chemistry. session opener session objectives problems related to 1. mole concept 2. stoichiometry
TRANSCRIPT
Chemistry
Session Opener
Session Objectives
Problems related to
1. Mole concept
2. Stoichiometry
Simple Titrations
Find out the concentration of a solutionwith the help of a solution of knownconcentration.
1 1 2 2N V N V
For mixture of two or more substances
N1V1 + N2V2 + ……= NVWhere V=(V1 + V2 + …..)
Normality of mixing two acids
1 1 2 2
1 2
N V +N VN=
V + V
Normality of mixing acid and bases
1 1 2 2
1 2
N V - N VN=
V + V
2 2 1 1
1 2
N V - N Vor N=
V + V
Questions
Illustrative example 1
Find the molality of H2SO4 solution whose specific gravity(density) is 1.98 g/ml and 95% mass by volume H2SO4.
100 ml solution contains 95 g H2SO4.
95
98Moles of H2SO4 =
Mass of solution = 100 × 1.98 = 198 g
Mass of water = 198 – 95 = 103 g
Molality =
95 1000
98 103= 9.412 m
Solution:
Molality = moles of solute
mass of solvent in kg
Illustrative example 2A sample of H2SO4 (density 1.787 g/ml) is 86% by mass. What is molarity of acid? What volume of this acid has to be used to make 1 L of 0.2 M H2SO4?
d×10×%M=
Molecular mass
1.787×10×86= =15.68 molar
98
Let V1 ml of this H2SO4 are used to prepare 1 L of 0.2 M H2SO4.M1V1 = M2V2 15.68 × V1 = 0.2 × 1000
10.2×1000
V = =12.75 ml15.68
Solution:
Illustrative example 3A mixture is obtained by mixing 500ml 0.1M H2SO4 and 200ml 0.2M HCl at 250C. Find the normality of the mixture.
2 2 1 1
1 2
N V + N VWe know, N =
V + V
For the mixture, 500 0.1 2 200 0.2 1N 0.2
700
Solution:
Illustrative example 4500 ml 0.2 N HCl is neutralized with 250 ml 0.2 N NaOH. What is the strength of the resulting solution?
Equivalents of HCl 3500 0.2 10 eqv
Equivalents of NaOH -3= 250×0.2×10 eqv
Equivalence of excess HCl 3 3(500 0.2 10 250 0.2 10 eqv)
Normality of HCl (excess)-3 3500×10 ×10
= = 0.067 N750
Strength of HCl = .067 × 36.5 g/litre
= 2.44 g/litre
Solution:
2HCl NaOH NaCl H O
Solution
1 1 2 2
1 2
N V - N VN=
V + V
0.2 × 1 × 500 - 0.2 × 1 × 250N =
500 + 250
N = 2.44 N
Strength of HCl = .067 × 36.5 grams/litre
= 2.44 grams/litre
Normality of HCl (excess),
Illustrative example 5
Calculate the empirical formula of a mineralhaving the following composition CaO=48.0% ;P2O5=41.3% ; CaCl210.7%
9
3
1
48.0/56=0.857
41.3/142=0.291
10.7/111=0.096
56
142
111
48.0
41.3
10.7
CaO
P2O5
CaCl2
Simple ratio
Mol. Mass
%tageConstituents Relative no. of const.
%=
Mol. mass
Illustrative example 6
1.5 g of an impure sample of sodiumsulpate dissolved in water was treated with excess of barium chloride solutionwhen 1.74 g of BaSO4 were obtained as dry precipitate. Calculate the percentagepurity of the sample.
2 4 2 42 23 32 16 4 137 32 4 16
142gm 233gm
Na SO BaCl BaSO 2NaCl
4 2 4
4
233 gm of BaSO is produced from 142 gmNa SO
1421.74 gm of BaSO is produced from 1.74 1.06 gm
233
2 4
2 4
The mass of pure Na SO in 1.5gm ofi mpure sample is 1.06 gm
1.06%tage purity of Na SO 100 70.67%
1.5
Illustrative example 7
A solid mixture weighing 5.00 g containing lead nitrate and sodiumnitrate was heated below 6000C until the mass of the residue was constant.If the loss of mass is 28%, find the massof lead nitrate and sodium nitrate in themixture.
2
3 2 2 22(207 16) 4462[207(14 48) ]
662
2Pb(NO ) 2PbO 4NO O
3 2
3
Let the mass of Pb(NO ) x gm
Let the mass of NaNO 5 x gm
3
3
662gm of Pb(NO ) gives residue 446 gm
446x gm of Pb(NO ) gives residue x gm
6620.674 x gm
Solution
3 2 22(23 14 32) 138gm2[23 14 48]
170gm
2NaNO 2NaNO O
3
3
170gm of NaNO gives residue 138gm
138(5 x) gm of NaNO gives residue (5 x) gm
1700.812(5 x) gm
28
Actual residue obtained (5 ) 100 3.6 gm100
3 2
3
0.674x 0.812(5 x) 3.6
or 0.318x 0.46
or x 3.33 gm
Pb(NO ) in mixture 3.33gm
and NaNO in mixture 5 3.33 1.67 gm
Illustrative example 8What volume of oxygen isrequired to effect complete combustion of200 cm3 of acetylene and what would bethe volume of CO2 formed?
