chemistry icse0001

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Gas laws are certain rules which a gas follows when -

subjected to a change in temperature, pressure or volume.

B. GA§,J:.AWS

~: .

/

• The effect of changes of anyone of the parameters i.e.-

temperature, pressure or volume affects the other two parameters.

UNITS

VOLUME

RESSUREEMPERATURE

An indir£ltor of the Averageforceexertedby Spaceoccupiedbyafixed

average kinetic energy gas molecules on the mass of gas:

possessed by a molecule. walls per unit area.

Atmospheres

1 Atmosphere

=

76 cm or 760 mm Hg

Celsius temperature = °C

Kelvin temperature

=

K

0° C = 0 + 273 = 273 K

Litres, cubic centimetre etc.

1

Litre =

1 dm

3

= 1000cm

3

= = 1000 ml.

2. Study of the relation between pressure and volume of a gas -

[temperature constant] & temperature & volume of a gas [pressure constant]

Summarizing the Laws studied

• BOYLE S LAW -

'Temperature remaining constant, the volume of

a

given mass of

/drY gas is - inversely proportional to it's pressure

V

a

\ >

[T = constant].

: l

CHARLE S LAW - 'Pressure remaining constant, the volume of a given mass of .

dry gas is-diredlyproportional to its absolute [Kelvin]temperature'. VaT [p = constant]

• Applying Boyle's

&

Charle's Law we derive at. the-

PV

AS EQUATION: T K [constant] i.e.

=

= K T constant]

3. Volumes of gases are converted to -

standard temperature and pressure conditions [s.t.p.] and then compared

• Volumes of gases change with - temperature and pressure.

• Hence a standard value of temperature and pressure is chosen -

to which gas volumes are referred.

• The standard temperature and pressure values are:

Standard temperature = ooe = 273 K

Standard pressure ='760 mm. oilig_= 26 cm. of Hg =

1

atmospheric pressure.



CHAPTER 4

o l t

~OnC tp t

n

~toicbiom ttr ,

..I.J 1D~J-

SCOPE OF SYLLABUS - in and after MARCH 2013

O L E

CONCEPT AND STOICHIOMETRY

i j GAYLUSSAC'S LAW OF COMBINING VOLUMES; AVOGADRO'S LAW.

Idea of mole - a number just as dozen, a gross; Avogadro's Law - statement and explanation;

Gay Lussac's Law of Combining Volumes - statement and explanation, the mass of22.4litres of

anygas at

S.

T.P.

is

equal

to

its molar mass , (Questions will not set on formal proof but may be taught

r clear understanding) simple calculations based on the molar volume. .

DJREFERTO THE ATOMICITY OF HYDRC?GEN, OXYGEN, NITROGEN AND CHLORINE (PROOF NOT REQUIRED).

Theexplanation can be given using -equaiions for the formation of

He

I, NH

3

, and NO.

RELATIVEATOMIC MASSES (ATOMIC WEIGHT) AND RELATIVE MOLECULAR MASSES (MOLECULAR WEIGHTS): EITHER

H =1 OR 12C=12WILL BE ACCEPTED; MOLECULAR MASS = 2xVAPOUR DENSITY (FORMAL PROOF NOT REQUIRED).

DEDUCfION OF SIMPLE (EMPIRICAL) MOLECULAR FORMULA FROM THE PERCENTAGE COMPOSITION OF A

COMPOUND;THE MOLAR VOLUME OF A GAS AT S.T.P.; SIMPLE CALCULATIONS BASED ON CHEMICAL EQUATIONS;

BOTHREACTING WEIGHT VOLUMES.

Ideaof relative atomic mass and relative molecular mass - standard H atom, or 1/12thof carbon 12 atom.

Relatingmole and atomic mass - arriving at gram atomic mass and then gram atom; atomic mass

is

a number

dealingwith one atom; gram atomic mass is the mass of one mole of atoms.

Relatingmole and molecular mass - arriving at gram molecular mass and gram molecule - molecular mass

is

anumberdealing with a molecule, gram molecular mass

is

the mass of one mole of molecules.

Molecularmass =

2

x vapourdensity (questionswill notbesetonformalproofbutmaybetaughtfor clearunderstanding).

Simplecalculations based on the formula.

Deductionof simple (empirical) & molecular formula - from the percentage composition of a compound.

