gms of HCl= 1 mole, so 3.65gms of HCl= 1/10=0.1 mole

Since 0.1 mole of HCl is present in 1 litre solution, the molarity=0.1M(iii) 98gms of H

2SO

4=1 mole, so 0.7kg i.e 700gms of H

2SO

4=700/98=7.143moles

Since 7.143moles of H2SO4 is present in 1 litre solution, the molarity=7.143MSAQ 4: The bottle marked 2M H

2SO

4 means, there would be 2 moles(2X98=196gms) of H

2SO

4 per litre of the

solution. However in the bottle marked 0.5M, there would be 0.5mole(0.5X98=49gms)of H2SO

4 per litre of

solution. So the second bottle of H2SO

4 is more diluted and the first bottle is more concentrated. The first solution

is 4 time more concentrated(49X4=196) than the second solution.SAQ 5:(i) 1000ml of solution needs 40gms(1mole) of NaOH to prepare 1M solution

So 250ml of solution must need (40/1000)X250=10gms to prepare the same 1M solution(ii) 1000ml of solution need 98gms(1mole) of H2SO4 to prepare 1M solutionSo 100ml of solution will need (98/1000)X100=9.8gms to prepare the same 1M solution.

SAQ 6: (i) 1000ml of 0.5M solution contains 0.5mole i.e 0.5X98=49gms of H2SO

4

So 250ml of 0.5M solution must contain (49/1000) X 250= 12.5gms.(ii) 1000ml of 0.1M NaOH solution contains 0.1mole i.e 0.1X40=4gms of NaOH

So 2litres i.e 2000ml of 0.1M solution must contain (4/1000)X2000= 8gms.(iii) 1000ml of 4M solution of Na

2CO

3 contains 4 moles i.e 4X106=424gms

So 100ml of 4M solution must contain (424/1000)X100=42.4gms.(iv) 1000ml of 12M solution of HCl contains 12moles i.e 12X36.5=438gms of HCl

So 500ml of 12M solution must contain 438/2=219gms.SAQ 7: Here we shall have to find the mass of the solute. Let us first find the number of moles and then mass byusing the formula method discussed in the examples in the text.

(i) Number of moles= volume in litre X molarity=2X(1/20)=1/10=0.1Mass of H

2SO

4= 0.1X98=9.8gms.

(ii) Number of moles of KBr= 2X (1/20) = 1/10=0.1 [same as (i)]Mass of KBr = 0.1X(39+80)=0.1X 119 = 11.9gms.

SAQ 8:(i) M.M of NaOH=40, 2gms of NaOH=2/40=0.05 mole,volume = 100ml=0.1 l, So Molarity= 0.05/0.1= 0.5M.

(ii) Number of moles of Na2CO

3 = 5.3/106=0.05; volume=2litres

So Molarity = 0.05/2=0.025M.(iii) Number of mole of HCl=0.365/36.5= 0.01; volume=250ml=0.25 litre

So Molarity= 0.01/0.25=0.04M.SAQ 9: We know that No. of mmoles = Molarity X Volume in ml(cc)

(i)Number of mmoles= 100X0.5=50(ii)3000X0.04= 120 (iii)500X3=1500

You found that the number of mmoles is not a fractional value(more than 1).SAQ 10: (i)50mmoles of NaOH= 5X 10-3 mole = 0.005mole= 0.005X40=0.2gm.

(ii)120mmole of H2SO

4 = 120X10-3mole=0.12 mole = 0.12X98=11.76 gms.

(iii)1500mmoles of Na2CO

3= 1500X10-3=1.5mole = 1.5X106= 159gms.

SAQ 11: The no. of mmoles of Na2CO

3 present in the solution=100 X1/10 =10

10 mmoles= 10 X10-3 moles = 0.01mole1 mole of Na

2CO

3 contains 2 moles of Na+ ions

So the number Na+ ions present in 2 moles = 2X6.023X1023 =1.2046X1024 ions0.01 mole of Na

2CO

3 will contain 0.01 X1.2046X1024 = 1.2046X1022 ions of Na+.

SAQ 12: Since the stock(original) solution is same, the molarity will remain same. Suppose you collect 500ml ofsea water and distribute the same in several containers. Do you think that the salinity of the sea water will bedifferent in these different vessels? That means the vessel containing less solution will taste less salty and thevessel containing more solution will taste more salty? The answer is NO. They will all taste same, because theconcentration of the original(stock) solution is same. Remember that the concentration of a solution gives thequality of the solution i.e how much of solute present per a fixed volume of solution(say 1 litre). Once a solutionof a particular concentration is prepared, its concentration remains the same even if this stock solution is divided

36

Stoichiometry-I

Dr. S. S.Tripathy

into any number of parts.SAQ 13: (i) 500ml of the solution contains 25gms of NaOH

So 100ml of the solution must contain (25/500)X100=5gms of NaOHSo pecent strength by volume = 5%.

(ii) M.M of Na2CO

3=106, 1 mole= 106gms, so 0.005 moles = 106X0.005= 0.53gm

1 litre i.e 1000ml of solution contains 0.53gm of Na2CO3So 100ml of the solution must contain (0.53/1000)X100= 0.053gmSo percent strength by volume=0.053%

SAQ 14: (i) M.M of Na2CO

3=106; 3M solution means

1000ml of solution contains 3moles i.e 3X106=318gms of Na2CO

3

So100ml of solution must contain (318/1000)X100=31.8gms.So percent strength by volume= 31.8%

(ii) 0.5M solution means1000ml of the solution contains 0.5mole i.e 0.5X98= 49gms of H

2SO

4

So 100 ml of the solution must contain 49/10=4.9gmsSo the percent strength by volume is 4.9%

(iii) Here 500ml solution carries no relevance and it will not be used. We have to consideronly the molarity of the solution. 2M solution means

1000ml of KBr solution contains 2 moles i.e 2X(39+80)= 2X56=238gmsSo 100ml of the solution must contain 238/10= 23.8gms of KBrSo the percent strength is 23.8% by volume.

SAQ 15: The mass of the solution = 15 + 150 = 165gms.165gms of solution contains 15 gms of HNO3,100 gms of solution must contain (15/165) X 100 = 9.09 gmsSo its percent strength is 9.09%(w/w)

SAQ 16: Volume %:18M solution means 1000ml of solution contains 18moles=18X98=1764gms of H

2SO

4.

1000ml of the solution contains 1764gms of H2SO

4.

So 100ml of the solution must contain 1764/10=176.4gmsSo percent strength by volume=176.4%

Weight %:density= m/v; 1.87gm/ml= m/1000ml,⇒ mass= 1000 X 1.87=1870 gms, So mass of 1000ml of solution is 1870gms.1870gms of the solution contains 18mole i.e 18X98=1764gms of H

2SO

4

100gms of the solution must contain (1764/1870)X100=94.33gmsSo percent strength by weight is 94.33%.

SAQ 17: Let us consider 100gm of solution. d=mass/volume,So volume of 100gm solution = 100/1.787= 55.959ml .As it is 86%solution by weight.So the mass of solute(H2SO4)in present in 100gms of solution = 86gms.

Molarity:55.959ml of solution contains 86gms or 86/98moles i.e 0.8775 mole of H

2SO

4

1000ml of the solution must contain 0.8775X (1000/55.959) = 15.68molesHence the molarity of the solution= 15.68M

Percent strength by Volume:55.959ml of solution contains 86gms of H2SO4So 100ml of solution must contain (86/55.959)X100= 153.68%(w/v)

SAQ 18:NaHCO

3 + HCl ---------> NaCl + CO

2 + H

2O

8484 gms of NaHCO

3 = 1 mole ⇒ 16.8gms = 0.2mole = 0.2 X 1000=200mmole

According the balanced equation, 1 mmole of NaHCO3 will require 1mmole of HCl

So 200mmole of NaHCO3 will require 200 mmole of HCl according to the balanced

equation.We know that mmoles = volume in ml X Molarity ⇒ Volume = 200/3=66.67ml.

37

Stoichiometry-I

Dr. S. S.Tripathy

SAQ 19:Ba(OH)

2 + 2HBr ---------> BaBr

2 + 2H

2O

No. of mmoles of Ba(OH)2 = 25 X 0.02= 0.5

According the balanced equation, 1 mmole of Ba(OH)2 will need 2mmoles of HBr.

So 0.5mmole of Ba(OH)2 will need 2 X 0.5= 1mmole of HBr

So the volume of HBr solution = mmole/Molarity = 1/0.025 = 40ml.SAQ 20: NaOH + HCl ----------> NaCl + H

2O

No. of mmoles of NaOH = 4.5 X 3= 13.5; According to the balanced equation, the number ofmmoles of HCl required = 13.5; So the molarity = mmoles/ml = 13.5/2.5= 5.4MSAQ 21:

2KMnO4 + 8H

2SO

4 + 10FeSO

4 --------> K

2SO

4 + 2MnSO

4 + 5Fe

2(SO

4)

3 + 8H

2O

158(39+55+64) gms of KMnO4 = 1 mole; So 1.58gms= 0.01mole = 10 mmoles

From the balanced equation, 2 mmoles of KMnO4 requires 10 mmoles of FeSO

4

So 10 mmoles of KMnO4 will need 50 mmoles of FeSO

4

The molarity of FeSO4 = 1/10 M; So the volume = mmoles/M = 50/1/10=500ml.

SAQ 22: “20 volume” solution is more concentrated because 1mL of this solution will evolve 20mL of O2at NTP while 1mL of “10 volumes” solution will evolve 10mL of O2 at NTP.SAQ 23: H2O2 → H2O + 1/2 O2

34g 11.2L(NTP)11.2L of O2 is produced by 34g of H2O2So, 20L of O2 is produced by 60.7g of H2O2Hence 1L or 1000mL of solution contains 60.7g of H2O2So, 100mL solution contains 6.07g. So % strength by volume =6.06%(w/v)

SAQ 24: H2O2 → H2O + 1/2 O2100mL solution contains 12g of H2O21000mL(1L) solution contains 120g of H2O234g of H2O2 liberates 11.2 L of O2 at NTP120g of H2O2 liberates 39.5 L of O2 at NTPHence 1L solution liberates 39.5L of O2 at NTP. Hence its volume strength is “39.5 volumes”.

ANSWER TO PRACTICE QUESTIONSPRACTICE QUESTIONS

1. (i) No. of Moles = Molarity X Volume in litres = 2X1=2moles(ii)0.02X0.2= 0.004mole (iii)17X 3.5= 59.5 moles (iv) 0.1X0.5= 0.05 mole

2. (i)No. of moles= 5 X 0.75= 3.75moles, Mass = 3.75X(29+16+1) = 172.5gms.(ii)No. of moles= 1X2=2moles; Mass = 2X98=196gms.(iii)No. of moles = 0.002 X 0.2= 0.0004mole; Mass= 0.0004 X 36.5= 0.0146gm(iv)No. of moles= 18 X 0.02= 0.36mole, Mass = 0.36 X 98= 35.28gm.

3. (i) Number of mmoles = 1/40 X1000 = 25mmol; volume in ml= 50mlWe know the formula; Molarity = No. of mmoles/volume in ml = 25/50= 0.5M.(ii)No. of mmoles= (2.12/106) X 1000 = 20; volume in ml = 250So Molarity = 20/250 = 0.08M(iii)No. of mmoles= (196/98)X1000= 2000; volume=2500ml;Molarity=2000/2500=0.8MNote that you can solve these problems by unitary method also.

4. mmoles = (24/40) X 1000 = 600; volume =300ml, Molarity=600/300=2M5. mmoles= (84/40)X1000 = 2100; Molarity =3M, Hence Volume =mmoles/Molarity

Volume = 2100/3=700ml = 0.7litre.6. No. of mmole = 0.2 X1000= 200; Molarity=1.71M, volume= 200/1.71=116.95ml7. 50ml of solution contains 0.585gms, So 100ml must contain 1.17gms. So percent strength b yvolume is 1.17%8. 100ml of solution contains 4gms i.e 4/40=0.1 mole of NaOH

So 1000ml of solution must contain 0.1X10=1mole of NaOH, So the molarity=1M

38

Stoichiometry-I

Dr. S. S.Tripathy

9. Let us find the mass of 25ml of solution by using the density data.m/v=1.7gm/ml ⇒ m/25 = 1.7 ⇒ mass=25 X 1.7= 42.5gms42.5gms of the solution contains 9.8gms of H

2SO

4

So 100gms of the solution must contain (9.8/42.5)X100= 23.05gmsSo the percent strength is 23.05%(w/w)

10. 5gms of pure NaCl is present in 100gms solution.3gms of pure NaCl must be present in (100/5)X3= 60gms of solution.

11. The mass of NaOH needed = 19.7gms and mass of water needed=100 - 19.7= 80.3gms.12. Let the required volume = x ml, The mass of x ml = 1.11 X x gm

Since it is 19% by weigth, 100gms of solution contains 19gms of HNO3

So 1.11x gms of solution must contain (19/100) X 1.11x gms of HNO3

According to the question; this mass is 10gm. ⇒ 0.2109x = 10x = 47.4ml.

13. 37.9gms of pure HCl is present in 100gms of solutionSo 5gms of HCl must be present in (100/37.9)X5= 13.19gms of solution

14. Let us find the mass of 100gms of solution. 100/v=1.2 ⇒ v = 83.33ml83.33ml of the solution contains 28gms i.e 28/36.5 i.e 0.767 mole of HCl.1000ml of the solution must contain (0.767/83.33)X 1000 = 9.2 moles.Hence the molarity = 9.2M

15. Let us first find the moles of H2SO

4 from the second data.

Moles = 2 X3= 6moles = 6 X 98gms; (i)Let the required volume = x mls, The mass of x mls of solution = 1.83x gms;If 100gms of solution contains 96gms,So 1.83x gms of solution must contain (96/100) X 1.83x gm (ii)Equating the relations (i) and (ii), we get x= 334.7ml.

16. CaCO3 + H2SO4 → CaSO4 + CO2 + H2O100g 98g25g 24.5g Hence (80/100)x =24.5 ⇒ x = 30.62g

17. MnO2 + 4HCl `→ MnCl2 + Cl2 + 2H2O(55+32)g 4X36.5g 22.4L22.4L of Cl2 is produced by 87g of MnO2So, 2L of Cl2 is produced by 7.7g of MnO222.4L of Cl2 is produced by 4X36.5g of HCl2L of Cl2 is produced by 13.03g of HCl50g of pure acid is present in 100g solution13.03g of pure acid is present in 2606g solutiond=(26.06/V) = 1.2 ⇒ V = 21.7mL

18. MgCO3 + H2SO4 → MgSO4 + CO2 + H2O84g 98g 22.4LMass of H2SO4 solution= 7g; Mass of MgCO3 reacted = 3.5-0.5= 3g84g of MgCO3 reacts with 98g of H2SO4So, 3g of MgCO3 will react with 3.5g of H2SO47g of H2SO4 solution contains 3.5g of pure H2SO4100g of H2SO4 solution contains 50g of pure H2SO4So % strength = 50% by weight(w/w)Find the volume of CO2 formed from 3g of MgCO3 or 3.5g o H2SO4 from the above equation.

19. Na2CO3 + 2 HCl → 2NaCl + CO2 + H2O106g 2X36.5g 2X58.5g100g of Na2CO3 reacts with 73g of HClSo, 50g of Na2CO3 reacts with 34.43g of HClSince there is 50g of HCl, the limiting reactant is Na2CO3106g of Na2CO3 produces 2X58.5g of NaCl

39

Stoichiometry-I

Dr. S. S.Tripathy

50g of Na2CO3 produces 55.18g of NaClMass of solvent in the beginning = 100-50=50g; Mass of pure HCl left = 50-34.43 = 15.57gSo the mass of HCl solution left = 50+15.57 = 65.18g

20. ZnO + 2HNO3 → Zn(NO3)2 + H2O(65+16)g 2X63g81g of ZnO reacts with 126g of HNO3So, 2g of ZnO reacts with 3.11g of HNO365g of pure HNO3 is present in 100g solution3.11g of pure HNO3 is present in 4.78g of solution

21. 2NaCl + H2SO4 → Na2SO4 + 2HCl98g 2X36.5g

Mass of pure HCl = 42f%of 100Kg = 42Kg = 4200g2X36.5g of HCl is produced by 98g of H2SO44200g of HCl is produced by 56383.56g = 56.383Kg of H2SO490 Kg of pure H2SO4 is present in 100Kg solution56.383Kg of pure H2SO4 is present in 62.64Kg solution.

22. 2KMnO4 + 3H2SO4 + 5H2O2 → K2SO4 + 2MnSO4 + 5O2 + 8H2O1mL of H2O2 solution liberates 20mL; Hence 20mL of the solution will liberate 200mL5X22400mL of O2 is liberated by 2X158g of KMnO4So, 200mL of O2 is liberated by 0.564g of KMnO4So, 50mL of KMnO4 solution contains 0.564g of KMnO4.

23. mmoles of HCl = 250 X 0.85 = 212.5V X 0.25 = 212.5 ⇒ V= 850mL (note that the mmoles will remain the same after dilution)Hence volume of water added = 850-250 = 600mL

24. 2H3PO4 + 3Ba(OH)2 → Ba3(PO4)2 + 6 H2Ommoles of Ba(OH)2 = 45 X (0.85/10) = 3.8253 mmoles of Ba(OH)2 reacts with 2 mmoles of H3PO43.825mmoles of Ba(OH)2 reacts with 2.55mmoles of H3PO4So, molarity of H3PO4 solution = (2.55/25) = 0.022M

25. (a)mmoles of BaCl2 = 250X2 = 500So, mmoles of Ba2+ = 500; mmoles of Cl- = 1000[Ba2+] = (500/750) = 0.66M; [Cl-] = (1000/750) = 1.33M(b) 2NaOH + H2SO4 → Na2SO4 + H2Ommoles of H2SO4 = 400; mmoles of NaOH = 5050mmoles of NaOH will react with 25 mmoles of H2SO4Also 50mmoles of NaOH will produce 25 mmoles of Na2SO4Excess mmoles of H2SO4 = 400-25 = 375Hence mmoles of H+ = 2X375; Molarity = 750/500 = 1.5Mmmoles of SO42- = 375+25 Molarity= 400/500 = 0.8Mmmoles of Na+ = 2X25 =50 Molarity = 50/500 = 0.1M(c) mmoles of Fe(NO3)2 = 50 X 0.12 = 6; mmoles of FeCl2 = 100X0.1= 10mmoles of Mg(NO3)2 = 50X0.26 = 13mmoles of Fe2+ = 16; Hence [Fe2+] = 16/200 = 0.08Mmmoles of Cl- = 30; [Cl-] = 30/200 = 0.15Mmmoles of NO3- = 12 + 26 = 38; [NO3-] = 38/200 = 0.19M

26. First balance the equation by partial equation or ON method.K2Cr2O7 + 7H2SO4 + 3 (NH4)2C2O4 → K2SO4 + Cr2(SO4)3 + 3 (NH4)2SO4 + 6CO2 + 7H2Ommoles of (NH4)2C2O4 = 100 X 1/10 = 103 mmoles of (NH

4)

2C

2O

4 reacts with 1 mmole of K2Cr2O7

So, 10mmoles of (NH4)

2C

2O

4 reacts with 3.333 mmoles K2Cr2O7

The molarity of solution = 0.5M, Hence the volume of the solution = 3.333/0.5 = 6.666 mL27. BaO + 2HCl → BaCl2 + H2O

(137+16)g 2X36.5gCaO + 2HCl → CaCl2 + H2O

40

Stoichiometry-I

Dr. S. S.Tripathy

(40+16)g 2X36.5gLet the mass of BaO = x g and hence CaO = (35-x)g153g of BaO reacts with 73g of HClx g of BaO reacts with 0.477x g of HCl56g of CaO reacts with 73g of HCl(35-x) g of CaO reacts with 1.3(35-x)g of HClmmoles of HCl = 150X6 = 900; Hence mass of HCl = 900 X 10-3 X 36.5 = 32.85g0.477 x + 1.3(35-x) = 32.85 ⇒ x = 15.37g(BaO), % of BaO = 43.91%

28. 1mL contains 1.6g solution; Hence 1000mL contains 1600g100g solution contains 67g pure solute; Hence 1600g solution contains 1072g of pure soluteLet us add x mL of water. Hence volume after dilution = (1000 +x)mLMass of the solution = (1600+x) g; Hence the density = (1600+x)/(1000+x) = 1.1⇒ x = 5000mL = 5000g of waterSo the total mass of solution = 1600 + 5000 = 6600 g6600g of solution contains 1072g of solute100g of solution contains 16.24g of solute. Hence percent strength by weight = 16.24%(w/w)

SOLUTION STOICHIOMETRY(VOLUMETRIC ANALYSIS)

MOLARITY AND PERCENT STRENGTH

Concentration or strength of a solution:When a solution is prepared by dissolving a solute in a solvent, it is always necessary to know how much of

solute is present in how much of the solvent i.e their relative proportions in the mixture(solution). If more amount ofsolute is present in a given solution, it is called a concentrated solution and if less amount of solute is present, it iscalled the dilute solution. In order to know the concentration or strength of the solution i.e to know how concentratedor how diluted a solution is, different terms are used. These terms are Molarity, Molality, Normality and %strength. Note that the volume of a solution should not be confused with gaseous volume. Both the types of volumesare measured in the same unit i.e mL or cc and L. Gas laws are applicable to gaseous volume but not to volume ofsolutions.Preparation of a solution of desired concentration

An accurately weighed amount of solute(solid or liquid) is taken in a measuring flask of a fixed volume(capacity).Measuring flasks of various capacities such as 2 litres, 1 litre, 500 ml, 250 ml, 100 ml etc. are availabe in thelaboratory. Each flask has a mark at the upper part of the stem of the flask. The solute is taken in a measuring flaskfirst and slowly distilled water is added from a wash bottle and the solute is allowed to mix with the solvent by gentleshaking of the flask. Water is added in small amounts slowly with constant shaking upto the mark located in the upperpart of the flask. Finally the flask is stopperd and is shaken thoroughly. The volume of solution becomes exactly equalto the capcity of the flask which is written on the flask. Don't add more water so as to cross the mark on the glass. Ifyou do so then the total volume will be more and cannot be properly recorded. Note that the volume of the solution isthe total volume of solute and solvent. In case of solid solutes, particularly ionic solutes like NaCl, Na

2CO

3 etc. the

dissolution involves the formation of free ions which occupy the intermolecular space of the solvent molecules. Hencethe volume change that occurs when such a solute dissolves in a solvent is very small. Hence in such case the volumeof the solution is approximately equal to the volume of the solvent. Of course this is true for dilute solutions preparedfrom solid solutes. However when a liquid solute such as H

2SO

4 is used, the addition of water brings about change in

the volume of solution compared to the volume of solvent. This is because H2SO

4 exists in the liquid state and itself

possesses a lot of free intermolecular volume like the solvent molecules. So the volumes of solute and solventapproximately add to give the volume of the solution.

MOLARITY:Molar Solution:(1M or M solution)

If one mole of solute is present in 1 litre(1000 mL) of solution, it is called a Molar(1M or M) solution.One mole of H

2SO

4(98 gms) present in 1 litre solution is called 1M H

2SO

4.Similarly when 40gms(1mole) of NaOH is

used to prepare 1000 cc of solution, it is Molar(M) or 1M solution of NaOH.____________________________________________________________________________________SAQ 1: How much of solute in gm. will be required per litre of solution to prepare 1M solution in the following cases.

(i)Na2CO

3(ii)HCl (iii)KMnO

4(iv)FeSO

4

SAQ 2: The solutions have the following amounts of the solutes per litre of the solution. What are their molarities.(i)56gms of KOH (ii)106gms of Na

2CO

3(iii)98gms of H

2SO

4

____________________________________________________________________________________Molarity:

The number of moles of solute present in 1000 mL(1 litre) of the solution is called the molarity of a solution.

4gms(0.1 mole) of NaOH present in 1000 cc of the solution is called 0.1M solution or the molarity of thesolution is 0.1M.

2gms(0.05mole) of NaOH present in 1000cc of the solution is called 0.05M solution.80gms(2 moles) of NaOH present in 1000cc of the solution is called 2M solution and so on.

SAQ 3: Find out the molarity of the solutions if the following amounts are dissolved to make 1 litre solution in eachcase.

(i)49gm of H2SO

4(ii)3.65gms of HCl (iii)0.7kg of H

2SO

4

SAQ 4: A certain bottle of H2SO

4 is marked as 2M H

2SO

4 and another bottle of H

2SO

4 is marked 0.5M H

2SO

4.

What do they mean? Which is a diluted solution and which concentrated? How many times the one is more concentratedthan the other.

Let us consider another situation. Supposing you do not have a 1 litre measuring flask with you, in stead youhave a 500 mL flask with you. How can you prepare a 1M solution of H

2SO

4? This is very simple. If 1 litre(1000cc)

of solution requires 98gms of H2SO

4 to make 1M solution, 500ml of solution will need half of it i.e 49 gms. So 49gms

of H2SO

4 taken to make 500 cc solution has a molarity 1M.

SAQ 5: (i)How much of NaOH will be needed to produce a 1M solution by using 250ml measuringflask?

(ii)How much of H2SO

4 will be needed to produce a 1M solution by using 100ml measuring

flask?

Now let us consider still a different situation. Supposing you want to prepare 0.2M solution of NaOH and you have500ml measuring flask, how shall you do? This is again very simple.

If 1000ml solution needs 0.2 mole i.e 0.2×40=8gms of NaOH for making 0.2M solution.500ml solution will need 8/2 = 4gms of NaOH to make the same 0.2M solution.

So in all these cases you have to find out in an unitary method how much of the solute is required for the given volumeof the solution.SAQ 6: Calculate the amounts of solute present in each case of the following solutions.

(i)250ml of 0.5M H2SO

4(ii)2litres of 0.1M NaOH

(iii)100ml of 4M Na2CO

3(iv)500ml of 12M HCl

FORMULA METHOD:

Molarity =number of moles of solute

volume of solution in litre

⇒ Number of moles of solute = Volume of solution in litres × Molarity

Example: How many moles and what mass of each of the following solutes are required to prepare thefollowing solutions.

(i)500 ml of 0.8M H2SO

4(ii)100ml of 2M Na

2CO

3

(iii)2 litres of 0.1 M HCl (iv)250ml of 10M NaOHSolution: Unitary Method

(i) 1litre of 0.8M solution of H2SO

4 contains 0.8 mole of H

2SO

4

So 500ml or 0.8l of this solution must contain 0.5 ×0.8= 0.4 moleFormula method:

volume of the solution =0.5 L and molarity is 0.8M.So the number of moles of H

2SO

4= 0.5×0.8=0.4

Once you find the number of moles of the solute, it is an easy task to find the mass in gm by multiplying thenumber of moles with molecular mass.

mass of 0.4 mole of H2SO

4 = 0.4×98= 39.2gms.

(ii) Moles of Na2CO

3= volume in litre × molarity= 0.1 litre× 2= 0.2.

Mass= 0.2×106 (M.M)= 21.2gms.(iii) Moles of HCl = 2× 0.1 =0.2, so the mass of HCl= 0.2×36.5=7.3gms.(iv) Volume in litre= 0.25 l , Molarity=10, So number of moles of NaOH= 0.25×10=2.5

So the mass of NaOH=2.5×40=100gmsSAQ 7: Find out the mass of the following solutes required to prepare 2 litres of M/20 solution of

(i)H2SO

4(ii)KBr

____________________________________________________________________________________

DETERMINATION OF MOLARITY OF A SOLUTION:Example: Find the molarity of a H

2SO

4 solution, 7gms of which is present in 250 cc solution.

Solution: Unitary method:7gms of H

2SO

4 = 7/98=1/14 mole

250cc of the solution contains 1/14 mole1000cc of the solution contain [(1/14)1000]/250 = 4/14= 0.286 mole

So the molarity of the solution= 0.286MFormula MethodWe know the formula that

volume in litre= Number of moles Molarity

In the above example , we first have to find the number of moles of solute.7gms of H

2SO

4 = 7/98=1/14 mole, volume of solution in litre= 0.25 l

So Molarity = 1/14/0.25= 0.286M.(same as obtained by unitary method)SAQ 8: Find out the molarity of the following solutions.

