A mole is defined as the amount of substance containing the same number of discrete entities (atoms, molecules, ions, etc.) ... The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol)

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SOME BASIC CONCEPT OF CHEMISTRY

2. MISCELLANEOUS EXERCISES
Exercise 1: What mass of silver nitrate will react with 5.85 g of sodium chloride to produce
14.35 g of silver chloride and 8.5 g of sodium nitrate, if the law of conservation of mass is true?
Exercise 2: Elements X and Y form two different compounds. In the first, 0.324 g of X is combined with 0.471 g of Y.
In the second, 0.117 g of X is combined with 0.509 g of Y. Show that these data illustrate the law of
multiple proportions.
Exercise 3: Ammonia contains 82.35% of nitrogen and 17.65% of hydrogen. Water contains 88.90% of oxygen and
11.10% of hydrogen. Nitrogen trioxide contains 63.15% of oxygen and 36.85% of nitrogen. Show that
these data illustrate the law of reciprocal proportions.
Exercise 4: Carbon dioxide contains 27.27% of carbon, carbon disulphide contains 15.79% of carbon and sulphur
dioxide contains 50% of sulphur. Are these figures in agreement with the law of reciprocal proportions?
Exercise 5: The phosphorous trichloride contains 22.57% of phosphorous, phosphine (PH3) contains 91.18% of
phosphorous while hydrogen chloride gas contains 97.23% of chlorine. Prove by calculations, which law
is illustrated by these data.
Exercise 6: How many atoms and molecules of sulphur are present in 64.0 g of sulphur (S8)?
Exercise 7: Calculate the mass of (i) 0.1 mole of KNO3 (ii) 1  1023 molecules of methane and (iii) 112 cm3 of
hydrogen at SP.
Exercise 8: 0.5 mole of calcium carbonate is decomposed by an aqueous solution containing 25% HCl by mass.
Calculate the mass of the solution consumed.
Exercise 9: How much marble of 96.5 % purity would be required to prepare 10 litres of carbon dioxide at STP when
the marble is acted upon by dilute hydrochloric acid?
Exercise 10: Calculate the volume of air containing 21% oxygen by volume at STP, required in order to convert 294
cm3 of sulphur dioxide to sulphur trioxide under the same conditions.
Exercise 11: What is the molarity of a solution of sodium chloride (At. wts. Na = 23, Cl = 35.5) which contains 60 g of
sodium chloride in 2000 cm3 of a solution?
Exercise 12: Calculate the mole fraction of ethyl alcohol and water in a solution in which 46 g of ethyl alcohol and 180
g of water are mixed together.
Exercise 13: A 100 cm3 solution of sodium carbonate is prepared by dissolving 8.653 g of he salt in water. The
density of solution is 1.0816 g per millitre. What are the molarity and molality of the solution? (Atomic
mass of Na is 23, of Cl is 12 and of O is 16).
Exercise 14: 4.0 g of NaOH contained in one deciliter of a solution. Calculate the following in this solution.
(i) Mole fraction of NaOH (ii) Molality of NaOH (iii) Molarity of NaOH.
(At. wt. of Na = 23, O = 16; Density of NaOH solution is 1.038 g/cm3)
Exercise 15: The percentage composition (by weight) of a solution is 45% X, 15% Y, 40% Z. Calculate the mole
fraction of each component of the solution. (Molecular mass of X = 18, Y = 60, Z = 60).
ANSWER TO MISCELLANEOUS EXERCISES
Exercise 1: 17.0 g
Exercise 3: In NH3, 17.65 g of H combine with N = 82.35 g
 1 g of H combine with N
82.35
g 4.6 g
17.65
 
H
N O
H2O
N2O3
NH3
Exercise 4: Yes
Exercise 6: Molecular formula of sulphur = S8
 Molecular mass of sulphur (S8) = 32  8 = 256.0 
1 mole of sulphur molecules = 256 g = 6.023  1023 molecules of sulphur
Now, 256 g of sulphur contain 6.022  1023 molecules
 64 g of sulphur will contains
23
23
6.022 10 64
1.506 10 molecules
256
 
  
1 molecule of sulphur (S8) contains 8 atoms of sulphur
 1.056  1023 molecules of sulphur will contain sulphur atoms
= 8  1.506  1023
= 1.2048  1024 atoms

3. Exercise 7: (i) 1 mole of KNO3 = 101 g ( Formula mass of KNO3 = 1  39 + 1  14 + 3 
16 = 101 )
 0.1 mole of KNO3 = 101  0.1 = 10.1 g of KNO3
(ii) 1 mole of CH4 = 16 g = 6.022  1023 molecules
i.e., 6.022  1023 molecules of methane have mass = 16 g
 1  1023 molecules of methane have mass
23
23
16
10 2.657 g
6.022 10
  