Part A

LUSSAC S LAW AVOGADRO S LAW MOLE CONCEPT

INTRO U TION

of particles per .-E,.nit

GASES SOLIDS LIQUIDS

Gases have no definite Solids have a definite Liquids have a definite

volume an s ape and volume and shape and vo ume,nodefiniteshapeand

nave no rigidity are highly rigid are less rigid

• intermoleculars ace =betweenl- intermolecular space -between .- intermolecular space -between

articlesis maximum. particles is minimum. particles is more than solids.

• forceof attraction - between. force of attraction - between • force of attraction - between

particlesis ne li ·ble. particles is maximum. particles is less than solids.

Gasesexertpressure Solids exert pressure Liquids exert pressure

• on the walls of the container .• -. 2 ..ll y Rartiall downwards. • only partially downw_ards.

Gasesgenerally have low Solids generally have high Liquids have density less than

densities densities solids

Gaseshave.high Solids are seen to have no Liquids have slight

miscibilitY. 1liscibili } -- mise ility

• particlesrapidly diffuse with. particles do not diffuse with. particles may diffuse with

particlesof other gases. particles of another solid. particles of another liquid.



G A Y LUSSAC S LAw

.~

\.../

INTRODUCTION

1.

In

1801

-_Rovles and Charles had proved the gas laws showing that -

Nual volumes of all gases behave similarlv under simibl

..l

·i.Us...of-

temperature and pressure if systematically changed.

2. In 1800 - John Dalton in his atomic theory of matter stated that-

the smallest indivisible particle of matter is the atom which takes part in a chemical

reaction fnrming.molecules or compo]lnd atoma.When atoms combine they do so in

numbers which bears a simple whole number ratio to one another.

3.

In

1805 -

Lussac on the basis of his experimental work discovered -

. g l

relationship between the combining volumes of gases used and produced under the

same conditions of tem12erature and Fress~

4. In 1811 - Amedeo Avogadro was able to explain Boylesand Charles Law on the basis that-

equal volumes of all gases under the same conditions of temperature and pressure

contain equal number of molecules.

. 7urther proposed Avogadro's law which helped in correcting Dalton's atomic theory.

~Y LUSSAC'S LAW [La~ of combining. volumes o~gases} . C~ v\.l-~ ~ ~

When gases,react they do so In volumes whIch bears a sImple whole number ratio

to one angtqer and to the XQ]UIDesof the-products if gasennacpmxided.fhe

temperature andJn',ess.urebo£~acting gases.an.d..the.iLproducts.r.emainconstant .

Lussac's Law can be verified or illustrated by studying the following examples in

which a simple whole number ratio exists between the gaseous reactants and products.

der the sa me cond itions o f tem perature a nd pressure.

2

volumes of steam are formed when -

Two volumes of hydrogen reacts with one volume of oxygen.

~

@

-7

@ @

2H2 O

2

~ 2H

2

O

1vol. 1vol.

Ratio

2 1 :2

I

,

~

2 vols.

1 vol.

2

vols.

hydrogen

oxygen

steam

2

volumes of ammonia are formed when -

One volume of nitrogen combines with three volumes of hydrogen.

@

®

+~

-7 ~

~,

1 vol. 3 vols. 2 vols.

nitrogen hydrogen ammonia

[1

volume means any particular volume at the given pressure and temperature.]

Ratio

N2

+

3H2 ~ 2NH3

1 3 \ 2·

..--{



~ ~ ~ ~ ~ ~

Gr,J. LUSSACI,S LAW - PROBLEMS

50 cm

3

of carbon monoxide and 200 cm

3

of oxygen

Calculate the composition of the resulting mixture.

are mixed together and ignite;

Solution: 2CO (g) 02 (g) -7 2C0

2

(g)

ie. 2 vol :. 1 vol -7 2 vol [By Lussac's Law]

To calculate the amount of unused co. To calculate the amount of C92 formed.

CO: 02 02 CO

2

2 1 1 2

X . . 200 200: Xl

X

=

400 cm

3

Xl

=

400 cm

3

. . Amount of unused CO = 450-400 = 50 cm

3

. • Amount of CO

2

formed = 400 cm

3

Unused CO: 50 cm

3

CO

2

formed: 400 cm

3

[Total mixture: 450 cm

3

- Ans.]

2.

One volume of nitrogen combines with one volume of oxygen

to

form two volumes of nitric oxidl

Calculate the amount of each reactant required

to

produce 250ml. of nitric oxide.

