Mole concept & Some Basics of Chemistry
Time: 2 hours
Total Marks: 50
GENERAL INSTRUCTIONS
1.
2.
3.
4.
5.
6.
7.
All questions are compulsory.
The question paper consists of 24 questions.
Paper dividend into four section A, B, C and D.
Section A comprises of 8 questions of 1 mark each.
Section B comprises of 10 questions of 2 marks each.
Section C comprises of 4 questions of 3 marks each.
Section D comprises of 2 questions of 5 marks each.
SECTION - A
1 × 8 = 8 Marks
1. Define Accuracy & Precision
Sol. Accuracy is a measure of the difference between the experimental value or the mean value of a set of
measurements & the true value.
Precision refers to how closely two or more measurements of the same quantity agree with one
another.
2. Explain law of reciprocal proportions.
Sol. Law of reciprocal proportion:
When two different elements combine separately with the fixed mass of a 3rd element. The ratio in which
they do so well be the same or some simple multiple of the ratio in which they combine with each other
3. Give one limitation of the law of constant composition.
Sol. When isotopes of an element take part in the formation of a compound, then the same compound has
different ratios of the elements.
For example,
In CO2, with C – 12 isotope, the ratio of C : O : : 12 : 32
In CO2, with C – 14 isotope, the ratio of C : O : : 14 : 32
4. What is a limiting reagent?
Sol. The reactant which is present in a lesser amount than calculated by balanced chemical equation and
thus gets entirely consumed when a reaction goes to completion is called a limiting reagent.
5. Which aqueous solution has higher concentration : 1 molar or 1 molal solution of the same solute?
Give reason.
Sol. 1 M has higher concentration than 1 m. This is because 1 M solution means 1 mole of the solute in
1000 cc of the solution whereas 1 m solution means 1 mole of the solute in 1000 g of water (= 1000
c.c. of water). Total volume of 1 m solution > 1000 c.c. due to presence of extra 1 mole of the solute.
Hence number of moles/cc of 1 m solution will be less than that in 1 M solution. So 1 M is more
concentrated than 1 m.
6. How many moles and how many grams of sodium chloride (NaCl) are present in 250 mL of a 0.50 M
NaCl solution?
Sol. A 0.50 M NaCl solution contains 0.50 mol of NaCl in 1 L or 1000 mL of solution. Therefore, number of
0.50 mol  250 mL
moles of NaCl in 250 mL of solution =
= 0.125 mol
1000 ml
mass of NaCl =0.125 × 58.5
Molar mass of NaCl = 58.5 g mol-1
7. Express the 32.392800 number to four significant figure
Sol. 32.39
8. What is density of water in 1 kg/m3 if in CGS unit it is 1 gm/m
gm 10 3 kg
Sol. D 

 10 3 kg/m 3
m 10 - 6 m3
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Section – B
2 × 10 = 20 Marks
(a) What are the equivalent weights of each of these compounds assuming the formula weights of
these compounds are x, y and z respectively.
(i) Na2SO4
(ii) Na3PO4.12H2O
(iii) Ca3(PO4)2
(b) Which is more concentration 1 N or 1 M H2SO4
Sol. (a) Equivalent weight = Molecular weight/Total positive valency of metal atoms. Thus equivalent
weights of the above compounds are x/2, y/3 and z/6 respectively.
(b) 1 M more concentrated than 1 N as it will have 98 gm H2SO4. Whereas 1 N will have 49 gm.
10. If law of constant composition is true, what weights of calcium, carbon and oxygen are present in 1.5 g
of calcium carbonate? Given that the sample of calcium carbonate from another sample contains Ca =
40.0%, C = 12.0% and O = 48.0%
1.5  40
Sol.
Weight of Ca 
 0.6 g
100
1.5  12
Weight of C 
 1.18 g
100
1.5  48
Weight of O 
 0.72 g
100
11. Why must oxidation and reduction occur together in a reaction?
Sol. In a redox reaction, the oxidation and reduction must occur together because it is a reaction in which
electrons are transferred between species and the electrical neutrality of the system is conserved. In
oxidation, there is a loss of electrons by a species and in reduction there is a gain of electron by a
species.
12. (i) Can the reaction Cr2O72– + H2O
2CrO42– + 2H+ be regarded as a redox reaction?
Sol. O.N. of Cr in Cr2O72– = +6 ;
O.N. of Cr in CrO42– = + 6
As the O.N. of Cr remains unaltered, the above reaction cannot be regarded as a redox reaction.
(ii) Determine the oxidation number of O in the following:
OF2, Na2O2 and KO2.
Sol. (i) OF2
Let the oxidation number of O = x
Oxidation number of each F = - 1
x–2=0
x=+2
9.
(ii) Na2O2
Let the oxidation number of O = x
Oxidation number of each Na = +1
2 + 2x = 0
or
2x = - 2
or
x=-1
(iii) KO2
+1 + 2x = 0
2x = - 1
1
x= 
2
13. An organic compound on analysis gave the following data: C = 57.82%, H = 3.6% and the rest is
oxygen. Its vapour density is 83. Find its empirical and molecular formula.
Sol.
Element
Mass in g per 100 g
Number of moles
Divide by 2.41
Multiplication by 2
C
57.82
57.82
 4.82
2.0
4
12
H
3.6
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2
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O
100 – (57.82 + 3.6)
= 38.58
3.6
 3.60
1
38.58
 2.41
16
1
2
Hence, the empirical formula is C4H3O2
Empirical formula weight is 48 + 3 + 32 = 83
Molecular formula weight = 2 V.D. = 166
MFW 166