(i)2gms of NaOH present in 100ml solution(ii)5.3gms of Na

2CO

3 present in 2liters of solution

(iii)0.365gm of HCl present in 250ml of solution

MILLIMOLE(MMOL) METHOD:Note that in stead of solving these problems in terms of moles which often appears as a fraction, it is more

convenient to work in terms of millimoles(mmols). You know that1 mole=103 mmoles=1000 mmoles

volume in litre= Number of moles Molarity OR 1000)(

)(

mLVolume

molesnMolarity solute

Molarity Number of mmoles = volume in ml or ccVolume in litre × molarity = number of molesVolume in ml(cc) × molarity = number of mmoles

If we express volume of the solution in ml(cc) in stead of litre, we get millimoles(mmoles) in stead of moles.

Example: Find out the molarity of 200 ml solution containing 0.7 gm of H2SO

4

Solution: Unitary Method: 200 ml of solution contains 0.7/98 mole of H2SO

4

1000ml of solution contains 0.7/98 X1000/200 =5/140=1/28moleSo the molartiy of the solution is 1/28M

Formula method(1): volume in litre= Number of moles Molarity

The number of moles= 0.7/98= 1/140, volume of the solution=200ml=0.2 litreHence Molarity= 1/140/0.2= 1/28M

Formula method(2): Molarity Number of mmoles = volume in ml or ccThe number of moles= 0.7/98, Hence number of mmoles = 103 × no. of moles= 1000 ×0.7/98=100/14;

volume in ml= 200ml molarity =100

14 X200= 1/28 M

(You can adopt any method you like)SAQ 9: Find out the number of mmoles of the solute present in the following solutions.

(i)100ml of 0.5M NaOH solution (ii)3litres of 0.04M H2SO

4 solution

(iii)500ml of 3M Na2CO

3 solution

SAQ 10: Find out the masses of solute in each case of SAQ 9.

SAQ 11: Find out the number of Na+ ions present in 100ml of M/10 Na2CO

3 solution.

SAQ 12: You are given 500ml of 2M solution of H2SO

4. You are asked to divide this solution in two bottles, one

containing 200ml and the other containing 300ml. What will be molarity of H2SO

4 in the first and second bottles?

PERCENT STRENGTHLike molarity, there is another way to express the concentration of a solution. This is called percent strength.

There are two types of percentages used to find the concentration of a solution. These are(i)Volume percent(w/v) and (ii) weight percent(w/w)

(i) Volume Percent(w/v):The mass of the solute( in gram) present in 100ml of the solution is called percent strength of the

solution by volume. This pecentage is denoted by the symbol w/v i.e weight of the solute(w) and volume of thesolution(v) are considered. Say for example, 2gms of sugar used to prepare 100ml solution is a 2% solution byvolume(w/v). In molarity, we have to find out the number of moles in stead of the mass in grams and moreover thevolume of the solution is taken to be 1000ml to find molarity whereas it is 100ml in percent strength by volume.Example: Find the percent strength of a solution containing 49gms of H

2SO

4 in 10 litres of solution.

Solution: In 10 litres i.e 10,000 ml of solution, the mass of solute=49gmsSo in 100ml of solution, the mass of solute would be (49/10,000) ×100=0.49So percent strength by volume is 0.49% (w/v)

SAQ 13:Find the percent strength of the following solutions by volume.(i)NaOH solution containing 25gms of NaOH in 500ml solution(ii)Na

2CO

3 solution containing 0.005moles of Na

2CO

3 in 1 litre solution

(iv)H2SO

4 solution which has a molarity equal to 0.05M

FINDING PERCENT STRENGTH FROM MOLARITY OF SOLUTION:Example: Find the percent strength by volume of a NaOH solution whose molarity is 0.01M.

A 0.01M solution means, 1000ml of the solution contains 0.01mole of NaOH i.e 0.01×40=0.4gm.If 1000ml of the solution contains 0.4gm of NaOHThen 100 ml solution must contain (0.4/1000)×100= 0.04gm of NaOH.So the percent strength of the solution by volume= 0.04%

SAQ 14: Find the percent strength by volume in respect of the following solutions.(i)3M solution of Na

2CO

3(ii)0.5M solution H

2SO

4(iii)500ml of 2M KOH solution.

(ii) Weight(Mass) Percent (w/w)The mass of the solute in grams that is present in 100gms of the solution is called the weight(mass)

percent of the solution. Note that in this case, for both solute and solvent we consider the mass. Therefore it isdesignated as w/w i.e weight/weight method. Let us take an examle. Supposing we took 2gms of sugar and addedwith it 98gms of water, so as the make the total mass of the solution 100gms. Note that here we will not use anymeasuring flask which measures exact volume of the solution. In this case the percent strength by weight is 2%(w/w).Example: 500gms of H

2SO

4 is mixed with 1kg of water to prepare a solution of H

2SO

4. Find the percent

strength by weight.Solution: Mass of the solute=500gms, Mass of the solvent= 1kg=1000gm.

So the mass of the solution= 500+1000=1500gms.If 1500gms of the solution contains 500gms of H

2SO

4

Then 100gms of the solution must contain (500/1500)×100=33.33gmsSo the percent strength by weight is 33.33%.

SAQ 15: 15gms of pure HNO3 is mixed with 150gms of water to prepare a solution. What is its percent strength by

weight?

FINDING THE WEIGHT PERCENTAGE IF VOLUME OF THE SOLUTION IS KNOWN IN STEADOF MASS BY USIND DENISTY OF THE SOLUTION

For finding the weight percentage from the volume of the solution, we need to know the density of thesolution.Example: Find out the percent strength by weight of a H

2SO

4 which contains 49gms of H

2SO

4 in 500ml

solution. The density of the solution is 1.32gm/ccSolution: We know that density of the solution= (mass of the solution)/(volume of the solution) d=m/v;

Volume of the solution=500ml(data);so mass= d ×V =1.32gm/ml × 500ml= 660gm.Mass of the solution is 660gms and out of which 49gms is the mass of solute.660gms of the solution contains 49gms of H

2SO

4

So 100gms of the solution must contain (49/660) ×100= 7.42gmsSo the percent strength by weight is 7.42%(w/w).

SAQ 16: Find the percent strength by volume and by weight of 18M molar solution of H2SO

4. The specific

gravity of the solution is 1.87.SAQ 17: A bottle of commercial sulphuric acid has a density 1.787gm/ml. It is levlled as 86% by weight. Findthe molarity of the acid and also the percent strength by volume.

STOICHIOMETRIC CALCULATION BASED ON MOLARITY AND STRENGTH OF SOLUTIONIn the previous chapter we have discussed molarity and percentage strength of the solutions. Now we

shall use it in solving stoichiometric problems. Read this example.Example 1: How much zinc will completely react with 200ml of 2M H

2SO

4 solution? Also calculate the

volume of hydrogen gas evolved at 270C and 900mm pressure.(Zn=65)Solution: M.M of H

2SO

4 is 98.

1000ml of a 2M solution contains 2 moles i.e 2×98gms of H2SO

4

So 200ml of such solution must contain (2×98)/(1000)×200= 39.2gms.Alternative (mmole) method:

Number of mmole of H2SO

4 = volume in ml × molarity= 200×2=400

Mass of H2SO

4 = 400 × 10-3 × 98 = 39.2gms.

( Note that when mmole is converted to mol, a factor of 10-3 is multiplied)Now let us write the balanced equation.Zn + H

2SO

4 ZnSO

4 + H

2

65 g 98 g 22.4L98gms of H

2SO

4 completely reacts with 65gms of zinc

39.2gms of H2SO

4 must react with (65/98)×39.2= 26gms of Zn.

98 gms of H2SO

4 produces 22.4 litres of H

2 gas at NTP

So 39.2 gms of H2SO

4 must produce (22.4/98)×39.2= 8.96 litres of O

2 at NTP.

(Note that you can also take the data of zinc found above to calculate the volume of H2)

Now we have to convert the volume from NTP to the given conditions.

760 mm X 8.96 l273A

=900mm X V2

(273+27)A ⇒ V2 = 8.31 litres.

Example 2: What volume of 0.3M H2SO

4 is required to exactly neutralize 200ml of 0.5M NaOH solution?

Solution:Number of mmoles of NaOH= 200 × 0.5 = 100Let us write the balanced chemical equation.2NaOH + H

2SO

4 Na

2SO

4+ H

2O

2 moles 1 mole2 mmoles 1 mmole2 mmoles of NaOH reacts with 1 mmole of H

2SO

4.

So 100 mmoles of NaOH must react with 50 mmoles of H2SO

4.

mmoles = volume in ml × Molarity ⇒ volume = 50/0.3= 166.67 ml.

(You are advised also to solve this problem in terms of mass and check your answer. For that you have to convertthe mmoles of NaOH to its mass and then find out the mass of H

2SO

4 required from the equation. Finally the

volume of the 0.3M H2SO

4 solution required is found out as usual by the unitary method).

SAQ 18: How many ml of 3M HCl solution should be needed to react completely with 16.8gms of NaHCO3?

SAQ 19: What volume of 0.025M HBr solution is required to neutralize 25.0 ml of 0.02M Ba(OH)2 solution?

SAQ 20: Calculate the molarity of a HCl solution if 2.5ml of the solution took 4.5ml of 3M NaOH solution forcomplete neutralization.SAQ 21: What volume of M/10 solution FeSO

4 solution will be required to completely react with 1.58gms

of KMnO4 in dilute H

2SO

4 medium?

PROBLEMS BASED ON VOLUME STRENGTH OF HYDROGEN PEROXIDE SOLUTIONWhen we buy hydrogen peroxide from the market, we get them usually in dark coloured bottles which

are labelled as “10 Volumes” or “20 Volumes” or in general “x volumes”. In fact these are not pure H2O

2, rather

are aqueous solutions.“x Volumes” H

2O

2 solution means if 1 volume of such H

2O

2 solution(liquid) at NTP is decomposed,

‘x’ volumes of O2 gas are evolved. Note that here the two volumes are of different types. One is volume of

the solution(1 volume i.e 1L or 1 mL) and the other is gaseous volumes(i.e ‘x’ L or ‘x’ mL).“10 Volumes” of H

2O

2 means, 1 mL of such a solution on decomposition will produce 10 mL of O

2 gas

at NTP. This is called volume strength of H2O

2. Volume strength can be converted to percent strength by volume

or vice versa. Look to the following examples.Example 1: Calculate the percent strength by volume of a H

2O

2 solution if its volume strength is “22.4 Volumes”.

Solution: 2 H2O

2 → 2 H

2O + O

2

2(2+32)g 22.4L(NTP)“22.4 Volumes” solution means 1 mL of such a solution liberates 22.4mL of O

2 at NTP.

1mL of the solution liberates 22.4mL of O2 at NTP

So, 100mL of the solution will liberated 2240mL of O2 at NTP

From the balanced equation,22400mL of O

2 is liberated by 2×34g of H

2O

2

So, 2240mL of O2 is liberated by 6.8g of H

2O

2

Hence 100mL of the solution contains 6.8g of H2O

2. So the percent strength is 6.8%(w/v).

Example 2: 20mL of a solution of H2O

2 labelled “15 volumes” just decolorized 100mL of KMnO

4 acidified with

dil. H2SO

4. Calculate the mass of KMnO

4 in the given solution.(K=39, Mn=55)

Solution: “15 volumes” means 1mL of the solution liberates 15 mL of O2 gas at NTP.

So 20mL of this solution will liberate 20×15 = 300mL of O2 at NTP.

2KMnO4 + 3H

2SO

4 + 5 H

2O

2 → K

2SO

4 + 2MnSO

4 + 5O

2 + 8H

2O

2×158 5×22400mL(NTP)5×22400mL of O

2 is produced by 2×158g of KMnO

4

So, 300mL of O2 is produced by 0.846g of KMnO

4

Hence the mass of KMnO4 required is 0.846g.

SAQ 22: Which is more concentrated H2O

2 solution, “10 volumes” or “20 volumes” and why?

SAQ 23: Calculate the % strength by volume of a H2O

2 solution labelled as “20 volumes”

SAQ 24: Find the volume strength on the basis of available oxygen of a H2O

2 solution which is 12% by volume.

SAQ 25: 200 mL of 0.1M BaCl2 was mixed with 100 mL of 0.05M Na

3PO

4. What mass of barium phosphate

will be formed ? (Ba = 137)

PRACTICE QUESTIONS1. Find the number of moles of solute present in each case

(i)1 litre of 2M H2SO

4(ii)200ml of 0.02M HNO

3(iii)3.5litres of 17M HCl

(iv)500ml of 0.1M Na2CO

3

2. Find the mass of the solute required to prepare the following solutions.(i)750ml of 5M KOH (ii)2 litres of 1M H

2SO

4(iii)200ml of 0.002M HCl

(iv)20ml of 18M H2SO

4

3. Find the Molarity of the following solutions.(i)1gm of NaOH present in 50ml solution. (ii)2.12gms of Na

2CO

3 present in 250ml solution

(iii)196gms of H2SO

4 present in 2.5litres solution

4. What is the molarity of NaOH in a solution which contains 24.0gms of NaOH dissolved in 300ml solution?5. What volume of 3M NaOH solution can be prepared with 84.0g NaOH?6. What volume of 1.71M NaCl solution contains 0.2 mol NaCl?7. 0.585gm of NaCl is present in 50ml solution. What is the strength of the solution in percent by volume?8. What is the molarity of a 4% NaOH solution by volume?9. 9.8gms of H

2SO

4 is present in 25ml solution. The density of the solution is 1.7gm/ml. What is the percent strength

by weight.10. How many grams of a 5% NaCl solution by weight are necessary to yield 3.2g of NaCl?11. To prepare 100gm of 19.7% by weight solution of NaOH, how many g of each of NaOH and H

2O are

needed?12. What volume of dilute nitric acid of density 1.11g/mL which is 19% HNO

3 by weight contains 10g HNO

3?

13. How many gm. of conc. hydrochloric acid which is 37.9 % by weight will contain 5.0gms of HCl?14. An aqueous solution of HCl contains 28% HCl by weight and has a density of 1.2gm/ml .Find the molarity of

the solution.15. What volume of 96% H

2SO

4 solution(density 1.83gm/ml)is required to prepare 2.0L of 3M H

2SO

4 solution?

16. Calculate the mass of 80% sulphuric acid by weight required to completely react with 25g of CaCO3.(30.62g)

17. Calculate the (i)mass of MnO2 and (ii)the volume of hydrochloric acid having density 1.2g/mL and containing

50% HCl by weight needed to produce 2L of chlorine gas at NTP? (7.76g, 21.7mL)18. 3.5g of MgCO

3 were added to double its mass of H

2SO

4 solution. After complete reaction, 0.5g of MgCO

3

remained unreacted. Calculate percent strength of H2SO

4 by weight and volume of CO

2 formed at 270C and

760mm pressuire. (50%)19. Calculate the mass of NaCl formed when 50g of Na

2CO

3 reacted with 100g of 50% HCl by weight. Also

calculate the mass of excess reagent which remained unreacted. (55.18g, 65.57g of HCl)20. A solution of nitric acid contains 65% HNO

3 by weight. What mass of this acid will be necesary to dissolve

2g of zinc oxide. (Zn =65) (4.78g)21. Most commercial HCl is prepared by heating NaCl with conc. H

2SO

4. How much of sulphuric acid containing

90% H2SO

4 by weight is needed for the production of 100Kg of conc. HCl containing 42% HCl by weight.

(62.64Kg)22. 10mL of a solution of H

2O

2 labelled “20 volumes” just decolorizes 50mL of KMnO

4 solution acidified with dil.

H2SO

4. Calculate the amount of KMnO

4 present in the solution. (0.564g)

23. How many mL of water must be added to 250mL of 0.85M HCl to dilute the solution to 0.25M?(600mL)24. Calculate the molarity of H

3PO

4 solution if 25mL of this solution is completely neutralised by 45mL of

0.85 M/10 Ba(OH)2 solution. (0.022M)

25. Calculate the molar concentration of each of the ionic species in solution.(a)250mL of 2M BaCl

2 is diluted to 750mL

(b)200mL of 2M H2SO

4 is mixed with 100mL of 0.5M NaOH and the mixture diluted to 500mL

(c)50mL of 0.12M Fe(NO3)

2 + 100mL of 0.1M FeCl

2 + 50mL of 0.26M Mg(NO

3)

2

26. What volume of 0.5M K2Cr

2O

7 solution will react completely with 100mL of M/10 (NH

4)

2C

2O

4 solution in

acidic medium. (Cr =52, K=39) (6.66mL)27. Calculate the pecent of BaO in 35g of a mixutre of CaO and BaO which just reacts with 150mL of 6M

HCl.(Ba=137) (43.91%)28. A solution of density 1.6g/mL is 67% by weight. What will be % strength by weight of solution if it is diluted

to have density 1.1 g/mL? (16.24%)(Hints : V = 62.5 mL (100g) Let x mL water is added. (100+x)/(62.5+x) = 1.1 ; x = 312.5 mL, w/w% = (67/412.5)X100 = 16.24%)

RESPONSE TO SAQs(Solutions: Molarity)

SAQ 1: In each case we require 1 mole or gm molecular mass of the solute(i) M.M of Na

2CO

3 =106, so we need 106gms per litre

(ii) M.M of HCl = 36.5, and so we need 36.5 gms per litre(iii) M.M of FeSO

4 = 56+32+64=152, so we need 152gms per litre

SAQ 2: (i) 1M(one Molar or Molar) since 1 mole of KOH=39+16+1=56(M.M)(ii) 1M since M.M of Na

2CO

3 = 46+12+48=106(1 mole)

(iii) 1M since M.M of H2SO

4 =98(1 mole)

SAQ 3: (i) 98 gms of H2SO

4=1 mole, so 49 gms of H

2SO

4 = 0.5 mole.

Since 0.5mole of it is present in 1 litre solution, the molarity = 0.5M(ii) 36.5gms of HCl= 1 mole, so 3.65gms of HCl= 1/10=0.1 mole

Since 0.1 mole of HCl is present in 1 litre solution, the molarity=0.1M(iii) 98gms of H

2SO

4=1 mole, so 0.7kg i.e 700gms of H

2SO

4=700/98=7.143moles

Since 7.143moles of H2SO4 is present in 1 litre solution, the molarity=7.143MSAQ 4: The bottle marked 2M H

2SO

4 means, there would be 2 moles(2X98=196gms) of H

2SO

4 per litre of the

solution. However in the bottle marked 0.5M, there would be 0.5mole(0.5X98=49gms)of H2SO

4 per litre of

solution. So the second bottle of H2SO

4 is more diluted and the first bottle is more concentrated. The first solution

is 4 time more concentrated(49X4=196) than the second solution.SAQ 5:(i) 1000ml of solution needs 40gms(1mole) of NaOH to prepare 1M solution

So 250ml of solution must need (40/1000)X250=10gms to prepare the same 1M solution(ii) 1000ml of solution need 98gms(1mole) of H2SO4 to prepare 1M solutionSo 100ml of solution will need (98/1000)X100=9.8gms to prepare the same 1M solution.

SAQ 6: (i) 1000ml of 0.5M solution contains 0.5mole i.e 0.5X98=49gms of H2SO

4

So 250ml of 0.5M solution must contain (49/1000) X 250= 12.5gms.(ii) 1000ml of 0.1M NaOH solution contains 0.1mole i.e 0.1X40=4gms of NaOH

So 2litres i.e 2000ml of 0.1M solution must contain (4/1000)X2000= 8gms.(iii) 1000ml of 4M solution of Na

2CO

3 contains 4 moles i.e 4X106=424gms

So 100ml of 4M solution must contain (424/1000)X100=42.4gms.(iv) 1000ml of 12M solution of HCl contains 12moles i.e 12X36.5=438gms of HCl

So 500ml of 12M solution must contain 438/2=219gms.SAQ 7: Here we shall have to find the mass of the solute. Let us first find the number of moles and then mass byusing the formula method discussed in the examples in the text.

(i) Number of moles= volume in litre X molarity=2X(1/20)=1/10=0.1Mass of H

2SO

4= 0.1X98=9.8gms.

(ii) Number of moles of KBr= 2X (1/20) = 1/10=0.1 [same as (i)]Mass of KBr = 0.1X(39+80)=0.1X 119 = 11.9gms.

SAQ 8:(i) M.M of NaOH=40, 2gms of NaOH=2/40=0.05 mole,volume = 100ml=0.1 l, So Molarity= 0.05/0.1= 0.5M.

(ii) Number of moles of Na2CO

3 = 5.3/106=0.05; volume=2litres

So Molarity = 0.05/2=0.025M.(iii) Number of mole of HCl=0.365/36.5= 0.01; volume=250ml=0.25 litre

So Molarity= 0.01/0.25=0.04M.SAQ 9: We know that No. of mmoles = Molarity X Volume in ml(cc)

(i)Number of mmoles= 100X0.5=50(ii)3000X0.04= 120 (iii)500X3=1500

You found that the number of mmoles is not a fractional value(more than 1).SAQ 10: (i)50mmoles of NaOH= 5X 10-3 mole = 0.005mole= 0.005X40=0.2gm.

(ii)120mmole of H2SO

4 = 120X10-3mole=0.12 mole = 0.12X98=11.76 gms.

(iii)1500mmoles of Na2CO

3= 1500X10-3=1.5mole = 1.5X106= 159gms.

SAQ 11: The no. of mmoles of Na2CO

3 present in the solution=100 X1/10 =10

10 mmoles= 10 X10-3 moles = 0.01mole1 mole of Na

2CO

3 contains 2 moles of Na+ ions

So the number Na+ ions present in 2 moles = 2X6.023X1023 =1.2046X1024 ions0.01 mole of Na

2CO

3 will contain 0.01 X1.2046X1024 = 1.2046X1022 ions of Na+.

SAQ 12: Since the stock(original) solution is same, the molarity will remain same. Suppose you collect 500ml ofsea water and distribute the same in several containers. Do you think that the salinity of the sea water will bedifferent in these different vessels? That means the vessel containing less solution will taste less salty and thevessel containing more solution will taste more salty? The answer is NO. They will all taste same, because theconcentration of the original(stock) solution is same. Remember that the concentration of a solution gives thequality of the solution i.e how much of solute present per a fixed volume of solution(say 1 litre). Once a solutionof a particular concentration is prepared, its concentration remains the same even if this stock solution is dividedinto any number of parts.SAQ 13: (i) 500ml of the solution contains 25gms of NaOH

So 100ml of the solution must contain (25/500)X100=5gms of NaOHSo pecent strength by volume = 5%.

(ii) M.M of Na2CO

3=106, 1 mole= 106gms, so 0.005 moles = 106X0.005= 0.53gm

1 litre i.e 1000ml of solution contains 0.53gm of Na2CO3So 100ml of the solution must contain (0.53/1000)X100= 0.053gmSo percent strength by volume=0.053%

SAQ 14: (i) M.M of Na2CO

3=106; 3M solution means

1000ml of solution contains 3moles i.e 3X106=318gms of Na2CO

3

So100ml of solution must contain (318/1000)X100=31.8gms.So percent strength by volume= 31.8%

(ii) 0.5M solution means1000ml of the solution contains 0.5mole i.e 0.5X98= 49gms of H

2SO

4

So 100 ml of the solution must contain 49/10=4.9gmsSo the percent strength by volume is 4.9%

(iii) Here 500ml solution carries no relevance and it will not be used. We have to consideronly the molarity of the solution. 2M solution means

1000ml of KBr solution contains 2 moles i.e 2X(39+80)= 2X56=238gmsSo 100ml of the solution must contain 238/10= 23.8gms of KBrSo the percent strength is 23.8% by volume.

SAQ 15: The mass of the solution = 15 + 150 = 165gms.165gms of solution contains 15 gms of HNO3,100 gms of solution must contain (15/165) X 100 = 9.09 gmsSo its percent strength is 9.09%(w/w)

SAQ 16: Volume %:18M solution means 1000ml of solution contains 18moles=18X98=1764gms of H

2SO

4.

1000ml of the solution contains 1764gms of H2SO

4.

So 100ml of the solution must contain 1764/10=176.4gmsSo percent strength by volume=176.4%

Weight %:density= m/v; 1.87gm/ml= m/1000ml,⇒ mass= 1000 X 1.87=1870 gms, So mass of 1000ml of solution is 1870gms.1870gms of the solution contains 18mole i.e 18X98=1764gms of H

2SO

4

100gms of the solution must contain (1764/1870)X100=94.33gmsSo percent strength by weight is 94.33%.

SAQ 17: Let us consider 100gm of solution. d=mass/volume,So volume of 100gm solution = 100/1.787= 55.959ml .As it is 86%solution by weight.So the mass of solute(H

2SO

4)in present in 100gms of solution = 86gms.

Molarity:55.959ml of solution contains 86gms or 86/98moles i.e 0.8775 mole of H

2SO

4

1000ml of the solution must contain 0.8775X (1000/55.959) = 15.68molesHence the molarity of the solution= 15.68M

Percent strength by Volume:55.959ml of solution contains 86gms of H2SO4So 100ml of solution must contain (86/55.959)X100= 153.68%(w/v)

SAQ 18:NaHCO

3 + HCl → NaCl + CO

2 + H

2O

1 mole 1 mole84 gms of NaHCO

3 = 1 mole ⇒ 16.8gms = 0.2mole = 0.2 X 1000=200mmole

According the balanced equation, 1 mmole of NaHCO3 will require 1mmole of HCl

So 200mmole of NaHCO3 will require 200 mmole of HCl according to the balanced

equation.We know that mmoles = volume in ml X Molarity ⇒ Volume = 200/3=66.67ml.

SAQ 19:Ba(OH)

2 + 2HBr → BaBr

2 + 2H

2O

No. of mmoles of Ba(OH)2 = 25 X 0.02= 0.5

According the balanced equation, 1 mmole of Ba(OH)2 will need 2mmoles of HBr.

So 0.5mmole of Ba(OH)2 will need 2 X 0.5= 1mmole of HBr

So the volume of HBr solution = mmole/Molarity = 1/0.025 = 40ml.SAQ 20: NaOH + HCl → NaCl + H

2O

No. of mmoles of NaOH = 4.5 X 3= 13.5; According to the balanced equation, the number ofmmoles of HCl required = 13.5; So the molarity = mmoles/ml = 13.5/2.5= 5.4MSAQ 21:

2KMnO4 + 8H

2SO

4 + 10FeSO

4 → K

2SO

4 + 2MnSO

4 + 5Fe

2(SO

4)

3 + 8H

2O

158(39+55+64) gms of KMnO4 = 1 mole; So 1.58gms= 0.01mole = 10 mmoles

From the balanced equation, 2 mmoles of KMnO4 requires 10 mmoles of FeSO

4

So 10 mmoles of KMnO4 will need 50 mmoles of FeSO

4

The molarity of FeSO4 = 1/10 M; So the volume = mmoles/M = 50/1/10=500ml.

SAQ 22: “20 volume” solution is more concentrated because 1mL of this solution will evolve 20mL of O2

at NTP while 1mL of “10 volumes” solution will evolve 10mL of O2 at NTP.

SAQ 23: H2O

2→ H

2O + 1/2 O

2

34g 11.2L(NTP)11.2L of O

2 is produced by 34g of H

2O

2

So, 20L of O2 is produced by 60.7g of H

2O

2

Hence 1L or 1000mL of solution contains 60.7g of H2O

2

So, 100mL solution contains 6.07g. So % strength by volume =6.06%(w/v)SAQ 24: H

2O

2→ H

2O + 1/2 O

2

100mL solution contains 12g of H2O

2

1000mL(1L) solution contains 120g of H2O

2

34g of H2O

2 liberates 11.2 L of O

2 at NTP

120g of H2O

2 liberates 39.5 L of O

2 at NTP

Hence 1L solution liberates 39.5L of O2 at NTP. Hence its volume strength is “39.5 volumes”.