Solution: N2 g)

+

02 g)

-7

2NO g)

ie. 1 vol : 1 vol -7 2 vol

To calculate the amount of N2 required.

NO: N2

250:

X

: . X = 125 ml

. . Amount of N2 = 125 ml- Ans.

[By Lussac's Law]

To determine the amount of 02 required.

NO: 02

250: X

:. X =

125 ml

. . Amount 02 required = 125 ml- Ans.

. I

3. What volume of oxygen would be required

to

burn completely 400 ml of acetylene (C

2

HJJ,

Also calculate the volume of carbon dioxide formed.

400 ml.

2C

2

H

2

g) 502 g)

2vol : 5 vol

.. ,2

X

200, ,5

X

200,

400 ml.

?

Solution: '

According to Lussac's law

-7 4C0

2

g) 2H

2

0

-7 4 vol : 2 vol

,4x200, ,2x200,

?

1000 ml. of oxygen is required and 800 ml of carbon dioxide is formed - Ans.

4. 3000 cc. of oxygen was burnt with 600 cc. of ethane (C

2

H £ J.

Calculate the volume of unused oxygen.

600 ml. 3000 ml.

Solution: 2C2H

6

g)

702

g)

-7

4C02

g)

6H20

According to Lussac's law 2 vol : 7 vol -7 4 vol : 6 vol

.. ,2)(300, ?x300, ,4x300, ,6 x 300,

600 ml. ?

:. Volume of unused oxygen = 3000 - [7 x 300] 2100 = 900 cc. -:-Ans.

5. 60 cc; of oxygen was added

to 24

cc. of carbon monoxide and the mixture ignited.

Calculate the volume of oxygen used up and the volume of carbon dioxide formed.

24 cc 60 cc

Solution: 2CO g) + 02 g) -7 2C0

2

g)

According to Lussac' s law 2 vol : 1 vol -7 2 vol

. . ,2 X 12, ,I X 12, ,2 X 12,

24 cc. ? ?

Volume of oxygen used up is 12 cc. and the volume of carbon dioxide formed is 24 cc. - Ans.

7



Calculate the vol. of cO

2

formed on cooling to room temp. What other gas if .any may also bepresent.

ii] ~ 2H

z

O [g)

2 vol.

200 crn

3

200 crn

3

Solution:

2CO(g)

°z (g)

~

According to Lussac's law

2vol

:

1 vol

~

.. 2 x 100

1 x 100

2CO

z

(g)

2 vol

2 x 100

I I I I 1 I

200 cm

3

100 cm

3

?

Volume of carbon dioxide formed is 200 cm

3

& the other gas present is 100 cm

3

of oxygen. - Ans.

7 . Calculate the volume of oxygen required to burn completely a mixture of

22.4

dm

3

o] CH

4

and

11.2 dm

3

of H

2

• [all volumes measured at s.t.p.] [ldm

3

=

1

litre]. .

11.2 drn

3

2H2

2 vol.

2x5.6

. 200 ml. of C

2

H

4

is burnt in just sufficient air [containing 20% oxygen] as per the equation -

CJI4

302 ~

2C0

2

2H

2

0

[g].

Calculate the resultant mixture compositon [atlOOOC constant press.]

200 ml.

Solution: C

Z

H

4

+ 30

z

~ 2CO

z

+

According to Lussacs law 1 vol. 3 vol. 2 vol.

I 1 x 200 I I 3 x 200 I I 2 x 200 I

200m . 600 ml. 400 ml.

When oxygen is 600 ml. the nitrogen is 80 x 600

=

2400 ml.

20

Hence the composition of the resultant mixture is - Ans.

Carbon dioxide = 400 ml. Steam = 400 ml. [at 100°C steam has volume]

Nitrogen = 2400 ml. [Ethylene = (200 - 200) ml. = 0 ml. & Oxygen =(600 - 600) ml. = 0 ml. ]

. A mixture of 10cm

3

of CO, 60 cm

3

of

H2

and 25 cc. of CH

4

are mixed with 750cm

3

of air [containing

20% oxygen] and ignited. Calculate the composition of the resultant mixture on cooling to room temp.

Hence the composition of the resultant mixture after ignition is - Ans.