n

2
EFW
83
M.F. = C8H6O4
14. An impure sample of sodium chloride which weighs 0.50 g gave, on treatment with excess of silver
nitrate solution, 0.90 g of silver chloride as a white precipitate. Calculate the percentage purity of the
sample.
Sol.
AgNO3 + NaCl  AgCl NaNO 3
1 mol
1 mol
143.5 g or 1 mol of ACl will precipitate from 58.5 g NaCl
0.90
 0.90 g AgCl will require 58.5 × 58.5 
143 .5
= 0.37 g NaCl
0.37
Percentage purity of NaCl =
 100
0.50
= 74%
15. (i) How many of these can act as only oxidising agent & why?
H2SO4, KMnO4, K2Cr2O7, N2O5, NO2
(ii) Find equivalent weight of reactant in the given change [At. Wt. of Sb = 121.76; O = 16]
Sb2O3  Sb2O5
Mol. mass
121 2  3  16 290
Sol. (ii) eq. mass 
; equivalent mass 

 72.5
4
4
4
16. 23 g of ethanol CH3CH2OH() on reaction with ethanoic acid CH3COOH form 22g of ethyl ethanoate by
esterification in the presence of conc. H2SO4. What is the % yield in the reaction.
Sol. CH3CH2OH + CH3COOH  CH3COOC2H5 + H2O
1 mole ethanol = 1 mole ethyl ethanoate
46 g ethanol = 88 g
88
23g ethanol =
 88 = 44 CH3COOC2H5
46
Actual mass = 22gm
Actual yield
22
% yield =
 100  100% 
 100  50%
Threotrica l yield
44
17. 15 gm of metal on oxidation give metal oxide which further on hydrolysis gives 20 gm metal
hydroxide. Calculate the equivalent weight of metal
15
20
Sol.

E E  17 ; 15 E + 255 = 20 E;
255 = 15 E ;
E = 17
18. Vapour density of chloride is 77. Equivalent mass is 3. What is valency of metal in chloride MCl x
Molecular mass
Sol. x =
E  35.5
2  Vapour density
=
3  35.5
=4
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SECTION – C
3 × 4 = 12 Marks
19. Balance this reaction by oxidation number method.
K2Cr2O7 + KI + H2SO4  K2SO4 + Cr2(SO4)3 + I2 + H2O
Sol. ionic equation
Cr2O7–2 + I–  Cr+3
+ I2
Balance No. of atom & find increase & decrease in oxidation number
(1) Cr2O7–2
+
2I–  2Cr+3
+
I2
Inc. by 2
dec. by 6
(2) Balance Increase & decrease in oxidation number
Cr2O7–2 + 6I–  2Cr+3 + 3I2
(3) Balance charge by adding H+ ion
Cr2O7–2 + 6I– + 8H+  2Cr+3 + 3I2
(4) Balance hydrogen by adding H2O
add spectator ion Cr2O7–2 + 6I– + 14H+  2Cr+3 + 3I2 + 7H2O
(5) K2Cr2O7 + 6KI + 7H2SO4  Cr2(SO4)3 + 3I2 + 7H2O + 4K2SO4
20. A sample of coal gas contained CO, CH4 and H2. 20 mL of this mixture was exploded with 80 mL
oxygen. After cooling the volume of the gas mixture was found to be 68 mL. On shaking with KOH
solution there was a contraction of 10 mL in volume. Find out percentage composition of the coal
gas.
Sol.
Let CO = x mL, CH4 = y mL,
H2 = (20 – x – y) mL, O2 = 80 mL
(1) 2CO  O2  2CO 2
1 v ol
x mL
(3)
2H2
1 v ol
(20 - x - 4)