ANSWERS PRACTICE QUESTIONS

1. (i) No. of Moles = Molarity X Volume in litres = 2X1=2moles(ii)0.02X0.2= 0.004mole (iii)17X 3.5= 59.5 moles (iv) 0.1X0.5= 0.05 mole

2. (i)No. of moles= 5 X 0.75= 3.75moles, Mass = 3.75X(29+16+1) = 172.5gms.(ii)No. of moles= 1X2=2moles; Mass = 2X98=196gms.(iii)No. of moles = 0.002 X 0.2= 0.0004mole; Mass= 0.0004 X 36.5= 0.0146gm(iv)No. of moles= 18 X 0.02= 0.36mole, Mass = 0.36 X 98= 35.28gm.

3. (i) Number of mmoles = 1/40 X1000 = 25mmol; volume in ml= 50mlWe know the formula; Molarity = No. of mmoles/volume in ml = 25/50= 0.5M.(ii)No. of mmoles= (2.12/106) X 1000 = 20; volume in ml = 250So Molarity = 20/250 = 0.08M(iii)No. of mmoles= (196/98)X1000= 2000; volume=2500ml;Molarity=2000/2500=0.8MNote that you can solve these problems by unitary method also.

4. mmoles = (24/40) X 1000 = 600; volume =300ml, Molarity=600/300=2M5. mmoles= (84/40)X1000 = 2100; Molarity =3M, Hence Volume =mmoles/Molarity

Volume = 2100/3=700ml = 0.7litre.6. No. of mmole = 0.2 X1000= 200; Molarity=1.71M, volume= 200/1.71=116.95ml7. 50ml of solution contains 0.585gms, So 100ml must contain 1.17gms. So percent strength by volume is

1.17%8. 100ml of solution contains 4gms i.e 4/40=0.1 mole of NaOH

So 1000ml of solution must contain 0.1X10=1mole of NaOH, So the molarity=1M9. Let us find the mass of 25ml of solution by using the density data.

m/v=1.7gm/ml ⇒ m/25 = 1.7 ⇒ mass=25 X 1.7= 42.5gms42.5gms of the solution contains 9.8gms of H

2SO

4

So 100gms of the solution must contain (9.8/42.5)X100= 23.05gmsSo the percent strength is 23.05%(w/w)

10. 5gms of pure NaCl is present in 100gms solution.3gms of pure NaCl must be present in (100/5)X3= 60gms of solution.

11. The mass of NaOH needed = 19.7gms and mass of water needed=100 - 19.7= 80.3gms.12. Let the required volume = x ml, The mass of x ml = 1.11 X x gm

Since it is 19% by weigth, 100gms of solution contains 19gms of HNO3

So 1.11x gms of solution must contain (19/100) X 1.11x gms of HNO3

According to the question; this mass is 10gm. ⇒ 0.2109x = 10x = 47.4ml.

13. 37.9gms of pure HCl is present in 100gms of solutionSo 5gms of HCl must be present in (100/37.9)X5= 13.19gms of solution

14. Let us find the mass of 100gms of solution. 100/v=1.2 ⇒ v = 83.33ml83.33ml of the solution contains 28gms i.e 28/36.5 i.e 0.767 mole of HCl.1000ml of the solution must contain (0.767/83.33)X 1000 = 9.2 moles.Hence the molarity = 9.2M

15. Let us first find the moles of H2SO

4 from the second data.

Moles = 2 X3= 6moles = 6 X 98gms; (i)Let the required volume = x mls, The mass of x mls of solution = 1.83x gms;If 100gms of solution contains 96gms,So 1.83x gms of solution must contain (96/100) X 1.83x gm (ii)Equating the relations (i) and (ii), we get x= 334.7ml.

16. CaCO3 + H

2SO

4 → CaSO

4 + CO

2 + H

2O

100g 98g25g 24.5g Hence (80/100)x =24.5 ⇒ x = 30.62g

17. MnO2 + 4HCl `→ MnCl

2 + Cl

2 + 2H

2O

(55+32)g 4X36.5g 22.4L22.4L of Cl

2 is produced by 87g of MnO

2

So, 2L of Cl2 is produced by 7.7g of MnO

2

22.4L of Cl2 is produced by 4X36.5g of HCl

2L of Cl2 is produced by 13.03g of HCl

50g of pure acid is present in 100g solution13.03g of pure acid is present in 2606g solutiond=(26.06/V) = 1.2 ⇒ V = 21.7mL

18. MgCO3 + H

2SO

4 → MgSO

4 + CO

2 + H

2O

84g 98g 22.4LMass of H

2SO

4 solution= 7g; Mass of MgCO

3 reacted = 3.5-0.5= 3g

84g of MgCO3 reacts with 98g of H

2SO

4

So, 3g of MgCO3 will react with 3.5g of H

2SO

4

7g of H2SO

4 solution contains 3.5g of pure H

2SO

4

100g of H2SO

4 solution contains 50g of pure H

2SO

4

So % strength = 50% by weight(w/w)Find the volume of CO

2 formed from 3g of MgCO

3 or 3.5g o H

2SO

4 from the above equation.

19. Na2CO

3 + 2 HCl → 2NaCl + CO

2 + H

2O

106g 2X36.5g 2X58.5g100g of Na

2CO

3 reacts with 73g of HCl

So, 50g of Na2CO

3 reacts with 34.43g of HCl

Since there is 50g of HCl, the limiting reactant is Na2CO

3

106g of Na2CO

3 produces 2X58.5g of NaCl

50g of Na2CO

3 produces 55.18g of NaCl

Mass of solvent in the beginning = 100-50=50g; Mass of pure HCl left = 50-34.43 = 15.57gSo the mass of HCl solution left = 50+15.57 = 65.18g

20. ZnO + 2HNO3 → Zn(NO

3)

2 + H

2O

(65+16)g 2X63g81g of ZnO reacts with 126g of HNO

3

So, 2g of ZnO reacts with 3.11g of HNO3

65g of pure HNO3 is present in 100g solution

3.11g of pure HNO3 is present in 4.78g of solution21. 2NaCl + H

2SO

4 → Na

2SO

4 + 2HCl

98g 2X36.5gMass of pure HCl = 42f%of 100Kg = 42Kg = 4200g2X36.5g of HCl is produced by 98g of H

2SO

4

4200g of HCl is produced by 56383.56g = 56.383Kg of H2SO

4

90 Kg of pure H2SO

4 is present in 100Kg solution

56.383Kg of pure H2SO

4 is present in 62.64Kg solution.

22. 2KMnO4 + 3H

2SO

4 + 5H

2O

2 → K

2SO

4 + 2MnSO

4 + 5O

2 + 8H

2O

1mL of H2O

2 solution liberates 20mL; Hence 20mL of the solution will liberate 200mL

5X22400mL of O2 is liberated by 2X158g of KMnO

4

So, 200mL of O2 is liberated by 0.564g of KMnO

4

So, 50mL of KMnO4 solution contains 0.564g of KMnO

4.

23. mmoles of HCl = 250 X 0.85 = 212.5V X 0.25 = 212.5 ⇒ V= 850mL (note that the mmoles will remain the same after dilution)Hence volume of water added = 850-250 = 600mL

24. 2H3PO

4 + 3Ba(OH)

2 → Ba

3(PO

4)

2 + 6 H

2O

mmoles of Ba(OH)2 = 45 X (0.85/10) = 3.825

3 mmoles of Ba(OH)2 reacts with 2 mmoles of H

3PO

4

3.825mmoles of Ba(OH)2 reacts with 2.55mmoles of H

3PO

4

So, molarity of H3PO

4 solution = (2.55/25) = 0.022M

25. (a)mmoles of BaCl2 = 250X2 = 500

So, mmoles of Ba2+ = 500; mmoles of Cl–- = 1000[Ba2+] = (500/750) = 0.66M; [Cl–] = (1000/750) = 1.33M(b) 2NaOH + H

2SO

4 → Na

2SO

4 + H

2O

mmoles of H2SO

4 = 400; mmoles of NaOH = 50

50mmoles of NaOH will react with 25 mmoles of H2SO

4

Also 50mmoles of NaOH will produce 25 mmoles of Na2SO

4

Excess mmoles of H2SO

4 = 400-25 = 375

Hence mmoles of H+ = 2X375; Molarity = 750/500 = 1.5Mmmoles of SO

42- = 375+25 Molarity= 400/500 = 0.8M

mmoles of Na+ = 2X25 =50 Molarity = 50/500 = 0.1M(c) mmoles of Fe(NO

3)

2 = 50 X 0.12 = 6; mmoles of FeCl

2 = 100X0.1= 10

mmoles of Mg(NO3)

2 = 50X0.26 = 13

mmoles of Fe2+ = 16; Hence [Fe2+] = 16/200 = 0.08Mmmoles of Cl– = 30; [Cl–] = 30/200 = 0.15Mmmoles of NO

3– = 12 + 26 = 38; [NO

3–] = 38/200 = 0.19M

26. First balance the equation by partial equation or ON method.K

2Cr

2O

7 + 7H

2SO

4 + 3 (NH

4)

2C

2O

4 → K

2SO

4 + Cr

2(SO

4)

3 + 3 (NH

4)

2SO

4 + 6CO

2 + 7H

2O

mmoles of (NH4)

2C

2O

4 = 100 X 1/10 = 10

3 mmoles of (NH4)

2C

2O

4 reacts with 1 mmole of K

2Cr

2O

7

So, 10mmoles of (NH4)

2C

2O

4 reacts with 3.333 mmoles K

2Cr

2O

7

The molarity of solution = 0.5M, Hence the volume of the solution = 3.333/0.5 = 6.666 mL27. BaO + 2HCl → BaCl

2 + H

2O

(137+16)g 2X36.5gCaO + 2HCl → CaCl

2 + H

2O

(40+16)g 2X36.5gLet the mass of BaO = x g and hence CaO = (35-x)g153g of BaO reacts with 73g of HClx g of BaO reacts with 0.477x g of HCl56g of CaO reacts with 73g of HCl(35-x) g of CaO reacts with 1.3(35-x)g of HClmmoles of HCl = 150X6 = 900; Hence mass of HCl = 900 X 10-3 X 36.5 = 32.85g0.477 x + 1.3(35-x) = 32.85 ⇒ x = 15.37g(BaO), % of BaO = 43.91%

28. 1mL contains 1.6g solution; Hence 1000mL contains 1600g100g solution contains 67g pure solute; Hence 1600g solution contains 1072g of pure soluteLet us add x mL of water. Hence volume after dilution = (1000 +x)mLMass of the solution = (1600+x) g; Hence the density = (1600+x)/(1000+x) = 1.1⇒ x = 5000mL = 5000g of waterSo the total mass of solution = 1600 + 5000 = 6600 g6600g of solution contains 1072g of solute100g of solution contains 16.24g of solute. Hence percent strength by weight = 16.24%(w/w)

EQUIVALENT MASS OF COMPOUNDSWe know that all chemical reactions take place in equivalents. One g. equivalent mass of any reactant(element

or compound) reacts completely with one g. equivalent mass of the other reactant to form one g. equivalent massof the product. This principle is applicable both for elements and compounds as well.

Let us know now how the equivalent masses for various types of compounds such as acids, bases andsalts, oxidising and reducing agents are determined.Equivalent Mass(EM) of Acids:

=EM of an acidMolecular Mass

basicity (where basicity = number of replaceable H atoms)

Example: EM(HCl) = 36.5/1 = 36.5 ( basicity of HCl =1)EM(H

2SO

4) = 98/2 = 49 (basicity of H

2SO

4 =2)

EM(H3PO

4) = 98/3=32.66 (basicity is 3)

Alternative defintion: Equivalent mass of an acid is the mass of an acid in gms that will furnish Avogadro’snumber of H+ ions i.e 1 mole of hydrogen ions.

For example, one mole of H2SO

4 having mass 98g contains 2X N

A number of H+ ions

So the mass of H2SO

4 furnishing N

A H+ ions is 49(EM)

Note that equivalent mass of any substance is determined for a particular reaction. If no reaction is mentioned,then it is to be presumed that the acid has undergone complete neutralisation.Example: Find the equivalent mass of the underlined acids in the following reactions.

(a) NaOH + H2SO

4 → NaHSO

4 + H

2O

(b) 2NaOH + H3PO

4 → Na

2HPO

4 + 2H

2O

Solution: (a) EM(H2SO

4) = 98/1 = 98 (here only 1 H atom has been replaced. So basicity is 1)

(b) EM(H3PO

4) = 98/2 = 49 (here only 2H atoms have been replaced. So bascity is 2)

SAQ 1: Find out the equivalent masses of the following acids. HNO3, CH

3COOH, H

2CO

3, H

3PO

3, H

3PO

2, H2C2O4,

CO2, SO2Hint: Note that for an acidic oxide, the basicity is the number of H+ ions that one molecule of the oxide produces byreacting with water.

SO2 + H2O → H2SO3 (so the basicity is 2)Equivalent mass of a base:

(a)Hydroxide base:

=EM of an base Molecular Massacidity (where acidity is the number of OH- )

Example : EM(NaOH) = 40/1=40 (acidity= 1) EM[Ca(OH)2] = 74/2 = 37 (acidity =2)Alternative definition: Equivalent mass of a base can be defined as teh mass of the base in grams that furnishesAvogadro’s number of OH- or 1 mole of OH- in the reaction.

(b)Oxide base:

=EM of an base Molecular Mass

Number of OH- produced by one molelcue of oxiEM(Na2O) = MM/2 (since one molecule of Na2O furnishes 2 OH- by reacting with H2O)Na2O + H2O → 2 NaOH; EM(CaO ) = MM/2 = 56/2=28 [CaO + H2O → Ca(OH)2]EM(Al2O3) = MM/6 [Al2O3 + 3H2O → 2Al(OH)3]

SAQ 2: Find out the equivalent masses of the following bases. Refer periodic table for atomic masses.KOH, Mg(OH)

2, MgO, Li

2O, BaO, NH

4OH, NH

3

Equivalent mass of salt:

EM of salt =molecular mass

total +ve or -ve valency or charge

Example:(2 X 23 + 32 + 4X16)

1X2EM( Na2SO4 ) = = 71

There are 2 Na+ radicals and eeach has a valency or charge equal to 1, so the total charge of basic radical is1 × 2 = 2. We could also consider total charge or valency of the acid radical. Here sulphate ion has valency 2 andthere is one sulphate ion. So total valency or charge of acid radical is also 2.

EM(CaCO3) = 100/2= 50; EM(NaCl) = 58.5/1=58.5

EM[Al2(SO

4)

3] = MM/6 = (2 × 27 +3 × 32 +12 ×16)/6 = 57 (here the total +ve or -ve valency is 3 × 2=6)

Note that the formulae of the determination of equivalent masses of acids, bases and salts that we discussedabove hold good only when these substances are involved in metathesis(non-redox) reactions wherethere is no change in oxidation number of atoms. For redox reaction, the above formulae do not holdgood.SAQ 3: Find the equivalent masses of the following: salts involved in metathesis reactions.Na

2CO

3, CaCl

2, AlBr

3, BaSO

4, Na

3PO

4, FeSO

4

Equivalent mass of oxidising agent(OA) and reducing agent(RA):

EM of OA or RA = molecular masschange in ON per molecule or ion =

molecular massno. of electrons lost or gained per molecule or ion

(Note that this change is also called VALENCY FACTOR)Example: EM of KMnO

4(in acidic medium):

KMnO4 + H

2SO

4 + RA → K

2SO

4 + MnSO

4 + ........... + H

2O

We know that KMnO4 in the acidic medium(H

2SO

4) is reduced to MnSO

4. The ON of Mn changes

from +7 to +2. So the change of ON is 5 per OA molecule since one KMnO4 molecule contains one Mn atom.

We can also find that in the balanced ion-electron equation, the number of electrons gained by one MnO4

– ion is5. So EM is molecular mass divided by 5.

MnO4– + 8H+ + 5e → Mn2+ + 4H

2O

EM(KMnO4) = (39+55+64)/5 = 158/5 = 31.6

EM of K2Cr

2O

7:

K2Cr

2O

7 + H

2SO

4 + RA → K2SO

4 + Cr

2(SO

4)

3 + ......... + H

2O

The ON of Cr changes from +6 to +3, hence the change of ON per Cr atom is 3. But there are 2 Cratoms in the OA molecule.

Henc the total change of ON per molecule = 2 × 3 = 6EM(K

2Cr

2O

7) = (2×39+2×52+112)/6 = 294/6 = 49

EM of FeSO4

We know that FeSO4 is reducing agent and it is always oxided to Fe

2(SO

4)

3. The ON of Fe changes

from +2 to +3. Hence the change of ON per molecule is 1.FeSO

4 → Fe

2(SO

4)

3

EM(FeSO4) = (56 + 32 + 64)/1 = 152 (EM is same as its MM)

SAQ 4: Find the equivalent masses of KMnO4 when it reacts in (a)acidic (b)alkaline and (c)neutral medium with

any reducing agent.SAQ 5(A): Find the equivalent masses of the boldfaced species in the following reactions.

(i) NaOH + H2SO

4 → NaHSO

4 + H

2O

(ii) Ca(OH)2 + H

3PO

4 → CaHPO

4 + H

2O

(iii) CuSO4 + KI → Cu

2I

2 + I

2 + K

2SO

4

(iv) Fe2O

3 + CO → Fe + CO

2

(v)NaNO3 + Zn + NaOH → Na

2ZnO

2 + NH

3 + H

2O

SAQ 5(B): Find the equivalent masses of the boldfaced species in the following reactions.(i) KMnO

4 + H

2SO

4 + H

2C

2O

4 → K

2SO

4 + MnSO

4 + CO

2 + H

2O

(ii) I2 + Na

2S

2O

3 → Na

2S

4O

6 + NaI

(iii) Br2 +KOH → KBrO

3 + KBr + H

2O

(iv) Fe3O

4 + KMnO

4 + H

2O → Fe

2O

3 + KOH + MnO

2

(v) N2 + H

2 → NH

3

(vi) KBrO3 + KBr + HCl → Br

2 + KCl + H

2O

(vii)K2Cr

2O

7 + H

2SO

4 + H

2O

2 → K

2SO

4 + Cr

2(SO

4)

3 + O

2 + H

2O

(viii)2H2O

2 → 2H

2O + O

2

CONVERSITON OF MASS INTO EQUIVALENTS AND VICE VERSA:The mass of a given substance can be converted to number of equivalents by dividing the mass by the

equivalent mass. This is similar to determining the number of moles from mass.

No of equivalents(eqs.) = masseqivalent mass

Mass = No. of equivalents × × × × × equivalent massSince 1 equivalent = 1000(103) milliequivalents,

No. of milliequivalents(meq.) = 103 × no. of equivalents(eq.)No. of equivalents(eq.) = 10-3 × no. of milliequivalents

Example: Find the number of eqs. and meqs. of the following substances from their masses.(a)7g of H

2SO

4(b)1.06g of Na

2CO

3(c) 15.8g of KMnO

4( used as OA in acidic medium)

Solution: (a) No. of eqs. = 7/49 = 1/7; No. of meqs. = 1000 × 1/7 = 142.85(b) eq = 1.06/53 = 0.02; meqs = 1000 × 0.02 = 20(c) eq = 15.8/31.6 = 0.5; meqs = 1000 × 0.5 = 500

Example: Find the mass of the following from the eqs and meqs given.(a) 100 meqs of H

2SO

4(b) 500 meqs CaCO

3(c) 0.05 eq of KMnO

4(in acidic medium)

Solution: (a) mass = 100 × 10-3 × 49 = 4.9g (b) mass = 500 × 10-3 × 53 = 26.5g(c) mass = 0.05 × 31.6 = 1.58 g

SAQ 6 :(a)Find the number of equivalents and milliequivalents in case of the following.(i)4g of NaOH (ii)1Kg of CaCO

3(iii)1.52g of FeSO

4 (used as RA)

(b)Find the mass of the following(i)1000 meqs of K

2Cr

2O

7(OA) (ii) 0.5 eq. of H

2SO

4(iii)20 meq of HCl

NORMALITYLike molarity that we have discussed before, normality is another way to express concentration of a solution. It isdefined as the number of gm equivalents of solute present in 1000mL or 1L of the solution. The differencebetween molarity and normality is that in molarity we take the number of moles of solute and in normality we takenumber of gm. equivalents of solute. But the common aspect of both is the equal volume of the solution of 1L.Unit of Normality: gm equivalent/litre which is express as (N).NORMAL(1N) SOLUTION: If one gm. equivalent of solute is present in 1L solution, it is called a normal or Nsoultion.Example: (i) 49g of H

2SO

4(1 g equivalent mass) present in 1L solution = 1N H

2SO

4

(ii) 40g of NaOH(1 g eq.) present in 1L solution = 1 N NaOH(iii)53g of Na

2CO

3(1 g eq.) present in 1L solution = 1N Na

2CO

3 and so on.

Note that like molarity, in normality too the total volume of solute and solvent is considered and for that purposemeasuring flasks of different capacities like 100 mL, 250 mL, 500 mL, 1 L etc. are used for preparing solutions ofdesired normality.Example: What mass of H

2SO

4 is required to make a 1N or normal solution by using a 250mL measuring flask?

Solution: 1N means 1000mL of solution contains 49g of H2SO

4

Hence 250mL of solution contains 12.25g of H2SO

4

So we shall have to take 12.25g of pure H2SO

4 in a 250mL measuring flash and add water upto the mark and

shake well to make a normal H2SO

4 solution.

Example : 4g of NaOH is present in 1L solution. Calculate the normality of the solution.Solution: Equivalent mass(EM) of NaOH = 40/1=40

40g of NaOH = 1 g. equivalent = 1 eq.So, 4g of NaOH = 0.1 eq.0.1 eq. is present in 1L solution, hence according to definition, the normality is 0.1N

Example : 14g of H2SO

4 is present in 500 mL of solution. Calculate its normality.

Solution: EM(H2SO

4) = 98/2 =49

49g of H2SO

4 = 1 eq., Hence 14g of H

2SO

4 = 14/49 = 2/7 eq.

500mL of solution contains 2/7 eq.;1000mL of solution contains 4/7 eq. Hence normality = 4/7 N

FORMULA METHOD:Normality can be expressed by using the following formulae.

Normality = No. of equivalents of solut

volume of solution in LHence; No. of equivalents of solute = Normality ××××× Volume of solution in L

Since 1 equivalent(eq.) = 103 milliequivalent (meq.) and 1L = 103 mL

Normality = 103 X no. of milliequivalents

103 X volume of solution in m

=no. of meq.

volume in mL

Hence, normality =no. of meq. volume in mL X

Example: Calculate the normality of a solution that contains 3g of HNO3 in 200mL of solution.

1st method(unitary method):EM(HNO

3) = 63/1 =63; 63 g of HNO

3 = 1 eq.; So, 3g of HNO

3 = 1/21 eq.

200mL of solution contains 1/21 eq.So, 1000mL of solution contains 5/21 eq. Hence the normality = 5/21 N

2nd method(formula method)No. of eqs = 3/63 = 1/21; voume in L = 200/1000 = 0.2LNormality = eq/L = 1/21/0.2 = 5/21N

3rd method(formula using meq.)No of eq. = 3/63 = 1/21; Hence no. of meqs = 1000 × 1/21 = 1000/21Normality - meq/mL = 1000/21/200 = 5/21 N

Note that since number of equivalents is often a fraction, it involves tideous calculation with the formula usingequivalents. Therefore it is often easier for calculation by the meq method.Example: What is the normality of a solution containing 10g of NaOH in 2000mL solution?What is its molarity?Solution: No. of meqs = 10/40 X 1000 = 250; Hence normality = meq/mL = 250/2000 = 1/8 NMolarity: No. of mmoles =10/40 X 1000 = 250; Hence molarity = mmole/mL = 250/2000 = 1/8 MIn this case normality is same as molarity. This is because the molecular mass of NaOH is equal to equivalentmass.Example: Calculate the normality and molarity of a solution containing 4.9g of H2SO4 in 250mL solution.Solution: Normality: No. of meqs = 4.9/49 × 1000 = 100; Hence normality = 100/250 = 0.4NMolarity: No. of mmoles = 4.9/98 = 50 Hence molarity = 50/250 = 0.2MHere we found that normality of the solution is 2 times greater than its molarity. This is because molecular massof H

2SO

4 is 2 times that of its equivalent mass. 98g of H

2SO

4(1mole) is equal to 2 equivalents. Here we can

formulate the relationship between molarity and normality of substances.

Normality of acid = basicity ××××× MolarityNormality of base = acidity ××××× molarityNormality of salt = total +ve valency ××××× molarityNormality of OA/RA = Change in ON per molecule/ion ××××× molarity

Remember that numerically normality is greater than normality. So the factor(acidity, basicity, total +ve/-vevalency/total change in ON per molecule/ion) is to multiplied with molarity to get normality.Example: Find the molarity of the following solutions from their normalities given.

(a)2.5N H2SO

4(b)0.01N Na

2CO

3(c)5N NaOH

(d)0.5N KMnO4(used as OA in acidic medium)

Solution: (a) Molarity = 2.5/2= 1.25M; (b)Molarity = 0.01/2 = 0.005M, (c) Molarity = 5/1 =5M(d)Molarity = 0.5/5 = 0.1M

Example: Find the normality of the following soltuions from their molarities.(a)18M H

2SO

4(b)0.1M FeSO

4(used as RA) (c)M/20 HCl (d) 2M K

2Cr

2O

7

Solution: (a) Normality = 18 X2 = 36N; (b) Normalty = 0.1 × 1=0.1N (c)Normality = 1/20 × 1 = 1/20 N(d) Normaltiy = 2 × 6 = 12N

Some commond units for Normality:decinormal solution = N/10; centinormal solution = N/100seminormal solution = N/2

SAQ 7: The normality of a KMnO4 solution is 1.5N. Find out its molarity if it takes part in the following reaction.

KMnO4 + MnSO

4 + H

2O → MnO

2 + K

2SO

4 + H

2SO

4

SAQ 8: Find the EM of HNO3 in the following reaction. Cu + HNO

3 → Cu(NO

3)

2 + NO + H

2O

SAQ 9: Find the normality of a 0.2M H2SO

4 if

(a)the solution is completely neutralised by an alkali(b)the solution reacts with some reducing agent to form H2S

SAQ 10: What is the normality of 0.25M K2Cr

2O

7 solution in g eq./litre if it reacts with a reducing agent in acidic

medium.SAQ 11: Find the normality of a Mohr salt solution [FeSO

4.(NH

4)

2SO

4.6H

2O] if 392 g of the salt is dissolved in

water to make 100 mL solution. Note that the salt solution is used as a reducing agent. (Fe=56, S=32)SAQ 12: 0.56 g of oxalic acid crystal(H

2C

2O

4.2H

2O) is dissolved in water to make 50mL solution. Find its

normality and molarity if it is oxided by any OA to CO2.

SAQ 13: Find the normality of the following solutions(a)0.05M KMnO

4 (used in acidic medium0 (b)5M FeSO

4 (used as RA)

(c)0.2M Na2S

2O

3 (reacts with I2 to form sodium tetrathionate)

SAQ 14: Find the number of milliequivalents in case of the following.(a)500mL of N/10 H

2SO

4(b) 2.5 L of N/20 NaOH

(c) 250mL of 2N H2C

2O

4 solution

SAQ 15: Find the number of milliequivalents for the following whose masses are given(a)1.06g of Na

2CO

3(b)7g of H

2SO

4(c)250 mg of NaOH

SAQ 16: Find the mass of the substances from the number of milliequivalents data.(a) 250 meq. of H

2SO

4(b) 100 meq of CaCO

3(c) 50 meq of Ca(OH)

2

SAQ 17: What is the normality of a solution containing 10g of Ba(OH)2 in 2000mL of solution? What is the

molarity? (Ba = 137)SAQ 18: (a)How much H

2SO

4 must be required to prepare 100mL of 0.5N solution.