Carbon monoxide = [10 -10] cm

3

= 0 cm

3

; Hydrogen = [60 - 60] cm

3

= 0 cm

3

Methane

=

[2S - 2S] cm

3

= 0 cm

3

; Carbon dioxide = [10

2S] cm

3

:= 3Scm

3

Oxygen = [ISO- (S

30

SO)]= 6Scm

3

[% of 0z in 7S0cm

3

of air

=

20 x 7S0

=

lS0cm

3

]

100

Solution: i]

Lussac's law

22.4 drn

3

CH

4

1 vol.

1 x 22.4

20z ~ CO

z

+ 2H

z

O

[g]

2 vol. 1 vol. 2 vol.

12 x 22.4 I

44.8 dm

3

Total oxyge~ required

=

44.8 dm

3

S.6 dm

3

=

SO.4dm

3

[lits.] - Ans.

10 crn

3

2CO

+

0z ~ 2CO

z

2 vol. 1 vol. 2 vol.

12x 51 l:J'<o.x5

1

12x 51

10cm

3

Scm

3

10cm

3

60 crn

3

ii] 2H

z

0z ~

2 vol. 1 vol.

12x 30

1

?

x 30

1

60cm

3

30cm

3

Solution: i]

Lussac's law

Nitrogen

= [7S0 - ISO] cm

3

= 600cm

3

°z

1 vol.

I

1

x

5.6 I

S.6 dm

3

2H

z

O

2 vol.

I 2 x 200 I

400 ml.

[When 0z is 20% the Nz in air is 80%]

25 crn

3

iii] CH

4

+

202 ~ CO

z

+ 2H

z

O

1 vol. 2 vol. 1 vol. 2 vol.

11x 25

1

12x 25

1

11x 25

1

25cm

3

50cm

3

25cm

3

73



RELATIVE ATOMIC MASS

[RAM]

RELATIVE MOLECULAR MASS

[RMM]

[Atomic Weight] [Molecular Weight]

3. R j:L A TlV E A TO M IC M A S S

[RAM]

RELA TIV E M O LEC ULA R M ASS

[RMM]

Atoms are extremely small and very light and hence - cannot be weighed directly.

Indirect methods have been utilized to- determine the absolute mass of an atom.

The relative mass of an atom or molecule is hence considered -

by consideringamass ofa light atom relating themass ofotheratoms ormoleculesto it.

Hydrogen atom [being the lighest element]was initially chosen -

~ the standard unit and masses of other atoms or mo[ecules70mpared to it.

Carbon

-12

atom [isotope of carbon] was later in

1961

considered - -

the standard for comparison ofmass of other atoms or molecules.

~_ [atomic weight] of an element - RMM[molecuIarweight]ofanelement/compound-

is the numbet_of _times one .atom is the number oCtim~s_ qtiemolecul

of an element is heavier than - of the substance is heavier than -

---- -- --- 2-

/ 1 , the mass of an atom of carbon [C].

1/12

the mass of an atom of,.lcarbon[C

I2

r- -

---

•

~--

-----_.:.....-

RMM =' Mass of one molecule of the substance

e

Mass of one atom of carbon [C12]

RAM= Mass of one atom of the element

e i

d

Mass of one atom of carbon [C12]

e

•

•

GR M TOMIC M SS GR M TOM

The relative atomic mass of an element The relative molecular mass of a substance

expressed in grams is known as - expressed in grams is known as -

gram atom ic mass orgram atom ofthat element. f lY a m m o le cu la r m a s s

Q r rnm 111olecu[g.Q£tbat~ent.

Atomic

molecular weights - of some common elements on the

C12

standard scale

Element

GR MMOLECUL RM SS GR MMOLECULE

RAM

[At. wt.]

RMM

[Mol. wt.]

ubstance

ymbol

Gram moleculeram atom

#omic weights are expressed in a.m., . Atomic weights are not whole numbers

• Atomic weights are expressed as simple· Natural elements are mixtures of - two or

numbers or in - atom ic m ass units [a.m.u.] more isotopes of constant composition.

• A tom ic m ass unit is defined as - • The atom ic w eight - is the weighted average

1/12 the mass of a carbon atom C12. of the atomicweight of its natural isotopes

• A tom ic w eight

0 1

O Y 1N cn -

is

16.000

a.m.u. ego Chlorine exists as

two

isotopes

:;0

35

37 .

h . 3 1

ie. the weight of an oxygenatom is

16.000 17

C

17Cl

In t e ratio : .

a.m.u.onthesca1eonwhichcarbonis12a.m.u. Hence the weighted average is35.43

a.m.u.