1
v ol
2
x
mL
2
(2) CH 4  2O 2  CO 2  2H2 O
1 v ol
x mL
1 v ol
x mL
2 v ol
2y mL
1 v ol
y mL
O2
 2CO 2
1
1 v ol
v ol
x mL
2
 20  x  4 

 mol
2


CO2 formed = x + y = 10 mL
Volume of O2 unused = 68 – 10 = 58
Volume of O2 used = 80 – 58 = 22
x + y = 10
0.5x + 2y +
20  x  y
= 22
2
…(i)
…(ii)
From equation (i) & (ii)
%age Composition:
y=8
x = 2 ml
Volume of H2 = 10 ml
2
 100  10%
20
8
CH4 = y = 8 =
 100  40%
20
10
H2 = (20 – x – y) = 10 =
 100  50%
20
CO = x = 2 =
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21. An aqueous solution of sodium chloride is marked 10% (w/w) on the bottle. The density of the solution
is 1.071 g/mL. What is its molality and molarity? Also, find mole fraction of each component in the
solution?
Sol. 10% (w/w) solution means 100 g of solution contains 10 g NaCl

wNaCl = 10 g and w H2O  90g
10
90
 0.17 and nH2O 
 5 mol
58.5
18
n
0.17
(i) Molality = B  1000 
 1000
wA
90
= 1.89 molal
100 g
(ii) Volume of solution 
1.071 g/mL
100

mL
1.071
1

L
10.71
M
0.17
Molarity  B  1000 
 1.071 1000
V(m)
100
nNaCl 
= 1.82
%  d  10
or M 
M.Wt .
(iii) Mole fraction of NaCl = xNaCl
nNaCl

nNaCl  n H2 O
0.17
 0.033
0.17  5.0
Therefore, the mole fraction of H2O = xH2O  1  0.033

= 0.967
22. A mixture containing 100 g of H2 and 100 g O2 is ignited so that water is formed according to the
reaction.
2H2 + O2  2H2O
(i) How much water is formed?
(ii) Which is the limiting reagent
(iii) Calculate the volume of the gas left unreacted at STP.
Sol. 2H2 + O2  2H2O
100 g H2 = 50 mol H2
100
100 g O2 
 3.125 mol O2
32
2 mol H2 + 1 mol O2  2 mol H2o
2 × 3.125 mol H + 3.125 mol O2  2 × 3.125 mol H2O
More H2 is present than required. Therefore, O2 is the limiting reactant
Amount of H2O formed
= 2 × 3.125 mol H2O
= 2 × 3.125 × 18
= 112.5 g H2O
Number of moles of H2 left unreacted = (50 – 2 × 3.125) = 43.75 mol H2
Volume occupied by 43.75 mol H2 at STP = 43.75 × 22.4 lit
= 980 L H2 at STP
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SECTION – D
5 × 2 = 10 Marks
23. (i) Balance the following redox reaction in basic medium using the half-reaction method:
OH
MnO4– + I–  MnO2 + IO 3
(ii) Balance the following ionic equations by using half reaction method:
I– + MnO4–  IO 3 + MnO2 (acidic medium)
Sol. (i) MnO 4 + I–  MnO2 + IO 3
Reduction Reaction
Oxidation Reaction
I IO 3  6e 
3e   MnO 4  MnO 2
7
( 1)
4
( 5)
Multiply by 2
6e– + 2MnO 4  2MnO2
Add two equation
2MnO 4  I  2MnO 2  IO 3Balance oxygen by adding H2O
2MnO 4 + I–  2MnO2 + IO 3 + H2O
Balance Hydrogen by adding H+
2MnO 4  I-  2H  2MnO 2  IO 3  H2O
Neutralise hydrogen by adding OH– ion
2MnO -4 + I- + 2H+ + 2OH -  2OH   2MnO 2  IO 3  H2O
2H2O
2MnO 4  I  H2O  2OH   2MnO 2  IO 3
 3e 
1
(ii)
7
5
4
I   MnO 4  IO 3  MnO 2
 6e 
Multiply the coefficient of MnO 4 by 2 we have
 6e 
1
7
5
4
I   2 MnO 4  IO 3  2MnO 2
 6e 
Balance O atom by adding one H2O to the right side.
I– + 2MnO4–  IO3– + 2MnO2 + H2O
We balance H atoms by adding 2H+ to the left. Hence, the complete balanced equation is
I– + 2MnO4– + 2H+  IO 3 + 2MnO2 + H2O
24. An aqueous solution of H3PO4 10% (w/v). The density of solution is 1.2 gm/ml. Calculate
(a) Molarity
(b) Molality
(c) Normality
(d) Mass %
(e) Mole fraction
Sol. M.Wt. = 98
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(a) Molarity (M) =
(b) Molality =
10 1000

 1.02 mole/lit
98 100
10
1000

98 Wsolv ent
Wsolvent = W solution – Wsolute
= Volume of solution × density – 10
= 100 × 1.2 – 10
= 110
m=
10 1000

98 110 = 0.092 M
(c) Normality = 3 × M = 3.06 N
(d) Mass % 

mass of solute
 100
Total mass
10  100
10

 100  8.33%
100  1.2 120
(e) Mole fraction
Mole of H3PO 4 
Mole of H2O =
H3PO4 
10
 0.102
98
110
 6.111
18
Mole of solute 0.102

 0.016
total mole
6.21
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