(b) 100 mL of a N/20 base solution is diluted to 300mL. What is the normality of the diluted solution?SAQ 19: 0.49g of H

2SO

4 is used to prepare a decinormal solution. What would be the volume of solution?

SAQ 20: Find out the molarity of the following solutions. Presume that the acids are completely neutralised.(a)0.25N H

2SO

4(b) 0.05N HCl (c)1.5N H

3PO

4(d)1N H

3PO

3

(e)2N H3PO

2(f)5N CH

3COOH (g)0.1N H

2C

2O

4

SAQ 21: Find the normality of the following solutions. Presume that the acids are completely neutralised.(a)5M H

2SO

4(b)0.2M H

3PO

4(c)0.05M HNO

3

SAQ 22: Find the normality of the following bases. Presume complete neutralisation(a)0.5M NaOH (b)2M Ca(OH)

2

Molality(m):

This is another way to express concentration of a solution. It is defined as the number of moles of solutepresent in 1000g or 1Kg of the solvent. Note that it is determined on the basis of mass of solvent and not thesolution. Its unit is moles/Kg and is denoted by the symbol ‘m’(small letter).

MOLAL SOLUTION:(1m): The solution in which one mole of solute is present in 1000g(1Kg) of solvent.Example: When 98g of H

2SO

4(1 mole) dissolves in 1000g of water, we get 1m solution. The volume of the

solution is obviously greater than 1000 mL. 1m aqueous solution prepared from different solutes will havedifferent volumes depending on the nature of the solute. For solid solutes like NaOH, Na

2CO

3 etc. the increase

in volume is small as the molecules or ions from the solid fit within the intermolecular free space of water. Butthe liquid solutes like H

2SO

4 the increase in volume is appreciably large as the liquid solute has also its intermolecular

space which increases the total free volume and hence total volume of the mixture. In case of molarity wediscussed before, the total volume of solution(solute+solvent) is adjusted to 1000mL by using a measuring flaskand hence the amounts of water needed for different solutes were different.

Molality is the number of moles of solute present in 1000 gm or 1 Kg of the solvent.

1000)(

)(

gmass

molesnMolality

solute

solute

)(

)(

Kgmass

molesnMolality

solute

solute

Example: Find the molaity of sulphuric acid prepared by dissolving 28 gms of pure H2SO

4 in 100 mL of pure

water.Solution: mass of 100 mL of water = 100 g (as density of water = 1 gm/mL).

Number of moles of solute = 28/98 = 2/7100 g of solvent contains (2/7) mole of solute,So 1000 g of solvent contains (2/7) 1000 = 2.85 mole; So MOLALITY = 2.85 m.(moleKg–1)

Formula Method: mMolality 85.21000100

9828

Mole fraction and Molality: Solutions having same molality have same mole fractions of solute and solventsince each has a fixed number of moles of solute and solvent. But solution having same molarity have differentmole fractions of solute and solvent because the moles of solvents are different althougth the number of molesof solute are same. Note that in all these cases the solvent considered is water (aqeuous solutions).

no. of moles of Ano. of moles of A + no. of moles of B

mole fraction of A=

Example : What are the mole fractions of solute and solvent in 0.5m aqeous solution of NaOH?Solutionl: 0.5 mole of NaOH is present in 1000gm of water.

No. of moles of NaOH = 0.5No. of moles of water = 1000/18 = 55.5So mole fraction of NaOH = 0.5/(55.5+0.5) = 0.0089Hence mole fraction of water = 55.5/(55.5+0.5) = 0.9911Alternatively mole fraction water = 1- 0.0089 = 0.9911 (since sum of mole fractions = 1)

Molality from molarity:Molality of solution can be determined if the molarity and density of the solution are known.Example 2: What is the molality of a 0.2M solution of sucrose(C

12H

22O

11) if the density of the solution is 1.0319

gm/ml ?Solution: Let the volume of the solution = 1L = 1000mL. This volume contains 0.2 mole of sucrose.

Mass of sucrose present in it = 0.2 × 342 = 68.4gLet us find out the mass of 1000mL of solution by using density data.⇒ m/v = 1.0319 ⇒ m/1000 = 1.0319 ⇒ m = 1031.0 g

Hence mass of solvent(water) = 1031.9 - 68.46 = 963.44 g963.44g of solvent contains 0.2 mole of soluteSo, 1000g of solvent contains 0.207 mole of solute.Hence the molality of solution = 0.207m

Formula Method: mMolality 207.0100044.963

2.0

Example: Find the molality of a 30% H2SO

4 solution by weight. If its density is 1.16 g/mL, find its molarity.

Solution: 30gms of H2SO

4 is present in 100g solution.

Hence mass of solvent = 100 -30 =70g70g solvent contains 30/98 moles of solute1000g of solvent contains 4.373 molesHence molality of the solution = 4.373m

To find the molarity, we shall have to use the density data. Let us first find the volume of 100g of solution.m/v = 1.16 g/mL ⇒ v = 100/1.16 = 86.2mL86.2 mL solution contains 30/98 moles of H

2SO

4

So, 1000mL solution contains 3.559 moles of H2SO

4

Hence the molarity of the solution = 3.559M

Formula Method : MMolarity 559.310002.86

9830

SAQ 23: Calculate the normality, molarity and molality of H2SO

4 solution which contains 4.9g of H

2SO

4 in 100mL

solution.(density of the solution = 1.15 g/mL)SAQ 24: 200mL of solution of Na

2CO

3 is prepared by dissolving 10g of the salt in water. The density of the

solution is 1.1 g/mL. What are the molarity, molality and normality of the solution.SAQ 25: Calculate the molarity, normality and molality of a 15% H

2SO

4 by weight having density 1.1 g/mL.

EFFECT OF TEMERATURE ON MOLARITY, NORMALITY AND MOLALITY:If we increase the temperature, the volume of the solution will increase due to thermal expansion. So

the solution gets diluted and molarity and normality change(decrease). However the molality of a solution remainsthe same as molality does not involve volume of the solution. Thus molality is independent of temperature whilemolarity and normality are inversely proportional to temperature.

QUANTITATIVE ANALYSISQuantitative analysis is the method in which the amount or concentration of a particular substance is

determined accurately. This is divided broadly into two types.(a)volumetric analysis (b)gravimetric analysis

Volumetric Analysis: Volumetric analysis is the quantitative analysis which involves the measurement of thevolume of a solution of unknown concentration required to react completely with a definite volume of a secondsolution of known concentration or definite mass of a second substance.

Suppose you are given an acid solution whose molarity/normality is not known. In order to find that youhave to take the help of a base soluion, e.g NaOH solution of known molarity/normality(say 0.1N). You haveto experimentally find out the volume of acid solution required to completely neutralise a given volume of (say25mL) of the base solution of known concentration. Suppose we find from the experiment that 15.5mL of theacid completely neutralises 25mL of 0.5N base solution. From this the normality/molarity of the acid solution canbe determined. This is done with the help of titration experiment. We shall study more about it later.Gravimetric Analysis: In gravimetric analysis, an insoluble solid is usually precipitated by the reaction of thesolution of substance(whose quantitative analysis is to be made) with another substance taken in excess. The solidformed is seprated by filtration, washed to make free of impurities and then dried to constant mass. The massof the solid is then stoichiometrically related to the mass of the original substance.

Suppose you have a sample of AgNO3 of unknown concentration. You add excess of HCl to get AgCl

white precipitate. From the mass of the dried AgCl, we can calculate the mass of AgNO3 present in the original

solution by simple stoichiometric method.

VOLUMETRIC ANALYSIS(TITRATION EXPERIMENTS)

Volumetric analysis by titration method is divided into three types.

(i)Acid-base Titration(acidimetry-alkalimetry)(ii)Redox Titration(iii)Precipitation Titration

ACID-BASE TITRATION:Let us first study the volumetric analysis of the simplest reaction i.e neutralisation reaction. An acid reacts

with a base in aqueous solution to form salt and water. This analysis is called acidimetry-alkalimetry.Acidimetry-alkalimetry:Example :

NaOH + HCl → NaCl + H2O

1mole 1 moleFrom the balanced equation, we fiind that 1 mole of NaOH(40g) reacts with 1 mole of HCl(36.5g). Since thereactants are allowed to react with each other in aqueous solutions, we have to make volumetric analsysis eitherbased on molarity or normality.

(a)Based on molarity: When we allow 1L of 1M NaOH soltuion to react with 1L of 1M HCl solution,complete neutralisation takes places as each solution contains 1 mole of the respective solute.

(b)Based on normality: When we allow 1L of 1N NaOH solution to react with 1L of 1N HCl solution,complete neutralisation takes palce as each contains 1 g eqivalent of solute which is also equal to 1 mole of eachreactant. Since normality is same as molarity in this case.Example :

2NaOH + H2SO

4 → Na

2SO

4 + 2H

2O

2×40g 98g2 mole 1mole

(a)Based on molarity: When we allow 1L of 2M solution or 2L of 1M solution of NaOH(each contain2 moles of NaOH) to react with 1L of 1M H

2SO

4(contains 1mole), complete neutralisation takes place according

to balanced equation.(b)Based on normality: When we allow 1L of 2N or 2L of 1N NaOH solution(each contains 2 g.equivalent

which equal to 2 moles) to react with 1L of 2N or 2L of 1N H2SO

4 solution(contains 2 g. equivalents which is

equal to 1 mole), complete neutralisation takes place.Note the striking difference in this case between the two methods. In molarity method, 2 moles of NaOH reactswith 1 mole of H

2SO

4 while in normality method, 2 g. equivalents of NaOH reacts with 2 g.equivalents of H

2SO

4.

In other words, ‘x’ g. equivalent of NaOH reacts with ‘x’ g. equivalents of H2SO

4. So remember that in

normality method a 1:1 ratio exists in the number of g. equivalents of the two reactants, whiel in molarity methodit may not be the case always with the number of moles of the reactants. Fortunately in some cases like (NaOH+HCl), 1: 1 mole ratio exists between the reactants. In molarity method, we have to balance the equation forpredicting mole-mole relationship, but in normality method, balancing of equation is not required, since both thereactants react in same number of g. equivalents - as per the famous slogan “ALL CHEMICAL REACTIONSTAKE PLACE IN EQUIVLENTS”

Example :3NaOH + H

3PO

4 → Na

3PO

4 + 3H

2O

3 mole 1 mole(a)Molarity method: 1L of 3M NaOH or 3L of 1M NaOH(contains 3 moles) react completely with 1L

of 1M H3PO

4(contains 1 mole) according to balanced equation.

(b)Normality method: 1L of 3N NaOH or 3L of 1N NaOH (contains 3 g. equivalents which is equal to3 moles) react completely with 1L of 3N or 3L of 1N H

3PO

4 solution(contains 3 g. equivalents which is equal

to 1 mole).CONCLUSION: Every chemical reaction takes place in equivalents. The number of equivalents or milliequivalentsof both the reactants must be same.Therefore for volumetric analysis normality method is more powerful and is strongly recommended.

Acid-base neutralisation on the basis of normality:We know that

No. of equivalents(eqs) = volume in L × Normality ( Eqs = L × N)No. of milliequivalents(meqs) = volume in mL × Normality( meqs = mL × N)

100mL of 2N solution of acid ≡ 100mL of 2N solution of base⇒ 100mL of 2N acid ≡ 50mL of 4N base (both contain 100×2 = 50×4 = 200 meq)

(note that the symbol ≡ stands for equivalent to i.e completely reacts with)

Example:: 50mL of N/10 acid ≡ V mL of N/100 base. Find the volume(V) of the base solution.The number of meqs of acid = 50 × 1/10 = 5; Hence the number of meqs of base =5So the volume of base solution = 5/1/100 = 500mL (No. of meqs of base = 500 X 1/100 =5)

Hence from the above discussion we can formulate the relationship

V1 ××××× N

1 = V

2 × × × × × N

2

Where V1 and N

1 are volume and normality of one solution and V

2 and N

2 are the volume and normlality of the

other solution.Example: Calculate the normality of NaOH solution, 25 mL of which required 20mL of N/10 H

2SO

4 solution for

complete neutralisation.Solution: Let x N is the normality of NaOH.

25 × x = 20 × 1/10 ⇒ x = 2/25 N = 0.08N

Calculation based on mass of one reactant and normality of the other reactant:

Example: Calculate the normality of a solution of H2SO

4 if 50 mL of the such a solution completely is neutralised

by 0.53 g of Na2CO

3.

Solution: Let us first calculate the meq of Na2CO

3 from its mass data.

No. of meq of Na2CO

3 = (mass/EM) × 1000 = 0.53/53 × 1000 = 10 (Eq. mass of Na

2CO

3=53)

Hence the number of meqs of H2SO

4 = 10

Volume in mL × normality = 10 ⇒ 50 × x = 10 ⇒ x = 10/50 = 0.2 N

SAQ 26: Calculate the volume of H2SO

4 of strength 2N required to completely neutralise 25 mL of N/10 NaOH?

SAQ 27: 37.5 mL of a HNO3 solution is required to neutralise 50 mL of 0.15N solution of Ba(OH)

2. Find the

normality of the HNO3 solution.

SAQ 28: Calculate the volume of 0.5 N of an acid solution completely neutralise 0.4 g of NaOH.SAQ 29: How many cm3 of 5N NaOH are required to neutralise 30 cm3 of 4N H

2SO

4?

STRENGTH IN g/L:The strength of the solution can also be expressed in terms of g/L in stead g eqivalent/L(normality). Let

the strength of the solution is x g per L. This means that 1 L of such a solution contains x gms of solute.The number of g. equivalent present in x g of the solute = x/E (where E = equivalent mass of the solute)If 1 L solution contains x/E equivalent, then its normality(N) is x/E.

N = x/E ⇒ x = N X EStrength in g/L = normality X equivalent mass

⇒ Normality = strength in g/L

equivalent mass NormalityMELgStrength .)/(

Example: Calculate the volume of NaOH solution having strength 4 gms per litre of solution required tocompletely neutralise 25mL of N/20 H

2SO

4 solution.

Solution: Let us first find out the normality of NaOH solution

Normality = strength in g/L

equivalent mass ⇒ Normality = 4/40 = 0.1N

Now let us use the formula; V1 × N

1 = V

2 × N

2

25 × 1/20 = x × 0.1 ⇒ x = 12.5 mL(requied volume of NaOH solution)

SAQ 30: What volume of 4N H2SO

4 is required to neutralise a solution containing 3.7g of Ca(OH)

2 per litre.

Also find how many gms of pure H2SO

4 is present in that volume?

SAQ 31: 0.25g of sample of a solid acid was dissolved in water. This solution was exactly neutralised by 50mL of 0.25N of a base. What is the equivalent mass of the acid?SAQ 32: 50 cm3 of a HCl solution requires 1.5 g of a pure CaCO

3 for complete neutralisation. Calculate the

normality of the acid.SAQ 33: (a) Calculate the normalilty of Na

2CO

3 solution if 0.53 g of it is present in 500 mL solution.

(b) Calculate the strength in g/L of N/100 H2SO

4 solution.

(c) Calculate the equivalent mass of an acid having normality 0.2N and strength in g/L equal to 9.8.SAQ 34: Calculate the normality of NaOH solution, 25 mL of which completely neutralises 50 mL of H

2SO

4

solution containing 0.7 g of H2SO

4 in it.

SAQ 35: A piece of aluminium weighing 2.7g is treated with 75 mL of H2SO

4(specific gravity 1.18, containing

24.7% H2SO

4 by weight). After the metal is completely dissolved, the solution is diluted to 500 mL. Caculate the

normality and molarity of this diluted H2SO

4 solution.

SAQ 36: Find the equivalent mass of a metal carbonate, 4 g of which react exactly with 200 mL of 0.377NH

2SO

4.

SAQ 37: 0.2g of Na2CO

3 required 40mL of an acid for complete neutralisation. Determine the normality of the

acid solution.SAQ 38: 0.84g of an acid is dissolved in 200 mL water and 25 mL of this solution required 14 mL of decinormalNaOH solution. Calculate the equivalent mass of the the acid. If its molecular mass is 150, what is its basicity?SAQ 39: 25 mL of a decinormal solution of KOH neutralises 27.5 mL of solution of dibasic acid containing 4.1g of acid per litre. Calculate the equivalent mass and molecular mass of the acid.SAQ 40: 30 mL of 2N H

2SO

4, 50 mL of 0.5N HCl, 75 mL of 5N HNO

3 and 1 litre of N/10 NaOH are mixed

in a container. Predict the whether the resulting mixture is acidic or alkaline. Calculate the normality of theresulting acid or alkali solution.

CALCULATION OF STRENGTH OF SALT FORMED AND STRENGTH OF EXCESS REACTANTIN ACID-BASE REACTIONExample: If 100 mL of N HCl solution is mixed with 50 mL of 2N solution, what is the normality of the resultingsalt solution?Solution: NaOH + HCl → NaCl + H

2O

100 meq 100 meq 100 meqThe number of meqs of HCl = 100 × 1 = 100The number of meqs of NaOH = 50 × 2 = 100100 meq of NaOH will react with 100 meq of HCl to form 100 meq of NaCl(salt).Note that in this case, no reactant will remain in excess.The total volume of mixture = 100 + 50 = 150 mLSo the normality of the salt solution = meq/mL = 100/150 = 0.667 NExample: If 100 mL of HCl and 100 mL of 0.8N NaOH solution are mixed together, what is the normality ofthe resulting solution? (Find the normality of salt and excess acid)Solution: meq of HCl = 100 × 1 = 100; meq of NaOH = 100 × 0.8 = 80The limiting reactant is NaOHSo 80 meq of NaOH will react with 80 meq of HCl to form 80 meq of NaCl(salt).No. of meqs of excess HCl = 100 - 80 = 20Total volume of the solution = 100 + 100 = 200 mLHence the normality of NaCl(salt) = meq/mL = 80/200 = 0.4NNomrality of excess HCl = 20/200 = 0.1NNote that the resulting solution contains both salt as well as acid(excess).

SAQ 41: 250 mL of N/10 NaOH solution is mixed with 100 mL of N/4 H2SO

4 solution. Find out the normaltiy

and molarity of the salt in the mixture.SAQ 42: 200 mL of N/2 Ba(OH)

2 solution is mixed with 100 mL of N/2 HCl solution. What will be the normaltiy

and molarity of the resulting solution?SAQ 43: 25 mL of 10N HCl is mixed with 20 mL of 18N H

2SO

4 and the mixture is made upto 1 litre with distilled

water. What is the normality of the mixture?

ACID-BASE TITRATIONThe experiment by which we can determine the volume of the acid solution (V

1) which exactly neutralises

a known volume of base(V2) or vice versa is called titration. If the normality of one of the solutions is known,

then the normality of the other solution can be found out from the relationship.V

1 × N

1 = V

2 × N

2

Preparation of a standard solution:The solution whose concentration(normality) is accurately known is called a standard solution. For

preparing a standard solution we have to use a measuring flask of fixed volume and shall have to take a fixedmass of the solute in it and add water slowly with constant shaking uptill the mark. Care should be taken to ensurethat the lower meniscus of water touches the mark of the flask while adding water. Then the flask is to bestoppered tightly and shaken well. For example, a N/10 solution of Na

2CO

3 can be preapred by any of the

following ways.(a) 5.3 g of Na

2CO

3 used to prepare 1 L solution or

(b)1.325 g of Na2CO

3 used to prepare 250 mL solution or

(c) 0.1325 g of Na2CO

3used to prepare 25 mL solution etc.

SAQ 44: How much of 98% H2SO4(by weight) required to prepare N/10 solution of H2SO

4 by using a 100 mL

measuring flask.SAQ 45: You are given to perepare a N/20 NaOH solution in a 50 mL measuring flask. How much of solidNaOH you will take in the flask?

For titration experiment the following apparatuses are required.(a)conical flask (b)pipette (c) burette (d)burette stand(e)indicator (f)acid solution (g)base or alkali solution

BURETTE: It a long calibrated glass tube attached with a stop-cock at the lower part which used to releasethe solution out through the pointed end. Usually in acid-base titration, acid is taken in the burette.PIPETTE: It is a narrow tube having a wider bulged out portion at the middle which ends with a pointed tip.There is a mark at the upper part of the pipette. Pipettes of different capacities such as 10mL, 25mL, 50mL etc.are available. The solution is drawn into the pipette by first dipping it inside the solution and sucking by mouthor using a drawing device called propipette. The level of the solution drawn has to be raised exactly upto themark(lower meniscus should touch the mark). The upper end of the pipette is carefully closed by the forefingerand is transferred into a clean conical flask and the solution is released inside it by removing the finger. Usuallyalkali is taken with the help conical flask into the conical flask.INDICATOR: Indicator is a substance that can exist in solution in two different forms with two differentcolours. It has a distinct colour in acidic solution and different but distinct colour in alkali or basic solution.Indicators are often organic dyes such as methyl oragne, methyl red, phenolphthalein etc. One to two drops ofindicator solution is added to the one of the solutions say alkali solutions taken in a conical flask. For instance,if one drop of methyl orange indicator is added to alkali solution, it renders the solution distinctly yellow.TITRATION AND END POINT:The conical flas containing alkali solution is kept below the burette. Acid is added from the burette slowly withperiodical shaking of the flask. The yellow colour gradually fades and at a particular point the colour changessharply to pink by the addition of one drop of excess acid. This is called End Point or Equivalence Point ofthe titration. A single drop of excess acid after complete neutralisation renders the solution acidic and the indicatorimparts pink colour. The addition of acid is stopped at this point and the volume of the acid added from the buretteis read from its calibrations.

Volume of acid added = Final burette reading - initial burette readingNote that the addtion of one drop of excess acid at the end point practically does not change the burette readings.The volume of solution noted from burette readings is called TITRE value. Note that the first titre value is usuallynot accurate because the end point is not known in advance. This is called rough reading. The expeirment isrepeated with the fixed volume of alkali each time until the volume of acid added(titre value) becomes same forat least consecutive experiments. These are called concordant readings.Let the volume of alkali taken in the conical flask = 25 mLThe volume of acid added from the burette till end point(titre value) = 20.5 mLLet the normality of the alkali = 0.9 N/10We can calculate the normality of acid solution by using the relationship; V

1 × N

1 = V

2 × N

2

20.5 × N1 = 25 × 0.9 × 1/10 ⇒ N

1 = 0.11 N

Hence the normality of the acid is 0.11N.

Similary if the normality of acid is known, we can find the normality of the base by using the same relationship.

A list of common acid-base indicators and their colours in acidic and alkaline medium is given below.

Indicators Colour in acid Colour in alkaliMethyl orange pink yellowmethyl red red yellowlitmus red bluephenolphthalein colourless red

STANDARDISATION OF SOLUTIONS AND DOUBLE TITRATION EXPERIMENTSA solution of desired normality is prepared by weighing the solute accurately in an electrical balance and

taking it in a measuring flask of a fixed capacity. Then distilled water is then added gradually upto the mark withperiodical shaking to get the solution of a desired normality. However there are some solutes which cannot beaccurately weighed because they react rapidly with the atmospheric constitutents such as water, carbon dioxide,oxygen etc. during the time of weighing. For example, when solid sodium hydroxide pellets are removed fromits closed container for weighing, it rapidly absorbs water and become moist. This property of NaOH is calleddeliquescence. So it becomes impossible to measure the accurate weight of pure NaOH as it is always contaminatedwith water. The approximate weight and hence approximate normality can be determined in such cases, neverthe accurate normality. To determine its accurate normality, you have to take the help of a titration experiment.This is called standardisation of a solution i.e to determine the accurate strength(normality) of a solution bytitration experiment.Primary Standards:

There are many other substances which are very stable in the atmosphere unlike NaOH. Their weightscan be measured accurately and so solutions with accurate normalities can be prepared from such solutes. Suchsolutions are called standard solutions. Succinic acid(C

4H

6O

4) belongs to this category and we can prepare a

standard succinic acid solution with accurate normality by weighing method. The stable solutes from whichstandard solutions can be prepared by weighing method are called primary standards. Such substancesshould have the following characteristics.

(i) They must not absorb or react with the components of atmosphere such as H2O, O

2, CO

2.

(ii)They should have high molecular mass so that they remain non-volatile and so there is negligible weightloss during weighing.

(iii)They should be soluble in the solvent like water(iv)They should be non-toxic.

Standardisation of solution: To determine the accurate normality of a solution whose normality is known onlyapproximately(NaOH type), we have to perform its titration with a standard acid solution(succinic acid type).Fromtitre value, the accurate strength of such solutions can be determined.Example: You are given a standard N/10 succinic acid solution. You are given a NaOH solution of unknownor approximate strength. Determine its accurate normality by titration method.

Solution: Let a given volume(say 25 mL) of the NaOH solution is taken in a conical flask with the help of apipette. A drop of methyl orange indicator is added to it. The solution turns yellow. Then the standard succinicacid solution is added gradually from a burette to the NaOH solution till the end point i.e the solution turns fromyellow to pink. Note the titre value from the burette readings.Let the volume of succinic acid solution required at the end point = 21.5 mL

Let us apply the relationship V1 × N

1 = V

2 × N

2

21.5 × 1/10 = 25 × x ⇒ x = 0.086 N (the accurate normlity of the NaOH soltution)

NORMALTIY OF BENCH ACID(DOUBLE TITRATION EXPERIMENTS)To determine the normality of acids(H

2SO

4, HCl etc) kept in laboratory benches, we have to first

standardise a base solution(NaOH) by titrating with a standard acid(succinic acid). Then the bench acid is titratedwith the the standardised base to determine the the normality of the bench acid. So two titration experiments haveto be perfomred for this. In double titration experiments, you will be given either two acids and a one base ortwo bases and an acid. The nomality of one among the three solutions will be given and you will be asked tofind out the normalites of the other two solutions. Study the following examples.Example: 25 mL of a solution of oxalic acid(H

2C

2O

4) containing 4.5 g of the acid per litre of solution neutralised

32.5 mL of a NaOH solution. 25 mL of this alkali required 20.2 mL of a H2SO

4 solution for complete neutralisation.

Calculate the normality and strength in g/L of the H2SO

4.

Solution: It is clear from the question, that the first acid(oxalic acid) is a primary standard and the other twosolutions are not standard solutions.

Equivalent mass of oxalic acid(H2C

2O

4) = 90/2 = 45

Normality of oxalic acid solution = 4.5/45 = 0.1N1st titration: 25 mL of 0.1N oxalic acid ≡ 32.5 mL of x (N) NaOH (let the normality of NaOH= xN)

25 × 0.1 = 32.5 × x ⇒ x = 0.0769N2nd titration: Once the normality of NaOH solution is determined(or is standardised), the solution can be usedto determine the normality of any other acid

25 mL of 0.0769N NaOH ≡ 20.2 mL of y (N) H2SO

4 ( let the normality of H

2SO

4 = y N)

25 × 0.0768 = 20.2 × y ⇒ y = 0.095NHence the normality of H

2SO

4 is 0.095N.

Strength in g/L = 0.095 × 49 = 4.65 g/L

SAQ 46: A 40 mL of an acid(A) solution completely neutralises 50 mL of Na2CO

3 solution. 25 mL of this Na

2CO

3

solution can completely neutralise 25mL of HCl solution having strength 3.6 g/L. Find the normality of theacid(A).SAQ 47: 25 mL of an approximately N/10 NaOH solution required 17 mL of 0.95 N/10 succinic acid solutionfor complete neutralisation. 25 mL of the same NaOH solution in another experiment required 27.5 mL of aH2SO4 solution. Find the strength in g/L and normality of the H

2SO

4 solution.