(iii) 1 mole of H2 = 2 g = 22400 cm3 at STP
i.e., 22400 cm3 of H2 at STP have mass = 2 g
 112 cm3 of H2 at STP will have mass
2
112 0.01 g
22400
  
Exercise 8: 146 Exercise 9: 46.26 gm Exercise 10: 700 cc
Exercise 11: 0.513 Exercise 12: 0.09, 0.91 Exercise 13: 0.816M, and 0.820 m
Exercise 14: (i) 0.0177 (ii) 1 M (iii) 1.002 ml Exercise 15: X = 0.732, Y = 0.073, Z = 0.19
Solved problem
Board Type Questions
Prob 1. Carbon dioxide contains 27.27% of carbon, carbon disulphide contains 15.79% of carbon and sulphur dioxide
contains 50% of sulphur. Are these figures in agreement with the law of reciprocal proportions?
Sol. 1 g C will combine with S =
84.21
5.33g
15.79

1 g C will combine with O =
72.73
2.67g
27.27

S O
C
CS2 CO2
SO2
 
15.79%  
27.27%
 
72.73%
 
84.21%
50% 50%
 Ratio of masses of S and O which combine with fixed mass of carbon (viz 1 g)
= 5.33 : 2.67
= 2 : 1
Ratio of masses of S and O which combine directly with each other = 50:50=1:1.
Thus the two ratio are simple multiple of each other.
Prob 2. Calculate the mass of iron which will be converted into its oxide (Fe3O4) by the action of 18 of steam on it.
Sol. The chemical equation representing the reaction is
2 3 4 2
3 56 4 18
168g 72
3Fe 4H O Fe O 4H
 
 
 
 
Now 72 g of steam react with 168 g of iron
 18 g of steam will react with
168
18 42g of iron
72
 
Thus the mass of iron required = 42 g.
Prob 3. What mass of slaked lime would be required to decompose completely 4 grams of ammonium chloride and
what would be the mass of each product?
Sol. The equation representing the decomposition of NH4Cl by slaked lime, i.e. Ca(OH)2 is,
 
       
4 2 3 2
2
2 14 4 35.5 40 2 35.5 2 2 1 16
2 14 3 1
40 2 1 16
111g
107g 36g
34g
74g
Ca OH 2NH Cl CaCl 2NH 2H O
     
 
 

 


 
  

4. (i) To calculate the mass of Ca(OH)2 required to decompose 4g of NH4Cl.
From the above equation,
107 g of NH4Cl are decomposed by 74 g of Ca(OH)2.
 4 g of NH4Cl will be decomposed by
 2
74
4 2.766g of Ca OH
107
 
Thus the mass of slaked lime required = 2.766 g
(ii) To calculate the mass of CaCl2 formed.
107 g of NH4Cl when reacted with Ca(OH)2 will produce
2
111
4 4.15g of CaCl
107
 
Hence the mass of CaCl2
= 4.15 g
(iii)To calculate the mass of NH3 produced.
107 g of NH4Cl when reacted with Ca(OH)2 give 34 g of NH3.
 4 g of NH4Cl when reacted with Ca(OH)2 will produce
3
34
4 1.271g of NH
107
 
Hence, the mass of NH3 produced
= 1.271 g
(iv)To calculate the mass of H2O formed
107 g of NH4Cl when reacted with Ca(OH)2 yield 36 g of H2O.
 4 g of NH4Cl when reacted with Ca(OH)2 will yield =
2
36
4 1.3458g of H O
107
 
So the mass of H2O formed = 1.3458 g
Prob 4. 1.5 g of an impure sample of sodium sulphate dissolved in water was treated with excess of barium chloride
solution when 1.74 g of BaSO4 were obtained as dry precipitate. Calculate the percentage purity of the sample.
Sol. 2
treated with
2 4 4
BaCl gave
Given that 1.5g of impure Na SO 1.74 g of BaSO


The chemical equation representing the reaction is
2 4 2 4
2 23 32 4 16 137 32 4 16
142g 233g
Na SO BaCl BaSO 2NaCl
      
 
 
 
Step 1. To calculate the mass of Na2SO4 from 1.74 of BaSO4. From the chemical equation:
233 g of BaSO4 are produced from 42 of Na2SO4.
 1.74 of it would be obtained form
142
1.74 1.06g
233
 