Molecular wt.of S02 is 64 a.m.u. ie. 1 molecule of S02 is 64 times as heavy as V

12

the mass of carbon atom C12.

Aluminium

Carbon

Chlorine

Hydrogen

Iron

Nitrogen

Oxyge~

Al

C

C

H

Fe

N

°

Nitrogen

Oxygen

Chlorine

Carbon dioxide

Sulphur dioxide

Sulphuric acid

26.98

12.0000

35.453

1.008

55.847

14.007

15.999

27g

12

g

35.5

g

19

56 g

14

g

16 g

28.014

31.998

70.906

4,3.998

64.062

98.076

28 g

32

g

71g

44g

64g

98g



TERMINES THE MOLECULAR FORMULA OF A GAS

Molecular formula - A chemical formula which sz1vesthe actual or exact number

0

atoms

of th~ elements present in one molecule oj a compound. -

Application of Avogadro s Law-

for determining the molecular formula of a gas.

It has been experimentally shown that -

One volume of hydrogen reacts with

1

vol. of chlorine to give

2

vols. of hydrogen chloride.

Hydrogen

1 vol.

1 molecule

2 x 1 atoms

Chlorine

1 vol.

1 molecule

2 x 1 atoms

-7 Hydrogen chloride[H2 Cl

2

-7 2HCI] .

2 vols. - [by Gay-Lussac sLaw]

2 molecules - [by Avogadro s Law]

2 molecules - [Hydrogen and chlorine - diatomic]

. p\(PPIJCATIONS OF AVOGADRO'S LAW ~

a] Determines the atomicity of the gas.

b] Determines the molecular formula of a gas.

c] Determines the relation between molecular weight and vapour density.

d] Explains Gay-Lussac's Law.

e] Determines the relationship between gram molecular weight and gram molecular volume.

JY TERMINES ATOMICITY OF A GAS .

Atomicity - The number, Qfatoms J2re~,nt- in one molecule of that element.

Mon~~mic : Elements which have 2e atom in their molecules. egoHelium, neon.

Diatgmic:

Elements which have

two atoms.in

their molecule. egoHydrogen, oxygen, chlorine.

(

Application of Avogadro s Law -

for determining the atomicity of nitrogen.

It has been observed that -

Onevolume of nitrogen reacts with

1

vol. of oxygen to produce two volumes of nitric oxide.

('Jitrogen

Oxygen

-7

Nitric oxide

[N2

O2 ~ 2NO]

, 1 vol. 1 vol. 2 vols. - [by Gay-Lussac s Law]

1 molecule Irnolecule 2 molecules - [by Avogadro s Law]

1/2

molecule

1/2

molecule 1 molecule - [An atom is indivisable. Therefore

1 atom 1 atom 2 atoms

1/2

molecule contains one atom.]

Therefore one molecule of nitrogen contains - 2 atoms of nitrogen.

Conclusion: A molecule of nitrogen contains two atoms is

therefore-.

DIATOMIC.

Similarly

Nitrogen

Hydrogen

-7

Ammonia

[Nz

3H

2

-7 2NH

3

]

1 vol. 3 vols. 2 vols. - [by Gay-Lussac s Law]

1~lecule 3molecules 2 molecules - [by Avogadro s Law]

A molecule of

hydrogen

and

chlorine

contain two atoms

are therefore - DIATOMIC.

..1atom of hydrogen combines with 1atom of chlorine to give 1molecule of hydrogen chloride.

Conclusion: Molecular formula of hydrogen chloride is -

Hel.

77



i\PPLICATIONS OF AVOGADRO'S LAW [Contd.]

. WEIGHT&VAPOURDENSI1Y

Molecular toeieht -

is the ratio i th ~ -

toeizh: o f

one m olecule of a substance to the w eight of one atom of hydrogen.

Molecular weight Mass of 1 molecule of a substance

=

[relative molecular mass]

Mass of 1 atom of hydrogen

Vapour density_- is the ratio of the -

.

m ass of a certain volum e ot vas or oapour to the m ass ot the sam e volum e of hydrogen.

[volumes measured under same conditions of temperature and pressure].