PROBMLEMS BASED ON DILUTION USING DENSITY OR SPECIFIC GRAVITY AND PERCENTSTRENGTH OF SOLUTION

Diltue solutions are prepared from concentrated solutions by the process of dilution. Calculated volumeof concentrated acid is taken in a measuring flask of known capacity. Distilled water is added upto the mark adnthe mixture is thoroghly shaken before use. The densities(specific gravities) and weight percent of the concentratedsolutions available commercially are known. For example, the density of commercial concentrated H

2SO

4 is 1.84

gm/mL(specific gravity= 1.84) which is 98% by weight(mass). From these data it is possible to calculate thevolume of the concentrated acid required to prepare a diluted acid of known normality. Study the followingexample.Example: What volume of conc. H

2SO

4 hving specific gravity 1.84 containing 98% H

2SO

4 by weight are

required to prepare 250 mL of 0.2N solution of H2SO

4?

Solution: Let us first find out the normality of the concentrated acid.Specific gravity = density = m/V = 1.84 g/mL

Let us find out the mass of 1000mL(1L) of the solution. of the conc. acid.mass = 1.84 X 1000 = 1840 g

Note that this amount is total mass of solute(H2SO

4) and solvent(water) in 1000mL solution.

Since it is 98% by weight100g of solution contains 98 g of H

2SO

4

So 1840g of solution contains (98/100) × 1840 = 1803.2 g of H2SO

4(pure solute)

So, 1000 mL solution contains 1803.2 gm of solute = 1803.2/49 = 36.8 g. equivalents(eqs) of H2SO

4

Hence the normality = 36.8NLet V mL of this acid is diluted to 250 mL whose normality becomes 0.2N. The number of meqs of H

2SO

4 will

remain the same.V × 36.8 = 250 × 0.2 ⇒ V = 1.358 mL

Hence 1.358 mL of conc. acid is used to prepare 250 mL of 0.2N acid.Alternative Method:

The meq of H2SO

4 present in 250mL of 0.2N solution = 250 × 0.2 = 50

Mass of H2SO

4 = 50 × 10-3 × 49 = 2.45g

In the conc. acid, 98 g of pure acid is present in 100g solutionSo, 2.45g of pure acid is present in (100/98) × 2.45 = 2.5 g of solutionm/V = 1.84 ⇒ V = m/1.84 = 2.5/1.84 = 1.358 mL

So the volume of conc.acid required is 1.358 mLWhich method you liked, the previous method or this method ? You can still find many other methods to tacklethis problem. You can select any one method which excites you most to solve a problem.

SAQ 48: What would be the normality of a solution prepared by diluting 100mL of 2.5N H2SO4 by adding 1000mL of water?SAQ 49: How many mL of a conc. HCl solution having density 1.17 g/mL and percent strength 33% by weightwill be required to prepare 200 mL of 1.5N solution?SAQ 50: Calculate the molarity and molality of a 15% H

2SO

4 solution by weight having density 1.1 g/mL. What

volume of water should be added to 200 mL of this acid in order to prepare 0.5N solution of H2SO

4.

SAQ 51: What is the purity of conc. H2SO

4 solution, having density 1.7 g/mL if 10 mL of it is completely

neutralised by 150 mL of 2N NaOH solution?

PROBLEMS BASED ON SUCCESSIVE REACTIONS INVOLVING BACK TITRATIONSuppose solution of A reacts with excess of solution B, so that we have some excess B at the end of

the reaction. The excess B present in the mixture is then titrated with a standard solution of C. Since the titrationis done at last, it is called back titration. From the back titration we can know the normality of the B(excess)in the mixture. From this we can know the amount of B reacting with A and then the amount of A present inthe solution. Suppose the sample A contains some impurities which does not react with B. In that case, theamount of A calculated from the back titration method cannot be equal to the whole of the mass of impure Ataken. If the mass of impure sample is 1 g and the mass of pure A calculated from back titration method is 0.8g,then the mass of impurity is 0.2g. Hence the percentage of purity of the sample A can be calculated as (08/1) × 100 = 80%. Let us study some numerical examples on back titration. Remember that such problems areapproached from the backward direction.

Example: A calculated quantity of calcium carbonate is treated with 80 mL of a decinormal H2SO

4 solution. The

excess of acid required 60 mL of 0.05N NaOH solution for complete neutralisation. Calculate the mass ofcalcuium carbonate taken in the first reaction.Solution: We know that CaCO

3 reacts with H

2SO

4 as follows.

CaCO3 + H

2SO

4 → CaSO

4 + CO

2 + H

2O

From the question it is clear that H2SO

4 is taken in excess and CaCO

3 is the limiting reactant. After the reaction

some H2SO

4 will be left out in the solution having volume 80 mL. It has now become a diluted acid

Let us find the normality of this diluted acid by the back titration data.80 mL of x N diluted acid ≡ 60 mL of 0.05N NaOH

80 × x = 60 × 0.05 ⇒ x = 3/80 NNote the normality of H

2SO

4 before the reaction was N/10 and now it is less.

The no. of meq of H2SO

4 present before the reaction = 80 × 1/10 = 8

The no. of meq of H2SO

4 present after the reaction(diluted acid) = 80 × (3/80) = 3

So the no. of meq of H2SO

4 consumed by CaCO

3 in the reaction = 8 - 3 = 5

But we know that chemical reactions take place in equivalents.meq of H

2SO

4 = meq of CaCO

3 = 5

So, mass of CaCO3 = 5 × 10-3 × 50 = 0.25g (Equivalent mass of CaCO

3 = 50)

Example: 0.21 g of the metal is treated with 200 mL of 0.25N H2SO

4 until the metal dissolved completely. The

residual acid required 65 mL of N/2 caustic soda solution for complete neutralisation. Calcuilate the equivalentmass of the metal.Solution: Let us start from the backward direction.The volume of residual acid = 200 mLNote that the volume acid solution does not change by the consumption of the acid in the reaction. Only theconcentration(normality) of the solution changes(becomes less than the original value).200 mL of xN residual acid ≡ 65 mL of N/2 caustic soda solution

200 x = 65 1/2 ⇒ x = 0.1625Nmeq of the excess acid = 200 0.1625 = 32.5meq of acid used original before the reaction = 200 0.25 = 50So, meq of acid consumed in the reaction = 50 - 32.5 = 17.517.5 meq of H

2SO

4 ≡ 17.5 meq of metal

Let the equivalent mass of the metal is x.The mass of the metal = 17.5 10–3 x = 0.21 (given)⇒ x = 12 (equivalent mass of the metal)

PROBLEMS BASED ON PERCENTAGE OF PURIY OF SAMPLESIt has already been explained before if the reactant A is an impure sample, say for example a mineral

which contains many impurities other than the main chemical substance. The mass of this main chemicalsubstance present in the impure sample can be determined by the back titration method as explained before. Herewe have to presume that the impurites do not react with the other reactant B during the reaction. Thus we canfind out the percentage of purity of the sample.Example: 0.5g of a sample of limestone(CaCO

3 mineral) is dissoved in 50 mL of N/10 HCl solution. After the

reaction is complete, the excess acid required 10 mL of 0.09N caustic soda for complete neutralisation. Find thepercentage of purity in the sample.Solution: CaCO

3 reacts with HCl and dissolves in it. After the reaction, some acid remains in excess. This excess

acid requires 10 mL of 0.09N NaOH solution for complete neutralisation.Let us start from backward direction,

The volume of the residual excess acid solution = 50 mLLet the normality of the residual acid solution = xN50 x = 10 0.09 ⇒ x = 09/50 N

Meq of excess acid in that solution = 50 × (0.9/50) = 0.9Meq of acid originally used before the reaction = 50 0.1 = 5So, the meq of acid which has been consumed by CaCO

3 = 5 – 0.9 = 4.1

Hence the meq of CaCO3 = 4.1

So, the mass of CaCO3 = 4.1 10-3 50 = 0.205g

But the mass of the impure limestone = 0.5 g

So percentage of purity = %411005.0

205.0

Example: 0.2g of limestone was dissolved in 25 mL of N/10 HCl and the solution was diluted to 100 mL withthe help of distilled water. 25 mL of this diluted solution required 5.0 mL of N/50 NaOH solution for completeneutralisation. Calculate the percentage of CaCO

3 in the limestone.

Solution: This example is similar to the previous one but there is a small difference. After CaCO3 present inthe sample has completely reacted with the acid, the residual acid(25mL) is not used as such for the titrationexperiment. It is diluted to 100 mL in a measuring flask by adding distilled water. Now the volume of the residualacid is 100 mL. This is called the stock solution. From this stock solution, 25 mL is pipetted out into a conicalflask for the titration experiment. There is no need to use the entire volume of stock solution(100 mL) for thetitration. 25 mL of this solution required 5 mL of N/50 NaOH solution.

Let the normality of the stock solution = x N

25 x = 5 × 1/50 ⇒ x = (1/250)NThis is the normality of the entire stock solution from which 25 mL was pipetted out.So the meq of acid present in the stock solution = 100 1/250 = 0.4Meq of acid used originally before the reaction = 25 1/10 = 2.5So, meq of acid consumed by CaCO

3 = 2.5 - 0.4 = 2.1

Meq of CaCO3 = 2.1 ⇒ mass of CaCO

3 = 2.1 10-3 50 = 0.105g

So the percentage of purity = %5.521002.0

105.0

Example: 1.5 g of impure sample of ammonium sulphate was boiled with an excess of caustic soda. Theammonia evolved was absorbed in 300 mL of N/5 H

2SO

4 solution. The volume of this partially neutralised acid

was made upto 500 mL with distilled water. 20 mL of this solution required 15 mL of N/10 solution of NaOHform complete neutralisation. Find the percentage of purity in the original sample.Solution:(i) Ammonium sulphate present in the impure sample reacts with caustic soda in the first step liberatingequivalent amount of NH

3 gas as per the equation.

(NH4)

2SO

4 + NaOH → Na

2SO

4 + NH3↑ + H

2O (not balanced)

(ii) All the ammonia gas evolved is carefully bubbled into a conical flask containing 300 mL of N/5 H2SO

4

where NH3 dissoves completey and react with acid.

NH3 + H

2SO

4 → (NH

4)

2SO

4

Note that the acid is taken in excess deliberately so that after its reaction with NH3 some acid will be left

unreacted(excess acid). Here NH3 is the limiting reactant.

(iii) This solution containing the excess acid(300 mL) is transferred to a 500 mL measuring flask and volumemade upto the mark with the help of distilled water. Now the volume of this diluted excess acid solution is 500mL. This is the stock solution.(iv) 20 mL of this stock solution containing excess acid is pipetted out to another conical flask and titratedagainst N/10 NaOH solution taken in a burette. The titre value is found to be 15 mL. Here we find that titrationhas been done at the end. Hence it is a back titration problem and we shall proceed to solve if from backwarddirection.(a)Titration experiment: 20 mL of x N diluted acid ≡ 15 mL of N/10 NaOH solution

20 x = 15 1/10 ⇒ x = 3/40 NThis is normality of the stock solution whose volume is 500 mL.(b)The meq of excess acid present in the stock solution = 500 3/40 = 37.5Meq of acid originally taken before reaction = 300 1/5 = 60Hence meq of acid consumed by NH

3 = 60 – 37.5 = 22.5

Hence meq of NH3 = meq of (NH

4)

2SO

4 = 22.5

This because in successive reactions, the meqs are same.Hence mass of (NH

4)

2SO

4 = 22.5 10-3 66 = 1.485g (equivalent mass of ammonium sulphate= MM/2=66)

percentage of purity = %8.921006.1

485.1

IMPORTANT: Remember that in percentage of purity problems, the mass of the impure sample is usedat the last. It is never used in the beginning while solving the problem.

SAQ 52: One gm of carbonate of an alkaline earth metal was dissolved in 50 mL of N/2 H2SO

4. The resulting

liquid required 50 mL of N/10 NaOH solution to neutralise completely. Calculate the equivalent mass of the metalcarbonate and metal.SAQ 53: 1.25g of a sample of Na

2CO

3 and Na

2SO

4 is dissoved in water and volume made to 250mL. 25 mL

of this solution required 20 mL of N/10 HCl form complete neutralisation. Calculate the % composition of thesample.SAQ 54: 1 g of impure Na

2CO

3 is dissolved in water and teh solutio is made upto 250 mL. To 50 mL of this

diluted solution, 50 mL of 0.1N HCl is added and the mixture after shaking well required 10 mL of 0.16N NaOHfor complete neutralisation. Calculate % purity of the sample of Na

2CO

3.

SAQ 55: 1 g of a sample of chalk(CaCO3) is dissolved in 230 mL of N/10 H

2SO

4. The resulting solution required

80 mL of N/20 solution form complete neutralisation. Calculate the % of CaCO3 in the sample.

SAQ 56: 1.13 g of an ammonium salt were boiled with 100 mL of N/2 NaOH till no more ammonia gas wasevolved. The excess of the alkali solution left required 60 mL of N/2 H

2SO

4 for complete neutralisation. Calculate

the % of ammonia and ammonium radical present in the salt.SAQ 57: A 12 mL of (NH

4)

2SO

4 was heated with an excess of NaOH solution. The ammonia gas evolved was

absorbed in 30 mL of 0.25N H2SO

4. The excess acid required 13.5 mL of 0.2N NaOH for complete neutralisation.

Calculate the strength of (NH4)

2SO

4 in the solution in g/L.

ANALYSIS OF MIXTURESWe shall divide this into three categories.

(a) Mixtures of an acid or a base with a salt(b) Mixture of two or more acids(c) Mixture of strong and weak bases or two weak bases(involving double indicators)

Mixture of an acid or base with a salt:This is analogous to purity problem we had discussed before. The salt present in the mixture acts like

and impurity and does not take part in the tiration reaction. Solve this SAQ.SAQ 59: 5 g of a mixture of NaCl and Na

2CO

3 were dissolved in water and volume made upto 250 mL. 25

mL of this solution required 50 mL of N/10 HCl for complete neutralisation. Calculat the % composition of theoriginal mixture.MIXTURE TO TWO OR MORE ACIDS

When we have a mixture of two or more acids, we have to add up the individual meqs. The volume ofthe mixture is to be found out. A given volume of this soluion is titrated against a standard alkali. From this thevolume or normality of any individual acid can be found out. Read this example.Eample: 10 mL of 5N HNO

3, 15 mL of 10 N HCl and a certain volume of 9M H

2SO

4 were mixed together

and the volume was made upto 2 litres. 25 mL of this diluted mixture required 35 mL of NaOH solution containing1.5 g of it in 5 mL of solution. Calculate the volume of original H

2SO

4 taken before mixing.

Solution: Meq of HNO3 = 10 5 = 50; Meq of HCl = 15 10 = 150

Meq of H2SO

4 = 18 x = 18x (let the volume of sulphuric acid is x mL)

Total Meq of acid = 50 + 150 + 18x = 200 + 18x (i)Titratiaon : 5 mL of NaOH solution contains 1.5g

So 1000 mL of solution contains 300gSo normality of the solution = 300/40 = 7.5N

25 y = 35 7.5 ( where y = normality of the acid mixture)⇒ y = 10.5N

Total Meq of acid = 2000 10.5 = 21000 (ii)From (i) and (ii), we have 200 + 18x = 21000 ⇒ x = 1155.5 mLSAQ 60: 15 mL of N/2 HCl, 25 mL of N/10 HNO

3 adn 50 mL of N/5 H

2SO

4 are mixed together. What will

be normality of the resulting mixture. This mixture is diluted to 250 mL in a measuring flask and mixed with 50mL of N/2 Ba(OH)

2 . What is the normality of the resulting mixture. Indicate whether the mixture will be acidic

or alkaline.

MIXTURE CONTAINING TWO MORE BASESCASE-I: ONE BASE ESTIMATED FROM MASS-VOLUME ANALYSIS

Example: 1.7 g of a mixture contaiing Na2CO

3, NaHCO

3 and NaCl when gently heated liberated 42 mL of CO

2

at NTP. Another 1.7g of the same mixture required 24.4 mL of 1N H2SO

4 for complete neutralisation.Caculate

the percentage composition of the mixture.Solution: You know that Na

2CO

3 and NaCl do not decompose on heating. Only NaHCO

3 decomposes to

produce NaCO3, CO

2 and H

2O.

2NaHCO3

→ Na2CO

3+ CO

2 + H

2O

2X84 g 22.4L(at NTP)22.400 mL of CO

2 at NTP is produced by 2 84 g of NaHCO

3

So, 42 mL of CO2 at NTP is prodiced by 0.315 g of NaHCO

3

Hence the mass of NaHCO3 in the mixture = 0.315 g

So the number of Meqs of NaHCO3 = (0.315/84) 1000 = 3.75

Titration: In the titration experiment, only Na2CO

3 and NaHCO

3 are bases and both react with H

2SO

4. NaCl

present in the mixture is a salt and will not react with the acid. It is as good as an impurity.Meq of acid = 24.4 1 = 24.4 = Meq of Na

2CO

3 + NaHCO

3

Meq of NaHCO3 has been already determined from the earlier experiment(3.75)

Hence Meq of Na2CO

3 = 24.4 - 3.75 = 20.65

So mass of Na2CO

3 = 20.65 10-3 53 = 1.09 g

Total mass of the Na2CO

3 and NaHCO

3 = 1.09 + 0.375 = 1.469

Hence mass of NaCl = 1.7 - 1.469 = 0.29Percentage composition can be determined.

CASE -II: MIXTURE OF A STRONG BASE AND A WEAK BASEOR MIXTURE OF TWO WEAK BASES SUCH AS CARBONATE AND BICARBONATE(NUMERICALS INVOLVING DOUBLE INDICATORS)

Example: A solution containing both NaOH and Na2CO

3 required 20 mL of N/10 HCl when titrated with

phenolphthalein as indicator. But the same volume of solution when titrated with methyl orange as indicatorrequired 30 mL of the same acid. Calculate the percentage composition of the mixture.Solution: Here NaOH is a strong base and Na

2CO

3 is a weak base. For estimating them in a mixture we have

to take the help of two indicators i.e methyl orange and phenolphthalein. When acid is added to this mixture, firstthe strong base(NaOH) is completely neutralised and then Na

2CO

3(weak base) is neutralised in to two steps.

In the first step it is half neutralised to NaHCO3 and in the second step NaHCO

3 is further half neutralised to

CO2, the salt and H

2O. So complete neutralisation of Na

2CO

3 involves the following two half neutralisations.

Na2CO3 HCl NaHCO3 N aCl (frst half neutralisation)+ +NaHCO3 H Cl NaCl C O2 H2O (second half neutralisatio

Na2CO3 HCl NaCl C O2 H2O+

+ + +

+ +2 2

Let the meq of HCl required for the 1st half neutralisation = pThe meq of HCl required for the second half neutralisation is also equal to pSo 2p meq of HCl are requierd to completely neutralise Na

2CO

3.

(SAQ: Can you tell the equivalent mass of Na2CO3 in the first reaction ?Answer : It is 106(not 53)

Phenolphthalein Indicator: When titration is done in the presence of phenolphthalein indicator the end point willreach(pink colour in alkaline solution disappears) when all strong base(NaOH) is neutralised and Na

2CO

3 is half

neutralised. The reason for this will be explained in the chapter “Theory of INDICATORS” in physical chemistry.Meq of HCl required for this end point = 20 1/10 = 2

Let the meq of NaOH in the solution = x and meq of Na2CO

3 = y in the orginal mixture.

So x + y/2 = 2 (1)Methyl Orange Indicator: When titration is done in a separate experiement with the same volume of thesolution using methyl orange as indicator, the end point is reached(colour changing from yellow to pink) whenall strong base(NaOH) is neutralised and Na

2CO

3(weak base) is completely neutralised. The reason for this also

will not be discussed here.Meq of HCl required for this end point = 30 1/10 = 3So x + y = 3 (2)

Solving equations (1) and (2), we get x = 1 and y = 2So mass of NaOH = 1 10-3 40 = 0.04 gMass of Na

2CO

3 = 2 10-3 53 = 0.106 g

So percentage of NaOH = (0.04/0.146) 100 = 27.39%; Percentage of Na2CO

3 = 72.61%

Example 2: A solution contains Na2CO

3 and NaHCO

3. 10 mL of the solution requires 2.5 mL of 0.1M H

2SO

4

for complete neutralisation using phenolphthalein as indicator. Methyl orange is then added when further 2.5 mLof 0.2M H

2SO

4 was required for the end point. Calculate the amount of Na

2CO

3 and NaHCO

3 present in one

litre of solution.

Solution: In this case, the second indicator(methyl orange) has been added to the same solution after the firsttitration using phenolphthalein indictor. In the previous numerical, the titrations were done separately using samevolume of the solutions. In this case the titrations are being done successively in the same solution.Phenolphthalein: When phenolphthalein is used as indicator, Na

2CO

3 is half neutralised while NaHCO

3 remains

as such without reaction.Let the meq of Na

2CO

3 = x ; and the meq of NaHCO

3 = y (in the original mixture)

The meq of H2SO

4 required = 2.5 0.2 = 0.5 (0.1M H

2SO

4 = 0.2N)

So, x/2 = 0.5 (1)Methyl Orange: When methyl orange is used as indicator, the 2nd half of Na

2CO

3 is neutralised and NaHCO

3

(present originally) is also neutralised.Meq of H

2SO

4 required = 2.5 0.4 = 1 ( 0.2M H

2SO

4 = 0.4N)

x/2 + y = 1 (2)Solving equations (1) and (2), we get;

x = 1 and y = 0.5 (meqs present in 10 mL of solution)Meq present in 1000 mL is Na

2CO

3 = 100 and NaHCO

3 = 50

Mass of Na2CO

3 per litre = 100 10-3 53 = 5.3 g/L

Mass of NaHCO3 per litre = 50 10-3 84 = 4.2 g/L

SAQ 61: A commercial sample(2.013g) of NaOH containing Na2CO

3 as an impurity was dissolved in water to

make 250 mL solution. A 10 mL portion of this solution required 20 mL of 0.1N H2SO

4 for complete neutralisation.

Calculate percentage by weight of Na2CO

3 in the sample.

SAQ 62: 20 mL of solution containing NaOH and Na2CO

3 when titrated against N/2 HCl using phenolphthalein

as indicator required 35 mL of HCl. The same volume of mixture when titrated against N/2 HCl using methylorange indicator required 40 mL of HCl. Calculate the amounts of NaOH and Na

2CO

3 per litre of solution.

SAQ 63: A certain quantity of a mixture of Na2CO

3 and NaHCO

3 was dissolved in water and the solution was

made upto 250 mL. 25 mL of this solution were titrated against N/10 HCl solution using phenolphthalein asindicator. It required 12.5 mL of the acid. When 25 mL of the same solution was separatey titrated with sameacid using methyl orange as indicator, 30 mL of the acid was required. Find the percentage composition of themixture.

REDOX TITRATIONThis involves a redox reaction. In this case, a solution of oxidising agent is titrated with a solution of

reducing agent. A fixed volume of solution of one of them is taken in the conical flask with the help of pipette.Suitable indicator is added if required to this solution. The other solution is addded from a burette to the solutionin the flask till the end point which is marked by a sharp colour change. From the volumes of two solutions andthe normality of any one of them, the normality of the other solution can be determined by the relation.

V1 N

1 = V

2 N

2

KMnO4– FeSO

4 Titration: Let us take 25 mL of a FeSO

4 solution in the conical flask with the help of a pipette.

We shall then add a few mL of dil H2SO

4 solution in this flask because this redox reaction is carried out in the

acidic medium. Note that the volume of acid added has no effect on effect on the titration because it is thereducing agent(FeSO4) that will react with oxidising agent(KMnO

4) in equivalent quantities. KMnO

4 solution is

added from a burette to the flask with constant shaking. The purple colour of KMnO4 vanishes as reaction

proceeds but at the end point the the purple colour persists.Therefore in this titration, no other indicator is usedas KMnO

4(OA) acts as a self indicator by change of colour from light green(FeSO

4) to purple. Note that in other

redox reactions selected indicators are used.Let the volume of FeSO

4 taken = V

1 = 25 mL

The normality of FeSO4 solution = N

1 = N/10

The volume of KMnO4 solution consumed till end point = V

2 = 30 mL(say)

Let the normality of KMnO4 solution = N2V

1 N

1 = V

2 N

2⇒ 25 N/10 = 30 N

2⇒ N

2 = N/12

Thus the normality of the KMnO4 solution is determined.

K2Cr

2O

7-FeSO

4 Titration:

N-phenyl anthranilic acid or diphenyl amine is used as indicator when K2Cr

2O

7 is used as OA. At the end point,

when K2Cr

2O

7 solution is added from burette, the green colour of the indicator changes to purple.

GUIDELINES FOR SOLVING PROBLEMS ON REDOX TITRATION:1. Write down the equation for the redox reaction and find the change in oxidation numbe per OA and RAmolecule. Then find the equivalent mass of the reagent required.2. Since the numbe of milliequivalents(meq) of OA and RA is same, meq method is also used here forcalculation as adopted in acid-base titration.3. Write “completely reacts with” in place of “completely neutralised by” in this case as this is not aneutralisation reaction.4. Back titration method is also followed here as done in acid-base titration.

Example : What volume of N/5 KMnO4 is required to oxidise 30 mL of N/10 FeSO

4 in dilute H

2SO

4 medium.

What mass of KMnO4 is present in that volume.

Solution: KMnO4 + H

2SO

4 + FeSO

4 → K

2SO

4 + MnSO

4 + Fe

2(SO

4)

3 + H

2O

The change in oxidation number of KMnO4 = +7 - (+2) =5Equivalent Mass of KMnO

4 = 158/5 = 31.6

Let the volume of KMnO4 requried = V mL

V1 1/5 = 30 1/10 ⇒ V

1 = 15 mL

The number of milliequivalent(meq) of KMnO4 = 15 1/5 = 3

Mass of KMnO4 = 3 10-3 31.6 =0.0948 g

Example : 1.5 g of a saturated solution of oxalic acid crystal(H2C

2O

4.2H

2O) were diluted to 100 mL. 20 mL

of this solution required 27 mL of N/10 KMnO4 for complete reaction. Find the solubility of oxalic acid crystal

in water in g per 100 gm of water.Solution: Let us first find out the normality of the diluted acid by the titration experiment.

27 1/10 = 20 x ⇒ x = 27/200 NMeq of oxalic acid in the entire stock solution of 100 mL = 100 27/200 = 13.5Equivalent mass of oxalic acid = M/2 = 126/2=63 as the change in ON per molecule in H

2C

2O

4 to CO

2 is 2.

Hence mass of oxalic acid = 13.5 10-3 63 = 0.85g1.5 g of saturated solution contains 0.85g of oxalic acid.Hence mass of water present in it = 1.5 - 0.85= 0.65g

0.65g of water contains 0.85g of oxalic acid crystal100 g of water contains 130.76g of acid(solubility)

SAQ 64: 10 mL of H2O

2 solution has a mass equal to 12g. It was diluted to 250 mL. 25 mL of this diluted solution

required 30 mL of N/10 KMnO4 solution. Find the mass of H

2O

2 per 100 mL solution and also the volume strength

of the solution.SAQ 65: 5.39g of a mixture of FeSO

4.7H

2O and anhydrous Ferric sulphate required 80 mL of 0.125N permanganate

solution for complete reaction to ferric sulphate. Calculate the masses of each component present in the originalmixture.SAQ 66: 1.5g of oxalic acid crystal were dissolved in water and volume was made upto 250mL. 10 mL of thissolution required 8 mL of N/10 KMnO

4 for complete reaction. Find the percentage of purity in oxalic acid.

(Do you know ?? : For the titration of Fe2+ with Cr2O

72–, n-phenyl anthranilic acid is used as indicator

which is added first to Fe2+ solution. The solution becomes green. At the end point, green colour turnsviolet-red.)

IODOMETRY-IODIMETRYThe redox reaction in which I

2 is produced is called iodometry and the redox reaction in which iodine

is consumed is called iodimetry. This phenomenon is called iodometry-iodimetry.Iodometry: Suppose we want to estimate CuSO

4 present in a sample. First it is allowed to react with execess

of KI in acidic medium to produce its equivalent amount of iodine. This is iodometry.CuSO4 + KI → K

2SO

4 + Cu

2I

2↓ + I

2

Cu2I

2 is formed as a white solution and I

2 forms a reddish brown solution. In this reaction if ‘x’ meqs of CuSO

4

is present in the sample, it will produce “x” meq of I2. Excess of the other reagent(KI) is taken for the complete

reaction CuSO4.