The mass of pure Na2SO4 present in 1.5 g of impure sample = 1.06 g.
Step 2. To calculate the percentage purity of impure sample.
1.5 g of impure sample contains 1.06 g of pure Na2SO4.
 100 g of the impure sample will contain
1.06
100 70.67
1.5
   g of pure Na2SO4.
Prob 5. A solution contains 25% water, 25% ethanol and 50% acetic acid by mass. Calculate the mole fraction of each
component.
Sol. Let us suppose the total mass of the solution is 100 g. Then
Amount of water = 25 g
Amount of ethanol = 25 g
Amount of acetic acid = 50g
 
Percentage bymass are given
25 g of water =
25
1.388 moles
18

 
Mol. mass of water 18


5. 25 g of ethanol =
25
0.543 moles
46

 
2 5
Mol. mass of ethanol C H OH 46
 

 
50 g of acetic acid =
50
0.833moles
60

 
3
Mol. mass of acetic acid CH COOH 60
 

 
 Mole fraction of water
1.388 1.388
0.503
1.388 0.543 0.833 2.764
  
 
0.543
Mole fraction of ethanol 0.196
2.764
 
and mole fraction of acetic acid =
0.833
0.301
2.764

IIT Level Questions
Prob 6. In an experiment, 2.4g of iron oxide on reduction with hydrogen yield 1.68g of iron. In another experiment 2.9g
of iron oxide give 2.03g of iron in reduction with hydrogen. Show that these data, illustrate the law of constant
composition.
Sol. In the first experiment
The mass of iron oxide = 2.4g
Mass of iron after reduction = 1.68
So mass of oxygen = Mass of oxide – Mass of iron
= 2.4 – 1.68 = 0.72
O: Fe = 0.72: 1.68 = 1:2.33
In second experiment mass of iron oxide = 2.9g
Mass of iron after reduction = 2.03g
Mass of oxygen = 2.9 – 2.03 = 0.87g
O : Fe = 0.87 : 2.03 = 1:2.33
This dated illustrate the law of constant proportion.
Prob 7. Certain non metal X forms two oxides I and II. The mass % of oxygen in 1st
4 6
(X O ) is 43.7, which is same as
that of X in nd
2 oxide. Find the formula of nd
2 oxide.
Sol. X Oxygen
Ist oxide 42.7 56.3
2nd oxide 56.3 43.7
Now 43.7 parts of oxygen corresponds to = 6 oxygen atoms
56.3 part of oxygen corresponds to
6
56.3 7.73
43.7
   oxygen atom
56.3 part of X in I correspond to = 4 X atoms
43.7 part of X in II will correspond
4
43.7 3.1 X atom
56.3
  
Now atomic ratio X: oxygen
3.1 7.73
: 1: 2.5 2 : 5
3.1 3.1
  
Formula of nd
2 oxide 2 5
= X O
Prob 8. Two oxides of a metal contain 27.6% and 30.0% of oxygen respectively. If the formula of the first oxide
is 3 4
M O , find that of the second.
Sol. In the first oxide, oxygen = 27.6, metal = 100-27.6 = 72.4 parts by weight.
As the formula of the oxide is 3 4
M O , this means 72.4 parts by wt. of metal = 3 atoms of metal and 4 atoms of
oxygen = 27.6 parts by weight.
In the second oxide, oxygen = 30.0 parts by weight and metal = 100-30 =70 parts by weight.
But 72.4 parts by weight of metal = 3 atoms of metal

6. 70 parts by weight of metal
3
70 atoms of metal
72.4
 
= 2.90 atoms of metal
Also, 27.6 parts by weight of oxygen = 4 atoms of oxygen
30 parts by weight of oxygen
4
30 atoms of oxygen
27.6
 
= 4.35 atoms of oxygen
Hence, ratio of M: O in the second oxide = 2.90: 4.35 = 1:1.5 = 2:3
Formula of the metal oxide is 2 3
M O .
Prob 9. A crystalline salt on being rendered anhydrous loses 45.6% of its weight. The percentage composition of the
anhydrous salt is
Aluminium = 10.50%; Potassium = 15.1%; Sulphur = 24.96%; Oxygen = 49.92%
Find the simplest formula of the anhydrous salt.
Sol. Step 1: To calculate the empirical formula of the anhydrous salt.
Element
Symbol
Percentage
of
elements
At.
mass
of
elements
Relative no. of atoms
=
Percentage
At. mass
Simplest
atomic
ratio
Simplest
whole
no.
atomic
ratio
Potassium K 15.1 39 15.10
0.39
39