Vapour density

=

Mass of a certain volume of gas or vapour

[of gas or vapour]

Mass of the same volume of hydrogen

Application of Avogadro's Law for -

determining the relationship between molecular weight and vapour density.

Avogadro's Law

-

Volumes at same temperature and pressure substituted by molecules

Vapour density

=

Mass of 1 molecule of gas / vapour

Mass of 1 molecule of hydrogen

.'. Vapour density

=

Mass of 1 molecule of gas/vapour

[Hydrogen is diatomic]

Mass of 2 atoms of hydrogen

Multiplying both sides by 2

2 x Vapour density

=

Mass of 1 molecule of gas/vapour

l = Molecular weight]

Mass of 1 atom of hydrogen

.'.2 x Vapour density

= Molecular weight [RMM]

Conclusion: Relative molecular mass of gas or vapour is twice its vapour density.

AINS - GAY-LUSSAC'S LAW

OF

COMBINING VOLUMES

. Application of Avogadro's Law for -

explaining Gay-Lussac's Law

Consider theformation of -

Two volumes of hydrogen chloride from one volume of hydrogen

one volume of chlorine.

Hydrogen

Chlorine -7 Hydrogen chloride

1

vol.

1

vol.

2 vols. [by Gay-Lussac's Law]

n molecules n molecules 2n molecules

[by Avogadro's Law 1volume = n molecules]

2 atoms 2 atoms 2 molecules [Hydrogen, chlorine-diatomic, 1molecule = 2 atoms]

.'. One molecule of Hel is formed from - one atom of hydrogen

one atom of chlorine.

Conclusion:

Thus Avogadro's Law that equal volumes of gases -

under the same conditions of temperature pressure contains the same number of molecules-

explains Gay-Lussac's Law of combining volumes.



. ~ICATIONS OF AVOGADRO S LAW

[Contd.]

DETERMINES THE RELATION$HIP BETWEEN -

GRAM MOLECULAR MASS & GRAM MOLECULAR VOLUME

Gram molecular mass -

is the relative m olecular m ass o f a substance -

ex ressed in

erams.

It is also called - g:ra m m olecule of

t h a t

element.

Gram molecular volume - th e volum e occupied by

-1

. m olecular w t. of a gas at s.t.p.

Application of Avogadro's Law for -

determining the above relationship.

-

According to Avogadro s L aw - equal volumes of all gases-

~der the same conditions

o f

temperature

Si

pressure .' the same number ofmolecules.

T here fo re e qu al n um ber of m olecules of different gases -

under the same conditions of temperature and pressure will occupy equal volumes.

B ut one m ole o f a g as c on ta ins - 6.023x 10

23

molecules occupying - the same volume at s.t.p.

as a mole of any other gas. That same volume is the

m olar volum e of the gas.

G ram m olecular volum e

Gram molecular weight

=

[MOLAR VOLUME]

Weight per litre of gas at s.t.p.

M olar volum e of

02=

32

[mol. wt.]

M olar volum e of

H2 =

2.016

[mol. wt.] =

22.4lits.

1.429g/l 0.09 g/l

C onclu sion: The gram molecular weight [1 mole] of any gas occupies -

one gram molecular volume [molar volume] which is equal to 22.4 litres at s.t.p.

~6LE - A mole is the amount of substance which

c on t ams..

-

the same no. of units as the no. of atoms in 12.000 g. of carbon -

12 [e

12

]

-

particles [atoms / molecules / ions] equal to 6.023~

1 0

23

i.e. Avogadro's no.

r

RELATING MOLE AND ATOMIC MASS

1 MOLE OF AN ATOM - weighs 1 gram atomic mass of the atom

e.g. 1 mole of oxygen atom, weighs gram atomic mass of oxygen i.e. 16g. [0

=

16].

-

GRAM ATOM

-

Mass in grams of element

Relative atomic mass [At. wt.]

-

RELATIVE ATOMIC MASS OF AN ELEMENT [RAM] -

Atomic weight

It is the no. of tim es an atom of a n elem ent is heavier than

l/12th

the m ass of a n atom of carbon

[C

12

].

RELATING MOLE AND MOLECULAR MASS

../

1 MOLE OF ANY SUBSTANCE [Molecule] -

weighs 1 g. molecular mass of the substance

e.g.1mole ofoxygen molecule,weighs gram molecular mass ofoxygen i.e.32g. [02

=

16x2].