Other iodometry reactions: H2O

2 + KI + H

2SO

4 → K

2SO

4 + I

2 + H

2O (not balanced)

HIO3 + HI → I

2 + H

2O

Iodimetry: The liberated iodine from the above iodometry reaction is titrated agaisnt sodium thiosulphate solutiontaken in a burette. Starch solution is used as indicator in this reaction. At the end point the blue colur of theindicator is discharged(colourless).

I2 + Na

2S

2O

3 → NaI + Na

2S

4O

6 (not balanced)

Here also x meq of I2 will consume x meq of Na

2SO

3.

Hence x meq of CuSO4 = x meq of I2 = x meq of Na

2S

2O

3

Hence quantitative estimation of CuSO4 or any other oxidising agents like H

2O

2, KMnO

4, K

2Cr

2O

7, MnO

2, PbO

2

etc. can be iodimetry method.Example: 50 mL of an aqeous solution of H

2O

2 was treated with an excess of KI solution in presence of dilute

H2SO

4. The liberated iodine required 20 mL of 0.1N Na

2S

2O

3 solution for complete reaction. Calculate the

concentration of H2O

2 in g/L.

Solution: H2O

2 + KI + H

2SO

4 → K

2SO

4 + I

2 + H

2O

The liberated I2 is titrated against Na

2S

2O

3 ( I

2 + Na

2S

2O

3 → NaI + Na

2S

4O

6)

Meq of Na2S

2O

3 = 20 0.1 = 2

Hence meq of I2 = meq of H

2O

2 = 2

Mass of H2O

2 = 210-317 = 0.034g(Eq. mass of H

2O

2 = M/2 as the change in ON per molecule is 2)

50 mL solution contains 0.034gHence 1000 mL solution contains 0.68 gm (concentration in g/L = 0.68 )

SAQ 67: 0.25g of a sample of pyrolusite was heated with an excess of HCl and the chlorine evolved was passedinto a solution of potassium iodide. The liberated iodine required 80 mL of N/30 Na

2S

2O

3 solution. Calculate the

percentage of pure MnO2 present in the ore.

SAQ 68: Calculate the percentage of copper in an ore, 0.2 g of which in solution as copper sulphate yieldedsufficient iodine with excess KI solution to completely react with 12 mL of hypo solution(Na

2S

2O

3) having

strength 10g of crystalline salt(Na2S

2O

3.5H

2O) per litre.

PRECIPITATION TITRATIONYou already know from the chapter “Types of Chemical Reactions” that precipitation reaction is one of

the metathesis reactions. Suppose we add AgNO3 solution taken in a burette with a fixed volume of NaCl solution

taken in a conical flask. The following reaction takes place.AgNO

3 + NaCl → AgCl ↓ + NaNO

3

A suitable indicator(K2CrO

4) is taken and the end point is marked by its colour change from yellow to red. CrO

42–

ion is yellow but at the end point Ag2CrO

4 precipitates out which is red. Since the number of milliequivalents(meqs)

of both the reactants are same, we can estimate one of the reactants if the normality of the other is known.Example: 5.85 g of chloride of an alkali metal were dissolved in water and the solution was made to 1 litre.20 mL of this solution required 25 mL of N/12.5 AgNO

3 solution for complete precipitation. Calculate the atomic

mass of the alkali metal.Solution: 20 mL of x N alkali metal chloride(MCl) diluted soluion ≡ 25 mL of N/12.5 AgNO

3 solution.

20 x = 25 1/12.5 ⇒ x = 0.1NHence the normality of stock solution(1L) = 0.1 NMeq of alkali metal chloride = 1000 0.1 = 100Hence mass of the metal chloride = 100 10-3 E = 5.85

(where E= equivalent mass of metal chloride)⇒ E = 58.5

Since it is an alkali metal chloride its molecular mass(M)= 58.1 (total positive valency =1)So atomic mass of the alkali metal = 58.5 - 35.5 = 23 (the alkali metal is sodium)

Example: 0.2 g of a mixture of NaCl and KCl were titrated with N/10 solution of AgNO3 and required 32.3

mL for complete reaction. Calculate the percentage composition of the mixture.Solution: AgNO

3 + NaCl → AgCl ↓ + NaNO

3

AgNO3 + KCl → AgCl ↓ + KNO

3

Meq of AgNO3 used = 32.3 0.1 = 3.23

Hence meq of NaCl + KCl = 3.23Let meq of NaCl = x; hence meq of KCl = 3.23 – xMass of NaCl = x 10-3 58.5; Mass of KCl = (3.23 - x) 10-3 74.5The mass of the mixture is given to be 0.2 g.

0.0585x + 0.0745(3.23 - x) = 0.2 ⇒ x = 2.5,Hence mass of NaCl = 2.5 10–3 58.5 = 0.146 g.Percentage of NaCl = (0.146/0.2) 100 = 73.1%, So pecentage of KCl = 26.9%

Solution Stoichiometry(Volumetric Analysis)PRACTICE NUMERICAL PROBLEMS (LEVEL-1)

ACID-BASE TITRATION1. Calculate the volume of conc. H

2SO

4(density 1.835 g/mL) which is 93.2% by weight required to prepare

500 mL of N/10 acid solution. (1.432 mL)2. Exactly 75 mL of Na

2CO

3 solution is equivalent to 48.5 mL of 1.8 N/10 H

2SO

4 solution. How

many grams of CaCO3 would be precipitated if an excess of CaCl

2 solution is added to 200 mL of this Na

2CO

3solution? (1.16g)3. 58.5 mL of a sample of HCl solution requires 0.85g of Na

2CO

3 for complete neutralisation. What is the

normality of the acid? (0.274N)4. What volume of N/2 and N/10 H

2SO

4 muxt be mixed to give 3 L of N/5 H

2SO

4? (0.75L and 2.25L)

5. 25 mL of 0.75N acid is required to dissolve completely 1.5g of an impure sample of quick lime. Whatis the percentage of purity in the sample? (35%)6. What is the purity of conc. H

2SO

4(having specific gravity 1.8) if 5 mL is neutralised by 84.6 mL of 2N NaOH?

(92.1%)7. 0.325g of an acid required 31.72 mL of 1.923 N/10 of a standard base for neutralisation. Calculate theequivalent mass of the acid. (53.28)8. Calculate teh mass of each of the following required to completely neutralise 50 milliequivalents of NaOH.

(a)H2SO

4(b)CO

2(c)NaHSO

4(d)KH

2PO

4(e) N

2O

5(2.45g, 1.1g, 6g, 3.4g, 2.7g)

9. What mass of CaCO3 of 90% purity would be required to neutralise 80 mL of 0.165N acid? (0.733g)

10. 5L of NH3 gas at 270C and 700mm pressure is neutralised by 180 mL of a solution of H

2SO

4. Calculate the

normality of the acid. (1.03N)11. What is the nature of mixture(acidic or basic) and normality of the acid or base solution when 0.5g ofNa

2CO

3.10H

2O is dissolved in 150 mL of N/10 H

2SO

4? (0.076N, acidic)

12. A 70% HNO3 by weight has a density equal to 1.42 g/mL Calculate its normality and the volume of water

added to 25 mL of this acid to prepare 2N solution. What volume of this diluted solution will be required tocompletely neutralise 1.2g of 90% pure washing soda? (15.78N, 172.25 mL and 3.98 mL)13. Find the normality of the following mixture,200 mL of 0.2N HCl + 150 mL of 0.1N HNO

3 + 250 mL of 2N H

2SO

4 + 50 mL of 6N NaOH.(0.392N)

14. 3 g sample containing Na2CO

3 and NaCl is dissolved in water to make 250 mL solution. 25 mL of this

solution required 45 mL of N/10 HCl for complete neutralisation. Find the percentage composition.(Na

2CO

3=79.5% rest NaCl)

15. A 8 g sample of ammonium chloride is boiled with excess NaOH and the resulting ammonia gas is passedinto 75 mL of 0.9N H

2SO

4. Exactly 12 mL of 0.5N NaOH solution is required to neutralise the excess acid.

Determine the percentage of NH4Cl present in the sample. (41.12%)16. 25 mL of solution of Na

2CO

3 having specific gravity 1.25 required 32.9 mL of a solution of HCl containing

109.5 g of acid per litre for complete neutralisation. Calculate the volume of 0.84N H2SO

4 that will be completely

neutralised by 125g of Na2CO

3. (470mL)

17. 2.5g of an impure sample of Na2CO

3 is dissolved in water and the solution is made upto 250 mL. To

60 mL of this solution, 60 mL of 0.2N HCl is added. The mixture after shaking well, required 24.5 mL of N/10 NaOH solution. Calculate the percentage of purity of the sample. (84.32%)18. Calculate the number of H

2SO

4 molecules present in 50 mL of N/10 solution. (1.5 1021)

19. 5 g of CaCO3 was dissolved in 250 mL of 1M solution of H

2SO

4 and the solution was boiled. What

volume of M/2 Ba(OH)2 solution will be required to completely neutralise the resulting solution? (400mL)

20. Upon heating 500mL of N/10 HCl solution, 0.5gof HCl is lost and the volume decreases to 400 mL.Calculate the normality of the resulting solution. (0.09N)21. 25 mLof a NaOH solutiion requires 28.5 mL of H

2SO

4. Again 10 mL of the above NaOH solution

required 12.5 mL of 1.01 N/10 solution of succinic acid for complete neutralisation. Find out the the strength ofH

2SO

4 in g/L. (5.41 g/L)

22. How much pure and dry NaHCO3 on heating produces enough Na2CO3 to prepare 250 mL of N/10

Na2CO

3 solution. (2.1g)

23. One gram of a metal carbonate was dissolved in 25 mL of N HCl. The resulting liquid required 50 mL

of N/10 caustic potash for complete neutralisation. Calculate the equivalent mass of the metal carboante. (50)24. 0.75g of a mixture of K

2CO

3 and Na

2CO

3 required 61.25 mL of N/5 HCl for complete neutralisation.

Find the percentage composition of the mixture. (K2CO

3 = 59.6% rest Na

2CO

3)

25. Find out the volume of 0.2N HCl required to completely react with 1g of a mixture of Na2CO

3 and

NaHCO3 containing equimolar amounts of the two compounds. (78.85mL)

26. 1.2g of sample containing a mixture of Na2CO

3 and Na

2SO

3 is dissolved and the volume is made to

250mL. 25 mL of this solution neutralises 20 mL of N/10 H2SO

4. Calculate the percentage of Na

2SO

3 in the

solution. (73.5%)REDOX TITRATION27. What is the normality of a solution of a solution of H

2SO

4 made by dissolving 14g of H

2SO

4 in water

to make 500mL solution of (a)the solution is to be completely neutralised by a base (b)H2SO

4 is reduced to H

2S.

(0.57N, 2.285N)28. Find the number of milliequivalents present in one mole of each of the following underlined species inthe respective reactions.

(a) MnO4

– → MnO2

(b) H2S → SO

2

(c) Cr2O

72- → Cr3+ (d) Sn + HCl + HNO

3 → SnCl

4 + NO + H

2O

(3000, 6000, 6000, 3000)29. What mass of KMnO

4 is required to make 250 mL of 0.2N solution for the reaction of KMnO

4 with KI

in scidic medium? (1.575g)30. What mass of H

3PO

4 is required to make 500 mL of 0.75N solution for

(a)complete neutralisation with an acid (b)oxidation of H3PO

4 to HPO

32-)

(12.25g and 18.37g)31. Calculate the mass of oxalic acid(H

2C

2O

4) which can be completely oxidised by 50mL of a KMnO

4solution, 10 mL of which can completely oxidise 25 mL of N/2 KI to I

2. (2.81g)

32. What volume of 2N H2SO

4 is needed to liberate 8 litres of H

2 at STP when treated with excess zinc.

(357.14mL)33. What volume of MnO

2 is reduced by 30 mL of 0.15N oxalic acid in sulphuric acid(Mn=55) (0.196g)

34. 25 mL of iodine solution is equivalent to 0.125g of K2Cr

2O

7. To what volume shall 500mL of this iodine

solution be diluted to make N/50 solution. (Cr=52) (2.55L)35. What mass of iodine is present in a solution which requires 50 mL of N/10 Na

2S

2O

3 solution to react

completely with it? (0.635g)36. An alloy of iron weighing 4.5g were dissolved in excess of dilute H

2SO

4 and the solution was made upto

1 litre. 20 mL of this solution required 22.5mL of a certain KMnO4 solution. 25 mL of acidified oxalic acid

containing 2 g of oxalic acid crystal per litre required 18.5 mL of the above KMnO4 solution for complete

oxidation. Calculate the percentage of iron in the alloy. (36.4%)***37. A mixture of KMnO

4 and K

2Cr

2O

7 weighing 0.3g was treated with excess KI solution in presence of

dil HCl. The liberated iodine was completely reacted 50 mL of N/10 Na2S

2O

3 solution. Find the percentage of

Mn and Cr in the mixture. (Mn=2.55% andCr=24.8%)38. 0.225 g of an acid requires 25mL of 0.04M KMnO

4 solution in sulphuric acid solution. Calculate the

equivalent mass of the acid. (45)39. Calculate the solubility of copper sulphate crystal at 270C from the following data.5g of saturated solution of the salt was diluted to 100 mL. 25mL of this solution was treated with excesds ofKI. The liberated iodine required 16.1 mL of 0.08N Na

2S

2O

3 solution. (Cu=63.5) (34.58g/100g water)

40. 100 mL of ozonised oxygen at STP were passed through KI solution and liberated iodine required 14.5mL of N/10 Na

2S

2O

3 for complete reaction. What is the volume of ozone in the above volume at STP?(16.2mL)

PRECIPITATION TITRATION41. 1g of a mixture of Na

2CO

3 and K

2CO

3 was dissolved in water and made upto 250mL. 25 mL of this

solution was completely neutralised y 18.5 mL of certain HCl solution. The neutralised solution required 17.5mLof N/10 AgNO

3 solution for complete precipitation. Calculate

(a)strength of HCl in g/L (b)molarity of HCl (c)%of Na2CO

3 in the mixture.

(3.45g/L, 0.0946M, 68.7%)42. 12 mL of a solution containing H

2SO

4 and HCl required 13.5 mL of N/10 caustic soda solution for

neutralisation. On adding excess of BaCl2 to 25mL of the same solution of mixture of two acids, it gives 0.2g

of BaSO4. What is the strength of HCl in the original mixture in g/L. (Ba=137) (1.6 g/L)

43. What volume of 0.085M NaOH is required for quantitative removal of Fe3+ from 750 mL of a solution thatis 87 ppm in Fe3+? (41.05mL)44. After mixing 45 mL of 0.25M lead nitrate solution with 25 mL of 0.1M chromic sulphate solution,

precipitation of lead sulphate takes place. How many moles of lead sulphate are formed? Also calculate the molarconcentration of the species left behind in the final solution. Assume that lead sulphate is completely insoluble.

[0.0075mole, 0.0535M, Pb(NO3)

2]

Practice Problems(Level-2)1. 1 g of limestone is dissolved in excess of dilute sulphuric acid and then excess of ammonium oxalatesolution is added to precipitate all the Ca2+ as CaC

2O

4. After filtering and washing the precipitate it required 82

m mL of 0.2N KMnO4 solution acidified with sulphuric acid for complete reaction. What is the percentage of

purity in limestone. Also calculate the percentage of lime in limestone. (82%, 45.9%)2. How much of BaCl

2 would be needed to make 250 mL of a solution having same concentration of Cl–

- ions as contained in 4.5g of NaCl per 200 mL of solution. (Ba =137) (9.984g)3. In acidic solution, IO

3– reacts with I- to form I

2. What will be the final concentration of IO

3–, I– and

I2 in a solution prepared by mixing 70 mL: of 0.01M KIO

3 with 30 mL of 0.05M KI solution.

( [I-] ≈ 0; [IO3-] = 0.004M; [I

2] = 0.009M)

4. What volume of KMnO4 having strength 2 g/L will be required to completely react with 25 mL of H

2O

2

solution having strength 3 g/L. What is the strength of H2O

2 solution in terms of available oxygen?

(Hint: strength in terms of available oxygen = volume strength = “ x volumes”) ( 69.7 mL, 0.988 volumes)5. 0.356g of an alloy of Zn and Cd is precipitated as ZnS and CdS by H

2S. The mixed precipitation is

dissolved in 5 millimoles of ferric sulphate in which sulphur gets precipitated. The filtrate is acidified and tehdivalent Fe2+ required 1.6 millimoles of KMnO

4. Find the percentage of Zn in the alloy.

(Zn=65.5, Cd=112) (36.5%)6. 2.18g of a sample containing a mixture of XO and X

2O

3 require 15 mmoles of K

2Cr

2O

7 to oxidise the

sample completely to form XO4– and itself reduced to Cr3+. If 18.7 mmoles of XO

4– is formed, what is the atomic

mass? Also find the mass of XO present in the sample. (95, 1.687g)7. In an ore the only oxidisable material is Sn2+. This ore is titrated with a dichromate solution containing2.5g of K

2Cr

2O

7 in 0.5 litre. A 0.4g of sample of the ore required 10 cm3 of the titrant to reach the equivalence

point. Calculate the percentage of tin in that ore. (Sn=119) (15.17%)8. 5g of bleaching powder was suspended in water and the volume made up to half a litre. 20 mL of thissuspension when acidified with acetic acid and treated with excess of KI solution, liberated iodine which required20 mL of a decinormal hypo solution for titration. Calculate the percentage of available chlorine in the bleachingpowder. (35.5%)9. In Kjeldahl’s method, the gas evolved from 1.325g of sample of a fertilizer is passed into 50 mL of0.2030N H

2SO

4. 25.32mL of 0.198N NaOH are required for titration of unused acid. Calculate the percentage

of nitrogen in the fertilizer. (5.42%)10. A mixture of KOH and Na

2CO

3 solution required 15mL of N/20 when titrated using phenolphthalein as

indicator but the same amount of alkali mixture when titrated using methyl organe as indicator required 25mLof same acid. Calculate the amounts of KOH and Na

2CO

3 present in the solution. (0.014g, 0.053g)

11. A solution contains Na2CO

3 and NaHCO

3. 10 mL of the solution requires 2.5mL of 0.1M H

2SO

4 for

neutralisation using phenolphthalein as an indicator. Methyl organe is then added when a further 2.5mL of 0.2MH2SO4 was required. Calculate the amount of Na

2CO

3 and NaHCO

3 in litre solution. (5.3g/L, 4.2g/L)

12. One litre sample of hard water contains 1 mg of CaCl2 and 1 mg of MgCl

2. Find out the total hardness

in terms of parts of CaCO3 per 106 part of water by weight. (1.95ppm)13. 0.1g of pyrolusite was boiled with 25 mL of N/10 Na

2C

2O

4 solution in presence of 50 mL of N H

2SO

4.

The excess of oxalic acid was then back titrated with N/10 KMnO4. This required 15 mL of permanganate

solution for complete reaction. Caculate the percentage of MnO2 and active oxygen in the ore.

(87%, 16%)14. 0.108 g of finely divided copper was titrated with excess of ferric sulphate solution until copper wascompletely dissovled. The solution after the addition of excess dil. H

2SO

4 required 33.7 mL of 0.1N KMnO

4 for

complete oxidation. Find the equation which represents the reactuion between copper and ferric sulphate.(Cu=63.5)(Cu + Fe

2(SO

4)

3 → CuSO

4 + 2 FeSO

4)

15. An aqueous solution containing 0.1g KIO3(formula weight = 214) was titrated with an excess of KIsolution. The solution was acidified with HCl. The liberated iodine consumed 45 mL of thiosulphate solution todecolorise teh blue starch-iodine complex. Calculate the molarity of the Na

2S

2O

3 solution. (0.062M)

16. H2O

2 solution(20 mL) reacts quantitatively with a solution of KMnO

4(20 mL) acidified with dil. H

2SO

4.

The same volume of KMnO4 solution just decolourises by 10 mL of MnSO

4 in neutral medium simultaneously

forming a dark brown precipitate of hydrated MnO2. The brown precipitate is dissolved in 10 mL of 0.2M sodium

oxalate under boiling condition in the presence of dil. H2SO

4. Write the balanced equation involved in the reactions

and calculate teh molarity of H2O

2. (0.1M)

17. 25 mL of a solution containing 6.1 g per litre of oxalate of formula KaH

b(C

2O

4)

c. nH

2O, required for

titration 18 mL of 0.1N NaOH and 24 mL of 0.1N KMnO4 solution for oxidation. Calculate a, b, c and n.

(a =1, b=3, c=2 and n=2)18. 0.5g of fuming H

2SO

4(oleum) is diluted with water. This solution is completely neutralised by 26.7 mL

of 0.4 N NaOH. Find the % of free SO3 in the sample. (20.6%)

19. A 1 g sample of Fe2O

3 solid of 55.2% purity is dissolved in acid adn reduced by heating the solution with

zinc dust. Teh resultant solution is cooled and made upto 100 mL. An aliquot of 25 mL of this solution required17 mL of 0.0167 M solution of an oxidant for titration. Calculate the number of electrons taken up bye the oxidantin the reaction of the above titratiopn. (6)20. 0.804 g sample of iron ore(magnetic oxide) was dissolved in acid. Iron was reduced to +2 state and itrequired 47.2 mL of 0.112 N KMnO

4 solution for complete titration. Calculate the percentage of Fe and Fe

3O

4

in the ore. (36.7, 50.67)21. One gram of commercial AgNO

3 is dissoved in 50 mL of water. It is treated wtih 50 mL of a KI solution.

The silver iodide thus precipiated is filtered off. Excess of KI in the filtrate is titrated with (M/10) KIO3 solution.

in the presence of 6M HCl till all I– ions are converted to ICl. It requires 50 mL of (M/10) KIO3 solution. 20

mL of this same stock solution of KI requires 30 mL of (M/10) KIO3 under similar conditions. Calculate the

percentage of AgNO3 in the sample.

Reaction: KIO3 + 2KI + 6 HCl → 3 ICl + 3 KCl + 3 H

2O) (85%)

22. A solution of 0.2g of a compound containing Cu2+ and C2O

42- ions on titration with 0.02M KMnO

4 in the

presence of H2SO

4 consumes 22.6 mL of the oxidant. The resultant solution is neutralised with Na

2CO

3, acidified

with dil. acetic acid and treated with excess KI. The liberated iodine requires 11.3 mL of 0.05M Na2S

2O

3 for

complete reduction. Find out the molar ratio of Cu2+ to C2O

42- in the compound. Writhe down the balanced redox

reactions involved in teh above titration.23. A 3 g sample containing Fe

3O

4, Fe

2O

3 and an inert impure substance is treated with excess of KI solution

in presence of dil. H2SO

4. The entire iron is converted into Fe2+ anong with the liberation of iodine. The resulting

solution is diluted to 100 mL. A 20 mL of the diluted solution requires 11.0 mL of 0.5 M Na2S

2O

3 solution to

reduce to the iodine present. A 50 mL of the diluted solution after complete extraction of iodine required 12.8mL of 0.25M KMnO

4 solution in dilute H

2SO

4 medium for the oxidation. Calculate the percentage of Fe

2O

3 and

Fe3O

4 in the original sample.

24. A mixture of H2C

2O

4 and NaHC

2O

4 weighing 2.02g was dissolved in water and the solution made upto

one litre. 10 mL of this solution required 3.0 mL of 0.1N NaOH solution for complete neutralisation. In anotherexperiment 10 mL of the same solution in hot dilute sulphuric acid medium required 4.0 mL of 0.1N KMnO

4 for

complete reaction. Calculate the amount of H2C

2O

4 and NaHC

2O

4 in the original mixture.

25. A sample of ferrous sulphate and ferrous oxalate was dissolved in dilute sulfuric acid. The completeoxidation of reaction required 40 mL of N/15 KMnO

4. After the oxidation, the reaction mixture was reduced

by Zn and dilute H2SO

4. On again oxidation by the same KMnO

4, 25 mL were required. Calculate the mole ratio

of Fe in ferrous sulphate and oxalate. (7/3)26. A sample of MnSO

4.4H

2O is strongly heated in air. The residue is Mn

3O

4.

(i) The residue is dissolved in 100 mL of 0.1N FeSO4 containing H

2SO

4.

(ii)The solution reacts completely with 50 mL of KMnO4 solution.

(iii) 25 mL of KMnO4 solution used in (ii) requires 30 mL of 0.1N FeSO

4 solution for complete reaction.

Find the amount of MnSO4.4H

2O in the sample.

HINTS TO PRACTICE QUESTION(LEVEL II)

11111 . Back titration: meq of CaC2O

4 = 82 × 0.2 = 16.4

In the titration of CaC2O

4 with KMnO

4, eq. mass of CaC

2O

4 = MM/2. (C

2O

42- → 2CO

2 + 2e)

In the reactuion of (NH4)

2C

2O

4 with CaSO

4, the eq. mass of CaC

2O

4 = MM/2 (valency of C

2O

42- =2)

So we can say that meq. of CaCO3 = 16.4, So mass of CaCO

3 = 16.4 × 10-3 × 50 = 0.82g

So % purity = 82%.Again 100g of CaCO

3 contains 56g of CaO, So 0.82g of CaCO

3 contains 0.459g of CaO.

So % of lime = 45.9%.2. m moles of NaCl = (4.5/58.5) × 1000 = 76.92; molarity of NaCl = 76.92/200 =0.384MSo mmoles of BaCl

2 = 0.384/2 × 250 = 96/2; Mass = 96 × 10-3 × (137+71) = 19.968/2g = 9.984 g

3. KIO3 + 5KI + 6 HCl → 6 KCl + 3I

2 + 3H

2O

mmole of KIO3 = 70 × 0.01 = 0.7; mmole of KI = 30 × 0.05 = 1.5

5 mmole of KI requires 1 mmole of KIO3, So 1.5 mmole of KI requires 0.3 mmole of KIO

3

So KI is the limiting reactant. So Molarity of I– = 0Excess KIO

3 = 0.7 - 0.3 = 0.4 mmole; Molarity of IO

3- = 0.4/100= 0.004M

5 mmole of KI produces 3 mmoles of I2; So, 1.5 mmole of KI will produce 0.9 mmole of I

2

So Molarity of I2 = 0.9/100 = 0.009M

4. Normality of H2O

2 = 3/17 N ( ON of O changes from -1 in H

2O

2 to 0 in O

2, so the total change in ON

per molecule is 2 × 1 = 2 );Normality of KMnO

4 = 2/31.6 N (ON of O changes from +7 in MnO

4- to +2 in Mn2+, so the change in ON

is 5)x × 2/31.6 = 25 × 3/17 ⇒ x = 69.7 mL

H2O

2 → H

2O + ½ O

2

34g 11.2 L(NTP)3g 0.989 L So volume stength = ‘0.988 Volumes”.

5. Let the mass of Zn = x g and mass of Cd = (0.356 - x) gZn2+ + H

2S → ZnS + 2 H+

65.5 97.5x 1.488 xCd2+ + H

2S → CdS + 2H+

112 144(0.356-x) 1.285(0.356 -x)ZnS + Fe

2(SO

4)

3 → ZnSO

4 + S + 2FeSO

4

CdS + Fe2(SO

4)

3 → CdSO

4 + S + 2FeSO

4

Meq of ZnS = (1.488x/48.75) × 1000 = 30.52 x ( S2- → S0 So, Eq. Mass = MM/2)Meq of CdS = [1.285(0.356-x)]/ 72 × 1000 = 17.84 (0.356 - x)Total meq of ZnS and CdS = meq of Fe

2(SO

4)

3 consumed

So, 30.52 x + 17.84(0.356 - x) cannot be equal to 5 × 2 (Ferric sulphate is excess reactant)( In the conversion Fe

2(SO

4)

3 → FeSO

4; the total change in ON is 1 × 2 = 2)

FeSO4 + KMnO

4 + H

2SO

4 → K

2SO

4 + MnSO

4 + Fe

2(SO

4)

3 + H

2O

Meq of KMnO4 = meq of FeSO

4 = 1.6 × 5 = 8 ( change in ON is from +7 to +2 = 5)

30.52x meq. of ZnS will form 30.52x meq of FeSO4

17.84(0.356 - x) meq of CdS will form 17.84(0.356 - x) meq of FeSO4

So, 30.52x + 17.84(0.356 -x) = 8x = 0.13 g; % of Zn = (0.13/0.365) × 100 = 36.5%.