2 2 2 2 22 volume 5 volume 4 volume2C H 5O 4CO 2H O
2 2 2
According to Gay lussac's law of gaseous volumes
2 volume of C H require 5 volume of O for complete combustion
3 32 2
5200 cm of C H require 200 500 cm at STP
2
2
3 32 2 2
2 volume of produce CO 4 volume
4200cm of C H produce CO 200 400 cm at STP
2
Illustrative example 9The formula weight of an acid is 82.0 in a titration.100 cm3 of a solution of this acid containing 39.0 g of the acid per litre were completely neutralised by 95.0 cm3 of aqueous NaOH containing40.0 g of NaOH per litre. What is the basicity of the acid?
Let requirement wt. of acid =E
Normality of acid 39 / E
Equivalent wt. of NaOH 40 / 1
Normality of NaOH 40 / 40 1
Solution:
Solution
1 1 2 2(acid) (NaOH)N V N V 39
100 1 95E
E 41.0
Formula wt. 82.0Basicity of acid 2
Eq. wt 41.0
Illustrative example 10
Calculate approximate molecular mass of dry air containing 78% N2 and 22% O2.
Solution:
78 22Molecular mass 28 32 28.88
100 100
Illustrative example 11
Calculate atomic mass of elementX and Y given thatComposition of X=9.76%Relative no. of atoms of X=0.406 Composition of Y=26.01Relative no. of atoms of Y=1.625
Solution:
%tage of element 9.76Atomic mass of X 24
Relative no. of Atoms of X 0.406
%tage of element 26.01Atomic mass of Y 16
Relative no. of Atoms in Y 1.625
Illustrative example 12The molality of a solution of ethyl alcohol (C2H5OH) in water is 1.55m.How manygrams of ethyl alcohol are dissolved in 2kgof water?
Solution:
2
2
1000gm of H O contain 1.55mol of ethyl alcohol
1.55 20002000gmof H O contain 3.10 mol
1000Mass of ethyl alcohol 3.10 46 142.6gm
Illustrative example 13
You are given 1 litre of 0.15 M HCl and1 litre of 0.40 M HCl. What is the maximumvolume of 0.25 M HCl which you can make from these solutions without adding any water?
Solution:Volume of 0.25 M HCl can’t be more than 2 litresbecause no water is added X litre of 0.40M HCl be added to 1L 0.15 M HCl
1 1 2 2 3 3M V M V M V 0.15 1 0.40 X 0.25 (1 X)
0.15X 0.10
0.10X 0.667litre
0.15
Total volume of 0.25M solution 1 X
1 0.667 1.667
Illustrative example 14
How many ml of H2SO4 of density 1.8 g/ml containing 92.5% by volume of H2SO4 should be added to 1 litre of 40% solution of H2SO4 ( density 1.30 g/ml) in order to prepare 50% solution of H2SO4
(density 1.4 g/ml).Molarity(M ) of solution containing 92.5% of H SO1 2 4
vol. of H SO Density 1000 92.5 1.8 10002 4 16.99M98 100 98 100
2 2 4Molarity(M ) of solution containing 40% of H SO
40 1.3 10005.31M
98 100
2 4Molarity(M) of solution containing 50% of H SO
50 1.4 10007.14M
98 100
Solution
1 1 2 2M V M V MV
16.99 V 5.31 1 7.14(1 V)
9.85V 1.83
1.83V 0.186L
9.85V 0.186L or 186 ml
1
2
Let VL of solution with molarity(M ) is added to 1L of solution
with molarity(M ) to prepare(1 V)L of solution with molarityM,
then
Illustrative example 15Gastric juice contains about 3.0 g of HCl per litre. If a person producesabout 2.5 litre of gastric juice per day, how many antacid tablets each containing400 mg of Al(OH)3 are needed to neutralizeall the HCl produced in one day?
3 3 23(35.1 1) 27 3(16 1)109.5 78g
3HCl Al(OH) AlCl 3H O
Amount of HCl produced ina day 2.5 3 7.5g
3Now 109.5g of HCl require Al(OH) 78g
378
7.5 g of HCl will require Al(OH) 7.5 5.34g109.5
5.34Number of tablets required 13.35 14 tablets
0.4
Solution:
Illustrative example 16
Calculate the number of Cl– ions in 100 ml of 0.001 M HCl solution.
Solution:HCl is a strong acid, it ionises completely
1000ml of 0.001M HCl solution contains Cl 0.001 mole
4
0.001 100100 ml of 0.001M HCl solution contains Cl
1000
1 10 mole
Conc. of HCl is equal to that of Cl– ions
23 4
19
No. of Cl ions 6.023 10 1.0 10
6.023 10
Helena kabr se farar
Group 1 elements (Alkali metals)
He
Li
Na
K
Rb
Cs
Fr
Be
Mg
Ca
Sr
Ba
Ra
Bear mugs can serve bar rats
Group 2 elements(Alkaline earth metals)
B
Al
Ga
In
Tl
Bob Allen Gave Indians Tennis Lessons
Group 13 elements (Boron family)
C
Si
Ge
Sn
Pb
Can sily or Genius snatch lead
Group 14 elements (Carbon family)
N
P
As
Sb
Bi
Never put arsenic in silver bullet bear
Group 15 elements (Nitrogen family)
O
S
Se
Te
Po
Oh, she sells tie Poles
Group-16 elements (Oxygen family)
F
Cl
Br
I
At
Fat Clyde bribed Innocent Atul
Group-17 elements (Halogen family)
Group-18 elements (The Noble Gas)
He
Ne
Ar
Kr
Xe
Rn
He needs our crazy Xerox repairman
Thank you