-

GRAM MOLECULE

=

Mass in grams of substance

Relative molecular mass [Mol. wt.]

-

RELATIVE MOLECULAR MASS OF AN ELEMENT [RMM] - Molecular weight

I t is th e n o. o f tim es o ne moleculeof the substance isheavierthan 1 /12th th e m a ss o f a n a to m o f ca rb on [C12].

1MOLE OFA GAS-weighs 1g.molecularmass-occupies22.4litres [molarvolumeats.t.p.]

MOLAR VOLUME -

Volume occupied by -1 g. molecular weight of a gas.

1 . h

g. mol. wt. .

22.4 Iits. or 22,400 cc.

. 6.023 X10

23

1 1 [

m/' ]

-- weIg s

occupIes contains . mo ecu es ato IOns

MOLE

WEIGHT VOLUME at s.t.p. PARTICLES



.LE CONCEPT f\ND AVOGADRO S NUMBER - PROBLEMS

alculate - i] The no. of moles ii] The mass ill] The vol. iv] The no. of molecules v] The gram molecular wl

1 MOLE weighs ? GRAMS occupies22.4lits.or 22,400 cc., contains 6.023 X 10

23

MOLECULES

[MOLE] [G. MOL. WT.] [VOLUME AT S.T.P.] [AVOG. NO.] [atoms/ions]

]? ? ? ? [atoms/

NO. OF MOLES WEIGHT [MASS] VOLUME NO. OF MOLECULES ions]

. Calculate the number of moles of nitrogen in

7

g of nitrogen {N=14J

Solution:

[1 mole of any substance

=

Igm mol. weight of it]

:. Molecular weight of N2

=

14 x 2

=

28

:. gram molecular weight of N2

=

28 g

:. 28 g of N2 = 1 mole of N2

-:.7 g of N2

=

1 x 7

=

0.25 moles

28

:. The number moles in 7 g of nitrogen is 0.25 moles. - Ans.

1 mole weighs

[MOLE]

? moles

a] 28 g. [N

2

]

[G. MOL. WT.]

7 g. [N

2

]

WEIGHT

b]

NO. OF MOLES

. Calculate the mass of

50

cc of

co

at s.t.p. [C=12, 0=16J

Solution:

[1mole = 1 gm mol. wt. occupies 22.4 lit. at s.t.p.]

gm mol. wt. of carbon monoxide

=

12

16

=

28 g

1 mole of CO

=

1 g. mol. wt., occupies 22400 cc [s.t.p.]

28 g of CO occupies 22400 cc [s.t.p.]

? g of CO will occupy 50 cc [s.t.p.]

28 x 50 = 0.0625 g

22400

:. Mass of 50cm

3

of CO at s.t.p. is 0.0625 g. - Ans.

a] 1

[MOLE]

b] ..

28 g. [CO] occupies 22,400cc. s.t.p.

[G. MOL. WT.] VOLUME

? g. occupied 50cc s.t.p.

WEIGHT by VOLUME

I

. Calculate the volume at s.t.p. occupied by 6.023 x 1(/22 molecules of a gas

X

Solution:

[1 mole of any substance contains 6.023 x 10

23

number of molecules [Avogadro's Number]]

a]

1 . 22.4 lits. t tai 6.023 x 10

23

M 1 1

occupies s.. p. con ams

0

ecu es

[MOLE] VOL. AVOG. NO.

b]

? t

.. .: 1 1

6.023

X

10

22

M

1

1 _ 22.4 x 6.023 x 10

2 2

-224lit

.. __ s.. p.

Wu

occupy

0

ecu es - -. s.

VOL. 6.023 X 10

23

:. The volume occupied at s.t.p. by 6.023 x 10

22

molecules of X

=

2.24 litres - Ans. I

. Calculate the number of molecules in 1kg of sodium chloride. {Na=23, CI=35.5J

Solution: [Molecular weight of NaCl = 23 + 35.5 = 58.5]

1000-g. [NaCI] contains ? Molecules

=

1000 x 6.023 x 1023

WT. 58.5

:. The number of molecules in 1 kg of NaCI is 17.1 x 6.023 x 10

23

molecules - Ans.

a]

1 weighs 58.5 g. [NaCI] contains

[MOLE] [G. MOL. WT.]

b]

I

6.023 x 10

23

M 1 1

o ecu es

AVOG.NO.

chemistry icse0001

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