Molarity method:Since Ferric sulphate has been taken in excess, we shall calculate the number of m.moles of FeSO4

formed from the 2nd titration.

2KMnO4 + 10FeSO

4 + 8H

2SO

4 → K

2SO

4 + 2MnSO

4 + 5Fe

2(SO

4)

3 + 8H

2O

2 mmoles of KMnO4 consumes 10 mmoles of FeSO

4

1.6mmoles of KMnO4 consumes 8 mmoles of FeSO

4

The mmoles of ZnS = mmoles of Zn = (x/65.5) × 1000 = 15.26xMmoles of CdS = mmoles of Cd = [(0.356-x)/112] × 1000 = 8.92(0.356-x)mmoles of FeSO

4 produced according to balanced equation(see before)

2× 15.26 x + 2× 8.92(0.356-x) = 8, ⇒ x = 0.13g(note that a numerical can be solved by both normality and molarity method. For simple problems, often

normality method is economical, but for more complex numericals molarity method is most dependable)6. Molarity method:

XO H2O XO4- H

+ e+ 56 6+ +

Cr2O7-2

H+ e Cr

+3 H2O+ + +14 6 2 75

Cr2O7-2

XO H+

XO4- Cr

+3 H2O+ + + +5 6346 10 17

3

X2O3 H2O XO4- H

+ e+ +

+Cr2O7-2

H+ e Cr

+3 H2O

+

+ +14 6 2 7

Cr2O7-2

X2O3 H+ XO4

- Cr

+3 H2O+++ 6

3 5 2 10 8

4

34 26+ 8 13

6 mmols of XO reacts with 5 mmoles of Cr2O

72-,

p 1000(q+ 16)

of XO reacts with 5/6 × p 1000(q+ 16)

mmoles of Cr2O

72-,

3 mmols of X2O

3 reacts with 4 mmols of Cr

2O

72-,

2.18 - p2q + 48 1000 of X

2O

3 reacts with 4/3 ×

2.18 - p2q + 48 1000 mmols of Cr

2O

72-,

Hence 5/6 × p 1000(q+ 16)

+ 4/3 × 2.18 - p2q + 48 1000 = 15 (1)

Again, 6 mmoles of XO produces 6 mmols of XO4- ;

p 1000(q+ 16)

of XO produces p 1000(q+ 16)

mmoles of XO4

- ;

3 mmols of X2O

3 produces 6 mmols of XO

4- ;

2.18 - p2q + 48 1000 of X

2O

3 produces 6/3 ×

2.18 - p2q + 48 1000 mmols of XO

4- ;

Hence p 1000(q+ 16)

+ 6/3 × 2.18 - p2q + 48 1000 = 18.7 (2)

Solving equations (1) and (2) simultaneously, we get, p=1.687g, q = 99(atomic mass)(you can try solving this numerical by normality method yourself and see which one is easier)

7. Normality = 5

294/6= 0.102N , meq of Sn2+ = 10 × 0.102 = 1.02

Mass of Sn = 1.02 × 10−3 × 119/2 = 0.0607 g ( Sn2+ → Sn4+ )% of Sn = (0.0607/0.4) × 100 = 15.17

8. Meq of I2 in 20 mL = 20 × 0.1 = 2; mmole of I2 = 1 ( Eq. Mass = M.M/2)

Ca(OCl)Cl + 2CH3COOH + 2KI → I

2 + Ca(CH

3COO)

2 + 2 KCl + 2H

2O

So the number of mmol of Ca(OCl)Cl in 20 mL solution = 12500 mL of solution contains 25 mmoles; Hence mass = 25 × 10−3 × 127 = 3.175gCa(OCl)Cl → CaO + Cl

2

127 71127g bleaching powder can produce 71g Cl

2

3.175g bleaching powder can produce 1.775g Cl2

Hence % of available chlorine = (1.775/5) × 100 = 35.5(IMPORTANT: In many cases, it is much easier to solve the titration problem by molarity method involvingbalanced equation rather than by normality method involving equivalents. This is because sometimes it becomesdifficult to find the equivalent mass of substances particularly in complex redox reactions. In such cases, molaritymethod using balanced equation is the sure sort method.)9. Back Titration: Meq of excess of acid = 25.32 × 0.198 = 5.013

Meq of acid initially used = 50 × 0.203 = 10.15So meq of acid consumed by NH3 produced from the fertiliser = 5.137;Mass of NH

3 = 5.137 × 10−3 × 17 = 0.0873 g

17g of NH3 contains 14g of N; So 0.0873 g of NH

3 contains 0.07189g of N

Hence % of N = (0.07189/1.325) × 100 = 5.42%10. Problem based on double indicators

meq of KOH + ½ (meq of Na2CO

3) = 15 × 1/20 = 0.75; (phenolphthalein)

x + y = 0.75 (1)meq of KOH + mq of Na

2CO

3 = 25 × 1/20 = 1.25 (methyl orange)

x + 2y = 1.25 (2)Solving the above two equations simultaneously , we get x = 0.25 and y = 0.5So meq of Na

2CO

3 = 2y =1; Mass of Na

2CO

3 = 1 × 10-3 × 53 = 0.053g

Meq of KOH = x = 0.25; Mass of KOH 0.25 × 10-3 × 56 = 0.014g11. Meq of 1/2 meq of Na

2CO

3 = x = 2.5 × 0.2 = 0.5 (phenolphthalein)

Meq of 1/2 (meq of Na2CO

3) + meq of original NaHCO

3 = x +y = 2.5 × 0.4= 1 (methyl orange)

Meq of Na2CO

3 = 2 × 0.5 = 1 ( in 10 mL solution)

Meq of original NaHCO3 = 0.5 (in 10 mL solution)

Meq of Na2CO

3 in 1 L = 100; Mass of Na

2CO3 = 5.3g/L

Meq of original NaHCO3 in 1 L = 50; Mass of NaHCO

3 = 4.2 g/L

12. Permanent hardness which is partly due to the presence of CaCl2 and MgCl

2 in water can be removed

by adding Na2CO

3 solution in the form of sparingly soluble CaCO

3 and MgCO

3 precipitate.Conveniently the

hardness is expressed in terms of parts of CaCO3 per million parts of water.

CaCl2 + Na

2CO

3 → CaCO

3 + 2NaCl

1 mmol 1mmol0.009mmol 0.009mmolMgCl

2 + Na

2CO

3 → MgCO

3 + 2NaCl

1 mmol 1mmol0.0105mmol 0.0105mmolIf we consider the precipiate in terms of CaCO

3 (presuming MgCO

3 to be equal to CaCO

3)

mmols of CaCO3 = 0.009 + 0.0105 =0.0195; Mass of CaCO

3 = 0.00195g

1 L(1000g) of water contains 0.00195g of CaCO3

106 g of water contains 1.95 g of CaCO3 (1.95 ppm )

13. MnO2 + Na

2C

2O

4 + H

2SO

4 → MnSO

4 + CO

2 + Na

2SO

4 + H

2O

Back Titration: Meq of excess acid = 15 × 0.1 = 1.5 = meq of excess Na2C2O4Total meq of Na

2C

2O

4 used = 25 × 0.1 = 2.5 ; Hence meq of Na

2C

2O

4 consumed by MnO

2=1=meq

of MnO2; Hence mass of MnO

2 = 1 × 10-3 × 87/2 = 0.0435 g; % of MnO

2= 0.0435/0.1 × 100 = 43.5%

MnO2 → MnO + ½ O

2

87 160.0435g 0.008g; So % of active oxygen = 8%

14. Cu + Fe2(SO

4)3 → Copper sulphate + FeSO

4 (since we do not know the valency of Cu)

meq of FeSO4 = 33.7 × 0.1 = 3.37 = meq of Cu; Hence mass of Cu = 3.37 × 10-3 × 63.5/x =0.107

So x =2 (valency of Cu in copper sulphate formed in the above rection).15. Back Titration: meq of I2 = 45 × x (where x is the normality of Na

2S

2O

3 solution)

meq of KIO3 = 45x; mass of KIO

3 = 45x × 10-3 × 35.6 = 0.1 Hence x = 0.062N = 0.062M

(note that equivalent mass of KIO3 = MM/6. IO

3- → I -; change in ON = 6. For Na

2S

2O

3, molarity

is equal to normality as change in ON of S = 1)Molarity Method:

2Na2S

2O

3 + I

2 → Na

2S

4O

6 + 2NaI

mmols of Na2S

2O

3 = 45 × x (let x = molarity); mmols of I

2 = 45x/2

KIO3 + 5KI + 6HCl → 3I

2 + 6KCl + 3H

2O

mmols of KIO3 = 45x/6; Mass = 45x/6 × 10-3 × 214 = 0.1; Hence x= 0.062M

Note that the molarity method gives the result easily in this case.16. Let the molarity of H

2O

2 = x;

Normality method: MnO2 + Na

2C

2O

4 + H

2SO

4 → MnSO

4 + CO

2 + Na

2SO

4 + H

2O

Meq of MnO2 = 10 × 0.2 × 2 = 4 ( C

2O

42- → CO

2; total change in ON =2)

Meq of KMnO4 = 4; Normality = 4/20; 20 × x = 20 × 4/40, ⇒ x = 0.2N

Molarity of H2O

2 = 0.2/2 = 0.1M ( since H

2O

2 → O

2, total change in ON =2)

Molarity Method: MnO2 + Na

2C

2O

4 + 2H

2SO

4 → MnSO

4 + 2CO

2 + Na

2SO

4 + 2H

2O

mmols of Na2C

2O

4 = 10 × 0.2 = 2; mmols of MnO

2 =2

2KMnO4 + 3MnSO

4 + 2H

2O → 5MnO

2 + K

2SO

4 + 2H

2SO

4 (balance by partial equation method)

mmols of KMnO4 = 2/5 × 2 = 4/5

2KMnO4 + 3H

2SO

4 + 5H

2O

2 → K

2SO

4 + 2MnSO

4 + 5O

2 + 8H

2O

mmols of H2O2 = 5/2 × 4/5 = 2; Hence molarity of H2O2 = 2/20 = 0.1M (answer)(The molarity method in this case appears to be more lengthy while normality method is very short. So whilesolving numericals you have to judge which method will be economical and easy)17. The oxalate is acting both as an acid having ‘b’ number replaceable H atoms and also a reducing agentcontaining ‘c’ number number of C

2O

42- ion in which the total change in ON of is 2c. There are two titrations

in this case- one acid-base and the other redox.Acid-base titration: 25 × N = 18 × 0.1 ⇒ x = 0.072N; Equivalent mass = 6.1/0.072 = 84.7

MM/b = 84.7 (1)Redox titration: 25 × N = 24 × 0.1 ⇒ x = 0.096N; Equivalent mass = 6.1/0.096 = 63.5

MM/2c = 63.5 (2) (since the total change of ON per molecule = 2c)a + b = 2c (3) (charge balance)Dividing eqn. (1) by (2), we get; 2c/b = 1.3338 ⇒ b = 1.5c (4)b:c = 3:2; Again a + b = 2c ⇒ a + 1.5c = 2c ⇒ a = 0.5c; So c:a = 2:1So a:b:c = 1:3:2; Hence the empirical formula is KH

3(C

2O

4)

2

(note that in ionic compounds like the above, the formula we talk about is really empirical formula, sinceit is an network solid, the actual molecular mass is infinitely large)

Empirical formula mass = 84.7 × b = 254.1; Part mass of of KH3(C

2O

4)

2 = 218;

So part formula mass containing nH2O = 254.1 - 218 = 36.1, So the number H

2O = 26.1/18 =2

So the empirical formula of the hydrate salt = KH3(C

2O

4)

2.2H

2O

18. SO3 + 2NaOH → Na

2SO

4 + H

2O; H

2SO

4 + 2NaOH → Na

2SO

4 + H

2O

Let SO3 = x g and H

2SO

4 = (0.5-x)g

x80/2 10

3 (0.5-x)

49

310 = 26.7 4+ ⇒ x = 0.103g (SO

3); % = 0.103/0.5 × 100 = 20.6

19. Mass of Fe2O

3 = 55.2/100 × 1 = 0.552g;

Fe2O

3 + H

2SO

4 → Fe

2(SO

4)

3 + H

2O; Fe

2(SO

4)

3 + Zn → FeSO

4 + ZnSO

4

Back titration: 25 × x = 17 × 0.0167 × y (where x = normality of FeSO4 solution and y = no. of electons

gained by the oxidant); ⇒ x = 0.0113 y (1)Meq of FeSO

4 in 100 mL solution = 100 × 0.0113y = 1.13 y = meq of Fe

2(SO

4)

3 (since the change

of ON is 1 for the reduction of Fe3+ to Fe2+)Mass of Fe

2O

3 = 1.13y × 10-3 × 160/2 = 0.0904y =0.552 (Fe

2O

3 → Fe2+, the total change in ON =2)

⇒ y = 6.1 = 6 ; Using the value of y in eqn. (1) we get x = 0.0678N20. Meq of Fe2+ = 47.2 × 0.112 = 5.28; Mass of Fe = 5.28 × 10-3 × 56 = 0.295g

So % Fe = 0.295/0.8 × 100 = 36.7Fe

3O

4 → 3Fe3+/2+

232 168;168g of Fe3+/2+ is produced by 232g of Fe

3O

4;

0.295g of Fe is produced by 0.407g of Fe3O

4; % of Fe

3O

4 = 50.67%

21. Blank titration(standardisation)The last titration is meant to find the strength of the stock KI solution.

Molarity method: KIO3 + 2KI + 6HCl → 3ICl + 3KCl + 3H

2O

mmols of KIO3 used = 30 × 0.1 = 3; Hence the mmols of KI consumed = 6

20 mL of KI solution contains 6 mmols of KI. So molarty = 6/20 MInvolved titration: (back titration)

mmols of KIO3 used = 50 × 0.1 = 5; mmols of KI consumed = 5 × 2 =10 = mmols of excess KI;

mmols of KI originally present in 50 mL of the stock solution = 50 × 6/20 = 15So mmols of KI consumed by AgNO

3 = 15 - 10 = 5

KI + AgNO3 → AgI + KNO

3; mmols of AgNO

3 = 5, Mass = 5 × 10-3 × (108+14+48) =0.85g

Hence % purity = 85%(Note that data 6M H

2SO

4 is not involved in calculation. It is easier to solve this numerical by molarity method)

Normality method:

Blank titration: 20 × x = 30 × 0.1 × 4 ( IO 3-

ICl+5 +1

; change in ON =4); x = 12/20 N

Involved titration: meq of excess KI = 50 × 0.1 × 4 = 20, meq of KI used = 50 × 12/20 = 30meq of KI consumed by AgNO

3 = 10; So mmols of KI used = 5 ( here I- → I+, change in ON=2)

AgNO3 + KI → AgI + KNO

3, So mmols of AgNO

3 = 5, mass = 0.85 (as before)

(note that meq of KI used in the reaction with KIO3 is not equal to meq of AgNO3, as in KI to ICl conversion,

the change in ON is 2, while the reaction of AgNO3 with KI, there is no change in ON(metathesis reaction) and

the equivalent mass is equal to molecular mass divided by total +ve/or -ve valency(1). We find here that normalitymethod can mislead us in getting the correct answer. So the molarity method is dependable in this case)22. 1st titration: meq of C

2O

42- = 0.2 × 5 × 22.6 = 2.26

C2O

42- + H+ + MnO

4– → CO

2 + Mn2+ + H

2O; mmols of C

2O

42- = 2.26/2 = 1.13(EM=MM/2)

2nd titration: meq of I2 = 11.3 × 0.05 × 1 = 0.565 = meq of Cu2+

Cu2+ + I– → Cu22+ + I

2, no of mmols of Cu2+ = 0.565 ( EM = MM/1)

mole ratio of Cu2+ : C2O

42- = 0.565 : 1.13 = 1:2

23. Fe3O

4 = FeO.Fe

2O

3 = x g; and Fe

2O

3 = y g (say), So impurity = (3-x-y) g

In Fe3O

4 there is equimolar amounts of FeO and Fe

2O

3

Let FeO = z g , So Fe2O

3 = x-z; z/72 = (x-z)/160 ⇒ z = 0.31 x

1st titration: Fe2O

3 + H+ + I– → Fe2+ + I

2 + H

2O

20 × x = 11 × 0.5 ⇒ x = 0.275N (normality of 100 mL of I2 solution)

meq of I2 = 100 × 0.275 = 27.5 = meq of Fe

2O

3; Mass of Fe

2O

3 = 27.5 × 10-3 × 0160 = 2.2g

( Fe2O

3→ Fe2+ and the change in ON per molecule is 2)

Total Fe2O

3 = y + x -z = 2.2 ⇒ y + x - 0.31x = 2.2;⇒ 0.69x + y = 2.2 (1)

2nd titration: Fe2+ + MnO4

- + H+ → Fe3+ + Mn2+ + H2O;

50 × x = 12.8 × 0.25 × 5 ⇒ x = 0.32N; Meq of Fe2+ = 0.32 × 100 = 32Mass of total Fe2+ = 32 × 10-3 × 56 = 1.792g(total iron)Let us find total Fe2+ in terms of x and y.

Fe3O

4 → 3Fe2+

232g of Fe3O

4 produces 3 × 56 =168g of Fe2+; So x g of Fe

3O

4 produces 0.724x g of Fe2+

Fe2O

3 → 2Fe2+

160g of Fe2O

3 produces 2 × 56 =112g of Fe2+ , So y g of Fe

2O

3 produces 0.7y g of Fe2+

0.724x + 0.7y = 1.792 (2)Solving equations (1) and (2) we get x = 1.04g and y = 1.48 g% Fe

3O

4 = 34.6 %; and Fe

2O

3 = 49.33%

24. Let the mass of H2C

2O

4 = x g; So the mass of NaHC

2O

4 = (2.02 - x) g

1st titration: meq of H2C

2O

4 in 1000mL solution = x/45 × 1000 = 22.22x (dibasic acid)

meq of NaHC2O

4 in 1000mL solution = (2.02-x)/112 × 1000 = 8.928(2.02-x) (monobasic acid)

In 10 mL, the meq of H2C

2O

4 = 0.222x and meq of NaHC

2O

4 = 0.089(2.02-x)

0.222x + 0.089(2.02-x) = 3 × 0.1, ⇒ x = 0.909g(H2C

2O

4), Mass of NaHC

2O

4 = 1.11g

2nd titration: We have already calculated the amounts of the two components. The second titration is superfluous.Sometimes to verify whether one experiment is correct or not, a second experiment is done. WE shall find thesame answer by using these data.

meq of H2C

2O

4 = x/45 × 1000 = 22.22x ( C

2O

42- → 2CO

2 + 2e)

meq of NaHC2O

4 = (2.02-x)/56 × 1000 = 17.856(2.02-x) (HC

2O

4– → 2CO

2 + H+ + 2e)

In 10 mL the meqs are 0.222x and 0.1785(2.02-x) respectively.0.222x + 0.1785(2.02-x) = 4 × 0.1 ⇒ x = 0.909g(H

2C

2O

4), Mass of NaHC

2O

4 = 1.11g

25. 1st titratration: Let the mmoles of FeSO4 = x; mmoles of FeC

2O

4 = y; In the first titration, Fe2+ from

both salts and C2O

42- from the second salt have reacted with KMnO

4.

meq of KMnO4 = 40/15 = 8/3 = x + y + 2y (1) (Since C

2O

42- → 2CO

2 + 2e)

x and y are meqs of Fe2+ from both the salts and meq of C2O

42- = 2x, but in case of Fe2+ the normality

is same as molarity)2nd titration: Only Fe2+(total) has been oxidised by KMnO

4;

x + y = 25/15 = 5/3 (2)Solving equations (1) and (2) simultaneously we get y = 1/2 and x = 7/6 Hence x:y = 7:3

26.

MnSO4. H2O 4 4+

3

MnSO4 H2O

MnSO4 Mn3O4 SO2 O2+ +3

Mn3O4 FeSO 4 H2SO4 MnSO4 Fe2(SO4)3 H2O+ + + +

+8/3 +2(1)

KMnO4 + FeSO

4 + H

2SO

4 → K

2SO

4 + MnSO

4 + Fe

2(SO

4)

3 + H

2O (2)

In the 2nd reaction, KMnO4 reacts with the excess FeSO

4 present after the 1st reaction.

3rd reaction is same as reaction (2), but this is standardisation of KMnO4 ( blank titration).

25 × x = 30 × 0.1 ⇒ x = 0.12N (normality of KMnO4)

meq of excess of FeSO4 = 50 × 0.12 =6; meq of FeSO

4 initially used = 100 × 0.1 = 10

meq of FeSO4 consumed by Mn

3O

4 = 4

Mn3O

4 + 4 H

2SO

4 + 2FeSO

4 → 3 MnSO

4 + Fe

2(SO

4)

3 + 4 H

2O

2 mmols of FeSO4 consumes 1 mmole of Mn

3O

4

4 mmols of FeSO4 consumes 2 mmols of Mn

3O

4

3MnSO4.4H

2O → Mn

3O

4

1 mmol of Mn3O

4 is produced by 3 mmols of MnSO

4.4H

2O

2 mmols of Mn3O

4 is produced by 6 mmols of MnSO

4.4H

2O

Mass of MnSO4.4H2O = 6 × 10-3 × 0(55+96+72) = 1.338g(Note that in the 2nd reaction, KMnO

4 might oxidise MnSO

4 along with the unreacted FeSO

4. But the question

does not speak about that. The data given says that KMnO4 only reacts with the unreacted FeSO

4. Moreover the

conversion of MnSO4 to MnO

2 can only occur in the presence of excess KMnO

4. This reaction can be skillfully done

till the green colour of Fe2+ ion is discharged. MnO2 is a sparingly soluble brown preciipate which can only be formed

with excess of KMnO4 with Mn2+).

VERY SHORT ANSWER QUESTIONS

1. Find the equivalent mass of Cu2S in the following reaction. Cu2S + O

2 → CuO + O

2 (Cu=63.5)

2. 2M solution of KBrO3 is ____N, in the reaction BrO

3– → Br–

3. In iodometry, iodine is liberated by a redox reaction while in iodimetry iodine is consumed by sodiumthiosulphate solution. (True or False)4. CuSO

4 reacts with KI to liberate ____ and _____ alongwith K

2SO

4.

5. The equivalent mass of hydrazine(N2H

4) in its conversion to N

2 is _____.

6. Correct the statement is there is any mistake. “Chemical reactions always takes place in equimolaramounts of the reactants”7. How much pure HCl is present in 1000g of 60% HCl by weight?8. Cr

2O

3 + Na

2O

2 → Na

2CrO

4; In the above hange 1 mole of Cr

2O

3 is how many equivalents of Cr

2O

3?

9. If 1500 mL of H2O

2 produces 45 L of O

2 gas at NTP, what is the volume strength of H

2O

2 on the basis

of available oxygen?10. What normality of 0.3M H

3PO

3?

11. 500g of water is mixed with 49g of pure H2SO

4. What is the molarity of the solution?

12. Find the volume of 2M H2SO

4 solution required to neutralise 10 mL of 0.5M NaOH.

13. 1 L of 18M sulphuric acid is diluted to 50 L. What is the normality of the diluted solution?14. What amount of NaHCO

3 required to prepare 100 mL of N/10 solution (MM=84)

SHORT ANSWER QUESTIONS

1. What is the mass of pure HNO3 required to convert 2 g of I2 to iodic acid?

2. For the reaction between SnCl2 and FeCl

3, find the normality of SnCl

2(Sn=119) solution prepared by

dissolving 19 g of SnCl2 in acidic solution and diluting to 500 mL with water.

3. What volume of 0.2M Na2S

2O

3 would be required to raect with iodine liberated by adding excess of KI

to 50 mL of 0.5M CuSO4 solution?

4. Find the equivalent mass of NaOCl in reaction; NaOCl + SO2 + H

2O → NaCl + H

2SO

4

5. Find the volume of 30%(by volume) HCl solution needed to dissolve oen mole of read lead.6. Find the mass of KMnO

4 required to prepare 250mL of N/10 solution. Is this question correct? If not

why?7. Find the volume of 0.2M KMnO

4 solution required to oxidise 200mg of FeC

2O

4 in acid solution.

8. Find the volume of O2 at NTP liberated by the action of 100mL of 0.1N KMnO

4 solution on excess of

H2O

2 in acid medium.

9. Find teh volume of N/10 K2Cr

2O

7 solution required to react completely with 0.34g of H

2S in acidic

medium.10. How many equivalents of reducing agent are present in 100mL of 0.2M Na

2S

2O

3 in the reaction

Cl2(g) + S

2O

32- → SO

42- + Cl–

11. What mass of MnO2(MM=87) is reduced completely by 20 mL of 0.2M oxalic acid in acidic solution.

12. Find the number of molecules of Na2CO

3 in 500 mL of 0.05N solution.

13. Find the total ionic strength(total molarity of all ions) of 0.2M ZnSO4 and 0.1M Al2(SO

4)

3

14. Find the normality of 10%(w/V) solution of ammonia(density = 0.34 g/cc).15. 0.16g of a dibasic acid requires 25mL of a decinormal NaOH solution for complete neutralisation. Findthe molecular mass of the acid.16. I

2 is violet but in solution in presence of KI it forms a dark brown solution. Explan.

17. Find the normality of acid solution obtained by mixing 50 mL of M/2 H2SO

4 and 100 mL of M/5 HCl

solution.18. Vinegar is 5.11% acetic acid by weight. Its density is 1.007g/mL. What is its molarity?

ANSWER TO SAQs

SAQ 1: H3PO

4 = MM/3 = 98/3=32.67 (tribasic acid)

HNO3 = MM/1; H

2CO

3= MM/2, H

3PO

2 = MM/1 = 66 (monobasic acid)

H3PO

3 = MM/2 = 90/2 = 45 (dibasic acid); H

2C

2O

4 = MM/2 = 90/2=45

SAQ 2: KOH = MM/1 = 56; Mg(OH)2 = MM/2 (diacidic)

MgO = MM/2 (MgO + H2O → Mg(OH)

2 , so 2 OH- are produced)

Li2O = MM/2 30/2=15 (Li

2O + H

2O → 2LiOH, so 2 OH- are produced)

BaO = MM/2(like MgO); NH4OH = MM/1;

NH3 = MM/1 ( NH

3 + H

2O → NH

4OH )

SAQ 3: Na2CO

3 = MM/2 = 106/2=53 (total +ve valency = 2 × 1=1)

CaCl2 = MM/1 = 111/2=55.5 ( total +ve valencfy = 1 × 2 = 2)

AlBr3 = MM/3; BaSO

4 = MM/2 Na

3PO

4 = MM/3

SAQ 4: Acidic medium: MnO4- → Mn2+ (change in ON = 5)

EM = MM/5 = 158/5= 31.6Alkaline medium MnO

4- → MnO

42-(manganate) (change in ON =1)

EM = MM/1 = 158Weakly alkaline or neutral medium : MnO

4- → MnO

2(manganese dioxide) change in ON = 3

EM = MM/3 = 158/3 = 52.675(A). (i) E(H

2SO

4) = MM/1=98 (as only 1 H atom has been replaced in the reaction)

(ii)E(H3PO

4) = MM/2 =98/2(as 2 H atoms have been replaced)

(iii)CuSO4 → Cu

2I

2(change in ON =1), So E(CuSO

4) = MM/1 = 159.5

(iv)Fe2O

3 → Fe (Change in ON per Fe atom = 3, for two Fe atoms present in Fe

2O

3, the total change

in ON = 6); Hence E(Fe2O

3) = MM/6 = 26.67

(v) FeSO4 → Fe

2(SO

4)

3; Here change in ON = 1 (note that only the reactant is to be considered)

E(FeSO4) = MM/1 = 152.

5(B). (i) H2C2O4 CO 2

+3 +4The change in ON per carbona atom = 1, So for 2 carbon atoms present

in the reducing agent, the total change in ON = 2; Hence EM of H2C

2O

4 = MM/2= 90/2 =45

You can also write the balanced ionic equation and take the number of electrons lost or gained permolecule to find the equivalent mass.

H2C

2O

4 → 2CO

2 + 2H+ + 2e; Hence EM = MM/2

(ii) I2 I0 -1

The change in ON per I atom = 1, so the total change for 2 I atoms = 2

So EM = MM/2 = (2×127)/2 = 127 (same as its atomic mass)

Na2S2O3 Na2S4O6

+2 +2.5; The change in ON per S atom = 0.5; So the total change per molecule =

1; Hence EM = MM =158(iii) Method for determining Equivalent Mass in Disproportionation Reaction::

EM =.MM/(No. of electrons lost or gained per molecule ) = MM/Valency FactorHere Valency Factor = No. of electrons lost or gained per molecule. This can be found out easilty by writing

the oxidation and reduction steps in ion-electron method or by electron-balance-diagram(EBD) method.Br

2 + 12OH– → 2BrO

3– + 6H

2O (oxidation)

[Br2 + 2e → 2Br–] × 5 (reduction)

_____________________________________________6Br

2 12OH– → 2BrO

3– + 10 Br– + 6H

2O

For 6 Br2 molecules, the number of electrons lost/gained = 10, So for 1 Br

2 molecule, the number of electrons lost

= 10/6. Hence the Valency Factor = 10/6 = 5/3EM = MM of Br

2/(5/3) = 160 X (3/5) = 96

(iv) Fe3O4 Fe2O3

83

++3

; The total change in ON per molecule = 3 × 1/3 = 1; So EM= MM=232

KMnO4 MnO 2

+7 +4; The change in ON per molecule = 3; Hence EM = MM/3 = 158/3 = 52.67

(v) N2 NH3

0 -3

; The total change for two N atoms in the molecule = 2 × 3 =6

So EM = MM/6 = 28/6 = 4.67

(vi) KBrO3 Br+5

; change in ON = 6; EM = MM/6 = 167/6 = 27.83

(Note that BrO3- is reduced to Br- (not Br

2), although there is no Br- in the product side. Since all Br- are oxidised

to Br2, we get Br

2 as the only product from both BrO

3- and Br-. Since 6 electrons are gained by one BrO

3- ion, 1

mole of BrO3- is equal to 6 equivalents. Often students commit mistake by taking its EM as MM/5)

(vii) H2O2 O2

-1 0; The total change in ON per molecule = 1 × 2 =2; Hence EM = MM/2 =34/2=17

(viii) H2O

2 → 2H+ + O

2 + 2e (oxidation)

H2O

2 + 2H+ + 2e → 2H

2O (Reduction)

______________________________2H

2O

2 → 2H

2O + O

2

For 2 H2O

2, the number of electrons lost/gained = 2, So for 1 H

2O

2 it is 1. Hence here the valency factor is 1.

EM = MM/1 = 34/1 =34 (Disproportionation reaction)(Note that in reactions in which H

2O

2 is either OA or RA, the valency factor is 2)

6.(a) (i) eqs = 4/40 = 0.1, meqs = 0.1 × 100 =100; (ii) eqs = 1000/50 =20; meqs = 20,000(iii) eqs = 1.52/152 = 0.01; meqs = 10(b) (i)mass = 1000 × 10-3 × 49 = 49g (ii) mass = 0.5 × 49 = 24.5g; (iii) 20 × 10-3 × 36.5 = 0.73g

7. EM of KMnO4 = MM/3; So 1 mole = 3 equivalents(eqs); In other words 1 eq = 1/3 mole

Molarity = 1.5 × 1/3 = 0.5M

8. HNO3 NO+5 +2

; Change in ON =3, So EM of HNO3 = MM/3 =63/3 = 21

9. (a) 2NaOH + H2SO

4 → Na

2SO

4 + H

2O (neutralisation); EM = MM/2; Hence Normalty =

Molarity × 2 = 4N

(b) H2SO4 H2S+6 -2

(redox reaction); Change in ON = 6 - (-2) = 8

So Normality = Molarity × 2 = 2 × 8 = 16N

10. K2Cr2O7 Cr+3+6

; The total change in ON per molecule = 2 × 3 =6; Normality = 0.25× 6=1.5N

11. MM of Mohr salt = 392; In the salt, only FeSO4 reacts while the other two components i.e (NH

4)

2SO

4 and

H2O do not take part in reaction.

FeSO4 Fe2(SO4)3

+2 +3 The change in ON per molecule = 1, So EM of Mohr salt = MM/1 =392

Normality = meq/mL = 1000/100 = 10N ( As 392g = 1 eq.)(note that for calculating the equivalent mass the molecular mass of the whole salt is taken although a part

of the molecule does not take part in the reaction)

12. H2C2O4 CO2

+3 +4; The total change in ON = 2 × 1 = 2; EM = MM of H

2C

2O

4.2H

2O

2 = 126/2= 63

50 mL of solution contains 0.56 g of oxalic acid crystal;So 1000 mL of the solution contains 11.2g of the crystal; No of eqs. = 11.2/63 = 0.178Normality = 0.178N; Molarity = Normality/2 = 0.178/2= 0.089M

13. (a) KMnO4 → Mn2+; change in ON =5; So Normality = 0.05 × 5 = 0.25N

(b) FeSO4 → Fe

2(SO

4)

3; change in ON = 1, So Normality = 5 × 1 = 5N

(c) Na2S2O3 Na2S4O6

+2 +2.5

; total change of ON per molecule = 2 × 0.5 = 1

So Normality = 0.2 × 1 = 0.2N14. (a) 500 × 1/10 = 50 (b) 2500 × 1/20 = 125 (c) 250× 2 = 50015. (a) meqs = 1.06/53 × 1000 = 20 (b) meqs = 7/49 × 1000 = 142.85

(c) meqs = 0.25/40 ×1000 = 6.2516. mass = meqs × 10-3 × EM (a) 250 × 10-3 × 49 = 12.25g (b) 100 × 10-3 × 50 =5g

(c) 50 × 10-3 × 37 = 1.85g17. meqs = 10/85.5 × 1000 = 116.96; [EM of Ba(OH)

2 = MM/2 = 171/2 = 85.5]

Hence normality = 116.96/2000 = 0.058NAlternatively: 2000 mL of solution contains 10/85.5 eqs

1000 mL of solution contains 0.058 eqs (normality = 0.058N)18. meqs = 100 × 0.5 = 50; mass = 50 × 10-3 × 49 = 2.45gAlternatively: 0.5N means 1000 mL of solution contains 0.5 equivalents = 0.5 × 49 = 24.5g

So 100 mL of the solution 2.45g19. meq = 0.49/49 × 1000 = 10; Volume = meq/N = 10/0.1 = 100 mLAlternatively:” 0.1 equivalent = 0.1 × 49 = 4.9g of H

2SO

4 is present in 1000 mL

So 0.49 g of H2SO

4 is present in 100 mL

20. Molarity = Normality/basicity (a) M = 0.25/2 = 0.125M (b) 0.05/1 = 0.05(c) 1.5/4 = 0.5M (d) 1/2 = 0.5M (H

3PO

3 is a dibasic acid) (e) 0.2/1 =0.1(H

3PO

2 is a

monobasic acid) (f) 5/1= 5M (g) 0.1/2 = 0.05M21. Normality = Molarity × bascity (a) 5× 2 = 10N (b) 0.2 × 3 = 0.6N

(c) 0.05 × 1 = 0.05N22. Normality = Molarity × acidity (a) 0.5 × 1 =0.5N (b) 2 × 2 = 4N23. meq = 4.9/49 × 1000 = 100; Normality = meq/mL = 100/100 = 1N; Hence Molarity = 1/2 = 0.5M[you can also find molarity first (M = mmols/mL) and then normality from molarity].Molilty: d = m/V = 1.15 g/mL ⇒ m/100mL = 1.15 ⇒ m = 115g; mass of solvent = 115 - 4.9 = 110.51g

110.51g of solvent contains 4.9/98 moles of H2SO

4

1000g of solvent contains 0.45 mole Hence molality = 0.45m24. Molarity: mmols = 10/106 × 1000 = 94.34; Molarity = 94.34/200 = 0.472MNormality = 0.472 × 2 = 0.944MMolality: d = 1.03 g/mL; m/200mL = 1.03; ⇒ m = 206gMass of solvent = 206 - 10 = 196g196g of solvent contains 10/106 moles of Na

2CO

3

1000g of solvent contains 0.481 moles of Na2CO

3; So the molality = 0.481m

(Note that for a dilute solution molarity and molality are values are close to each other)25. Molality: 15% H

2SO

4 by mass means; 100 g of solution contains 15 g H

2SO

4

So mass of solvent = 100-15 =85g;85g of solvent contains 15/98 moles of H

2SO

4

1000g of solvent contains 1.89 moles ; Hence molality = 1.8mMolarity: d = m/v = 1.1 g/mL ⇒ 100g/V = 1.1 ⇒ V = 90.9 mL;

90.9 mL of solution contains 15/98 moles1000 mL of solution contains 1.68 mole; Molarity = 1.68MNormality = 1.68 × 2 = 3.36 N

26. x × 2 = 25 × 1/10 ⇒ x = 1.25 mL27. 37.5 × x = 50 × 0.15 ⇒ x = 0.2N28. Mass of NaOH = 0.4g; EM = 40; meqs = 0.4/40 × 1000 = 10

meqs of acid = meqs of base =10; x × 0.5 = 10; ⇒ x = 20 mL29. x × 5 = 30 × 4 ⇒ x = 24 mL

30. MM of Ca(OH)2 = 74; EM = 74/2 = 37; 3.7g of Ca(OH)

2 = 3.7/37 = 0.1 eq

meqs = 0.1 × 1000 = 100; x × 4 = 100 ⇒ x = 25 mLmass of H

2SO

4 = 100 × 10-3 × 49 = 4.9g

31. Meqs of acid = 0.25/x × 1000 (x = EM of acid); meqs of base = 50 × 0.250.25/x × 1000 = 50 × 0.25 ⇒ x = 20 (EM of acid)

32. Meqs of CaCO3(base) = 1.5/50 × 103 = 30; 50 × x = 30 ⇒ x = 30/50 = 0.6N

33. (a) 500 mL solution contains 0.53g; So 1000mL solution contains 1.06gNormality = strength in g per litre/E = 1.06/53 = 0.02N

(b) Strength in g/L = EM × N = 49 × 1/100 = 0.49 g/L(c) EM = strength in g per L/ N = 9.8/0.2 = 4934. 50 mL of H

2SO

4 solution contains 0.7g

So 1000 mL of solution contains 14g; So strenght in g/L = 14; Normality = 14/49 N25 × x = 50 × 14/49 ⇒ x = 0.571N

Alternatively: meqs of H2SO

4 = 0.7/49 × 1000 = 14.285; 25 × x = 14.285 ⇒ x = 0.571N

(note that we did not use the volume of the acid(50 mL) in this method of calculation)35. density = 1.18 g/mL; m/100 = 1.18 ⇒ m = 118 g (mass of 100 mL solution which 24.7% by wt.)

100g of solution contains 24.7g of solute; So 118g of solution conains 29.146gSo 100 mL solution contains 29.146g or 29.146/49 eqs of H

2SO

4

1000 mL of solution contains 5.948 eqs. Hence Normality = 5.948 Nmeqs of Al reacted with acid = 2.7/9 × 1000 = 300Total meqs of acid present = 75 × 5.948 = 446.1; meqs of acid reacted with Al = 300So the meqs of excess unreacted acid = 446.1 - 300 = 146.1; Normality of the resulting acid solution =

146/500 = 0.292N( total volume after dilution is 500 mL36. Meqs of metal carbonate = 4/x ×1000 = 200 × 0.377 ⇒ x = 53 ( x = EM of the metal carbonate)37. Meqs of Na

2CO

3 = 0.2/53 ×1000 = 3.77; meq of acid = 40 × x ( x = normality)

40 × x = 3.77 ⇒ x = 0.094 N = 0.94 N/1038. Stock solution of acid = 200 mL; This contains 0.84g of acid

25 × x = 14 × 0.1 ⇒ x = 0.056 N (normality of the stock solution)(note that the concentration of a solution is not changed if a part of the stock solution is transferred to another

container. The relative proportions of solute and solvent remain the same until some change is not brought about)meqs of acid in stock solution = 200 × 0.056 = 11.2; Hence meq acid = 0.84/y × 1000 (where y=EM

of acid)0.854/y × 1000 = 11.2 ⇒ x = 75, Hence basicity = 150/75 =2

39. 25 × 0.1 = 27.5 × x ⇒ x = 0.091NEM = 4.1/0.091 = 45; Hence MM = 45 × 2 = 90

40. Meq of H2SO

4 = 30 × 2 = 60; meq of HCl = 50 × 0.5 = 25; meq of HNO

3 = 75× 5 =375

Total meq of acid = 460; meq of base(NaOH)= 1000 × 0.1 = 500After neutralisation, the resulting solution will be basic since meq of base is greater than acid.meq of excess base after neutralisation = 40, total volume of the mixture= 1155 mLSo the normality of the base solution = 40/1155 = 0.0346N

41. Meq of NaOH = 250 × 0.1 = 25; meq of H2SO

4 = 100 × 1/4 = 25

There will be exact equivalent and no excess of acid or base will be left.meq of salt formed = 25; total volume of mixture = 350normality of salt solution = 25/350 = 0.071N, Molarity = 0.071/2 = 0.0355M

42. Meq of Ba(OH)2 = 200 × 1/2 = 100; meq of HCl = 100 × 1/2 =50

So limiting reactant is HCl; meq of excess Ba(OH)2 = 50

Total volume of solution = 300; Normality of salt = 50/300 = 0.167N;Normality of base = 50/300 = 0.167N; Molarity of each = 0.167/2 = 0.0835M

43. Meq of HCl = 25 × 10 =250; meq of H2SO

4 = 20× 18 =360; total meq of acid= 610

Total volume of solution after dilution = 1000 mL, So normality of acid solution = 610/1000 = 0.61N44. Meq of H

2SO

4 needed = 100 × 0.1 = 10; So mass of H

2SO

4 = 10 × 10-3 × 49 = 0.49g

Since it is 98% H2SO

4 by weight; 98g of H

2SO

4 is present in 100g of solution

So 0.49g of H2SO

4 is present in 0.5g solution.

45. Meq of NaOH = 50 × 1/20 = 2.5; Mass of NaOH = 2.5 10-3 × 40 = 0.1g46. This is a problem involving double titrations. Two acids are given(one of known concentration and the otherunknown) and one base(unknown concentration) are given. Using the acid with known concentrtion, the normality ofthe base is found out from one titration. Using the normality of base, then the normality of the other acid is determinedfrom the other titration.

2nd titration data: strength of HCl in g/L = 3.65; its normality = 3.65/36.5 = 0.1N25 × 0.1 = 25 × x ⇒ x = 0.1N (normality of Na

2CO

3 solution)

1st titration data: 40 × y = 50 × 0.1 ⇒ y = 0.125N (normality of acid A)47. 1st titration: 25 × x = 17 × 0.95/10 ⇒ x = 0.646 N/10 (normality of NaOH solution)

2nd titration: 25 × 0.646/10 = 27.5 × y ⇒ y = 0.587 N/10 (normality of H2SO

4)

48. Meq of acid = 100 × 2.5 = 250; total volume after dilution = 1100 mLnormality = 250/1100 = 0.227N

49. Normality of conc. HCl:m/1000 = 1.17 ⇒ m = 1170g; 100g of solution contains 33 g, So 1170g contains 386.1gSo 1000 mL solution contains 386.1g i.e 386.1/36.5 = 10.578 eqs.; So normality = 10.578NThe volume of diluted acid = 200mL, normality of diluted acid = 1.5Nx × 10.578 = 200 × 1.5 ⇒ x = 28.36; So the volume of water that is to be added = 200-

28.36=171.64mL50. m/1000 = 1.1 ⇒ m = 1100g;

100g of solution contains 15g H2SO

4, So 1100g of solution contains 165g of H

2SO

4

Mass of solvent = 1100 – 165 = 935g935g of solvent contains 165/98 moles of H

2SO

4, So 1000g of solvent contains 1.8 moles(molality=1.8m)

1000 mL solution contains 165/98 moles of H2SO

4, So molarity = 1.68M

Normality = 2 × 1.68 = 3.36 N (since H2SO

4 is dibasic)

200 × 3.36 = x × 0.5; ⇒ x = 1344 mL, So the volume of water added = 1344-200 = 1144mL51. 10 × x = 150 × 2 ⇒ x = 30N; Strength in g/L = 30 × 49 = 1470g

m/1000 = 1.7 ⇒ m = 1700g; 1700g solution contains 1470g H2SO

4

So 100g solution contains 86.47g H2SO

4 , So % of purity = 86.47%(percent by weight).

52. Back tiration: 50 × x = 50 × 1/10 ⇒ x = 0.1N (normality of the diluted acid)meq of excess acid = 50 × 0.1 = 5; meq of acid taken for 1st reaction = 50 × 1/2 = 25;meq of acid consumed by metal carbonate = 25 - 5 = 20 = meq of metal carbonatemass = 20× 10-3 × x = 1 ⇒ x = 50 (equivalent mass of metal carbonate)EM of metal carbonate = EM of metal + EM of carbonate = y + (12+48)/2 ⇒ y = 20(EM of metal)

53. Na2SO

4 in the mixture is an impurity which does not react with acid. So only one reaction is considered.

25 × x = 20 × 1/10 ⇒ x = 0.08N (normality of Na2CO

3 in the stock solution)

meq of Na2CO

3 = 25 × 0.08 = 20; mass of Na

2CO

3 = 20 × 10-3 × 53 = 1.06g

% of Na2CO

3 = 1.06/1.25 × 100 = 84.8%; % Na

2SO

4 = 15.2

54. Back titration: The total volume of the mixture = 50 + 50 = 100mL (diluted excess acid solution)100 × x = 10 × 0.16 ⇒ x = 0.016N (normality of diluted excess acid solution)meq of excess acid = 100 × 0.016 = 1.6; meq of acid used in the reaction = 50 × 0.1 =5meq of acid consumed by 50 mL of stock solution of Na

2CO

3 = 5 – 1.6 = 3.4

In 50 mL the meq of Na2CO

3 = 3.4; So in 250 mL the meq of Na

2CO

3 = 17

mass of Na2CO

3 = 17 × 10-3 × 53 = 0.901g; % purity = 0.901/1 × 100 = 90.1

55. Back tiration: 80 × 1/20 = 230 × x ⇒ x = 0.0174N; meq of excess acid = 230 × 0.0174=4meq of acid taken in the 1st reaction = 230 × 1/10 = 23;So meq of acid consumed by CaCO

3 = 23 - 4 = 19 = meq of CaCO

3

mass of CaCO3 = 19 × 10-3 × 50 = 0.95g; % purity = 0.95/1 × 100 = 95%

56. Back titration: 100 × x = 60 × 1/2 ⇒ x = 0.3Nmeq of excess NaOH = 100 × 0.3 = 30; meq of NaOH taken for 1st reaction = 100 × 1/2 =50Hence meq of NaOH consumed by ammonium salt = 50-30 = 20 = meq of NH

4+ salt = meq of NH

3

mass of NH3 = 20 × 10-3 × 17 = 0.34g ( Since EM of NH

3 = 17);

mass of NH4+ = 20 × 10-3 × 18=0.36; % of NH

3 = 0.34/1.13 × 100 = 30%; % NH

4+ = 31.85%

57. Back titration: 30 × x = 13.5 × 0.2 ⇒ x = 0.09N(normality of diluted excess acid)meq of excess of acid = 30 × 0.09 = 2.7; meq of aacid used in the reaction = 30 × 0.25 = 7.5meq of acid consumed by NH

3 = 4.8 = meq of NH

3 = meq of (NH

4)

2SO

4

mass of (NH4)

2SO

4 = 4.8 × 10-3 × 66 = 0.3168g (EM of ammonium sulphate = MM/2=132/2=66)

12 mL of the original solution contains 0.3168g of (NH4)

2SO

4

1000 mL of the original solution contains 26.4g; So strength in g/L = 26.458. 20 × x = 15 × 1/10 ⇒ x = 0.075N(normality of diluted Na

2CO

3.10H

2O)

meq of Na2CO

3.10H

2O = 250 × 0.075 = 18.75;

mass of Na2CO

3.10H

2O = 18.75× 10-3 × 143 = 2.681g (EM of Na

2CO

3.10H

2O = MM/2=286/2=143)

% purity = 2.681/3.2 × 100 = 83.78%59. NaCl acts as impurity in this case and does not react with acid.

25 × x = 50 × 1/10 ⇒ x = 0.2N (normality of the stock solution)meq of Na

2CO

3 = 250 × 0.2 = 50; mass of Na

2CO

3 = 50 × 10-3 × 53 = 2.65g

% of Na2CO

3 = 2.65/5 × 100 = 53%; % of NaCl = 47%

60. Total meq of acids = 15 × 1/2 + 25 × 1/10 + 50 × 1/5 = 20Total volume = 90 mL; Hence normality of the acid mixture = 20/90 = 2/9 Nmeq of base added = 50 × 1/2 = 25, So the limiting reactant is acid(20 meq)meq of salt formed = 20, meq of excess base = 5Total volume after mixing the base = 300 mLNormality of the excess bas solution = 5/300 = 0.0167N; Normality of salt solution = 20/300 = 0.067N

61. Let the mass of NaOH =x and hence mass of Na2CO

3 = 2.013 – x

meq of NaOH = x/40 × 1000; meq of Na2CO

3 = (2.013 - x)/53 × 1000

Titration: 10 × x = 20 × 0.1 ⇒ x = 0.2N(normality of stock solution)meq of acid in the stock solution = 250 × 0.2 = 50 = meq of total base

1000x40 +

1000(2.013-x)

53= 50 ⇒ x = 1.97 g(NaOH); mass of Na

2CO

3 = 0.043g

62. Phenolphthalein: meq of acid = 35 × 1/2 = 17.5 = all NaOH + half neutralisation of Na2CO

3 to NaHCO

3

= x +yMethyl orange: meq of acid = 40 × 1/2 = 20 = all NaOH + complete neutralisation of Na

2CO

3 to NaCl

= x + 2yx = y = 17.5 and x + 2y = 20; Solving the two equations, we get y=2.5meq of Na

2CO

3 = 2y =5; mass of Na

2CO

3 = 5 × 10-3 × 53 = 0.265g

meq of HCl = x = 15; mass of NaOH = 15 × 10-3 × 40 = 0.6gMass of NaOH present per litre = 0.6/20 × 1000 = 30g;Mass of Na

2CO

3 present per litre = 0.265/20 × 1000 = 13.25g

63. Phenolphthalein: meq of acid = 12.5 × 1/10 = 1.25 = half neutralisation of Na2CO

3 to NaHCO

3 = y

Methyl Orange: meq of acid = 30 × 1/10 = 3 = NaHCO3(original) + complet neturalisation of Na

2CO

3 to NaCl

= x + 2ymeq of Na

2CO

3 = 2y = 2.5, meq of Na

2CO

3 in 250 mL = 2.5/25 × 250 = 25

Mass of Na2CO

3 = 25 × 10-3 × 53 = 1.325g

Putting the value of y in eqn 1, we get x = 0.5 = meq of NaHCO3(original)

meq of NaHCO3 present in 250 mL = 0.5/25 × 250 = 5

Mass of NaHCO3 = 5 × 10-3 × 84 = 0.42g; % of Na

2CO

3 =1.325/(1.325+0.42) × 100 =75.93%;

% of NaHCO3 = 24.07%

64. Titration: 25 × x = 30 × 1/10 ⇒ x = 0.12N(normality of 250 mL of stock solution of H2O

2)

meq of H2O

2 = 250 × 0.12 = 30; mass of H

2O

2 = 30 × 10-3 × 17 = 0.51g (EM of H

2O

2 = MM/2 as

change in ON per molecule is 2)10 mL of H

2O

2 solution contains 0.51g; So 100 mL of solution contains 5.1g (% w/v)

Volume Strength: H2O

2 → H

2O + 1/2 O

2(g)

34g 11.2L (NTP)34g of H

2O

2 liberates 11.2L of O

2gas at NTP; So 0.51g of H

2O

2 liberates 0.168L of O

2 at NTP

10 mL of H2O

2 solution liberates 168 mL of gas at NTP

1 mL of H2O

2 solution liberates 16.8 mL of O

2 gas at NTP; So volume strenght = “16.8 volumes”

(note the mass data i.e 12g is not used in this numerical)65. Fe

2(SO

4)

3 present in the mixture does not react with KMnO

4 as it is in its oxidised state. Only FeSO

4.7H

2O

reacts with KMnO4.

Let the mass of FeSO4.7H

2O = x g; Mass of Fe

2(SO

4)

3 = 5.39 – x

EM of FeSO4.7H

2O = MM/1 = 278 (Since Fe+2 → Fe+3, the change in ON =1)

meq of FeSO4.7H

2O = x/278 × 1000; Meq of KMnO

4 = 80 × 0.125

x/278 × 1000 = 80 × 0.125 ⇒ x = 2.78g(mass of FeSO4.7H

2O), mass of Fe

2(SO

4)

3 = 2.61g

66. Titration: 10 × x = 8 × 1/10 ⇒ x = 0.08N (normality of the stock solution)meq of oxalic acid crystal = 250 × 0.08 = 20; mass of oxalic acid = 20 × 10-3 × 63 = 1.26g(since in the reaction, C

2O

42- → 2CO

2 + 2e and oxalic acid crystal = H

2C

2O

4.2H

2O, EM=MM/2)

% purity = 1.26/1.5 × 100 = 84%67. Back titration: Meq of Na

2S

2O

3 = 80 × 1/30 = 8/3 = meq of I2 = meq of MnO

2

(pyrolusite is an ore containing MnO2 alongwith other impurities)

Mass of MnO2 = 8/3 × 10-3 × 43.5 = 0.116g (MnO

2 → Mn2+; Change in ON =2, EM = MM/2)

% purity = 46.4%68. EM of Na

2S

2O

3.5H

2O = MM/1 = 290 ( Na

2S

2O

3 + I

2 → Na

2S

4O

6 + NaI, the change in ON per molecule

of Na2S

2O

3 = 2 × 1/2 = 1)

Normality = 10/290 = 1/29 N; Meq of Na2S

2O

3 = 12 × 1/29 = meq of I

2 = meq of Cu2+

Mass of Cu2+ = 12/29 × 10-3 × 63.5 = 0.026g (CuSO4 + KI → Cu

2I

2 + I

2 + K

2SO

4, change in ON of

Cu is 1, So EM of Cu2+ = atomic mass/1)Mass of copper metal = mass of Cu2+ in CuSO

4 = 0.026g; % of copper = 0.026/0.2 × 100 = 13%

(Note that if you would like climb the ladder of meq upto copper metal, you can do so i.e meq of copper metalis 12, but in this case the EM of Copper is not to be taken as 63.5/2 although Cu has changed from 0 to +2. HereEM of Cu is to be taken as 63.5/1. The factor 1 is to be used from the involved titration of Cu2+ to Cu+, not fromthe dissolution of copper metal to Cu2+. To avoid the confusion, you are strongly recommended to use mole methodjust after the titration experiment to come to the root).