0.39
1
0.39

1
Aluminium Al 10.50 27 10.50
0.39
27

0.39
1
0.39

1
Sulphur S 24.96 32 24.96
0.78
32

0.78
2
6.39

2
Oxygen O 49.92 16 49.92
3.12
16

3.12
8
0.39

8
Thus the empirical formula for the anhydrous salt is K AlS2O8.
Step 2: to calculate the empirical formula mass of the anhydrous salt.
Empirical formula mass of the anhydrous salt (KAIS2O8).
= 1  39.0 + 1  27.0 + 2  32.0 + 8  16.0
= 258.0 u
Step 3: To calculate the empirical formula mass of the hydrated salt.
Let of weight due to dehydration = 45%
 Empirical formula mass of the anhydrous salt = 100 – 45.6 = 54.4 u
Now, if the empirical formula mass of the anhydrous salt is 54.4 that of hydrated is
= 100
 If the empirical formula mass of the anhydrous salt is 258, that of hydrated is
=
100
258 474.3u
54.4
 
Step 4: To calculate the number of molecules of water in the hydrated salt.
Total loss in mass due to dehydration
= 474.3 – 258.0 = 216.3 u
Loss in mass due to one molecule of water = 18.0 u
 No. of molecule of water in the hydrated sample =
216.3
12
18

Step 5: To calculate the empirical formula of the hydrated salt.
Empirical formula of the anhydrous salt = KAS2O8.
No. of molecules of water of crystallization = 12
 Empirical formula of the hydrated salt = KAIS2O8.12H2O.

7. Prob 10. A carbon compound contains 12.8% carbon, 2.1% hydrogen, 85.1% bromine. The molecular weight of the
compound is 187.9. Calculate the molecular formula.
Sol. Step 1: Percentage composition of the elements present in the compound.
C H Br
12.8 2.1 85.1
Step 2: Dividing with the respective atomic weights of the elements.
12.8
12
2.1
1
85.1
80
1.067 2.1 1.067
Step 3: Dividing by the smallest number to get simple atomic ratio.
1.067
1.067
2.1
1.067
1.067
1.067
1 2 1
The empirical formula is 2
CH Br .
Empirical formula weight 12 + (2 1) + 80 = 94

The molecular weight = 187.9 (given)
187.9
n 2
94
  
2
The molecular formula = (empirical formula)

2 2 2 4 2
(CH Br) = C H Br
Prob 11. Calculate the number of molecules present in 350 cm3 of NH3 gas at 273 K and
2 atomsphere pressure.
Sol. First of all, we have to determine the volume of the gas at STP.
Given conditions At STP
V1 = 350 cm3 V2 = ?
T1 = 273 K T2 = 273 K
P1 = 2 atomsphere P2 = 1 atm
1 1 2 2
1 2
P V P V
Applying gas equation :
T T

We get 2
1 V
350 2
273 273



or 3
2
350 2 273
V 700cm
273 1

  
By mole concept,
1 mole of NH3 = 6.022  1023 molecules = 22400 cm3 at STP.
Thus, 22400 cm3 of NH3 at STP contain 6.022  1023 molecules.
 700 cm3 of NH3 at STP will contain
23
6.022 10
700
22400

 
= 1.882  1022 molecules
Prob 12. In order to find the strength of a sample of sulphuric acid, 10 g were diluted with water and a piece of marble
weighing 7g placed in it. When all action had ceased, the marble was removed, washed, dried and was found
to weigh 2.2 g. What was the percentage strength of sulphuric acid?
Sol. Mass of marble taken = 7.0 g
Mass of marble left unused = 2.2 g
 Mass of marble reacted = 7.0 – 2.2 = 4.8 g
The chemical equation involved in the above problem is
3 2 4 4 2 2
2 32 4 16 98g
40 12 16 3 100g
CaCO H SO CaSO H O CO
   
   
 
  

8. Step 1. To calculate the mass of pure H2SO4 required to react 4.8 g of marble.
100 g of marble react with H2SO4 = 98 g
 4.8 g of marble will react with H2SO4
98
4.8 4.704g
100
  
Step 2. To calculate the strength of sulphuric acid.
10 gm of dil. H2SO4 contain pure H2SO4 =
4.704
100 47.04
10
 
Thus the percentage strength of sulphuric acid = 47.04.
Prob 13. Calculate the mole fraction of ethanol and water in sample of rectified spirit which contains 95 per cent of
ethanol by weight.
Sol. As the rectified spirit contains 95% of ethanol by weight, this means that 95 g ethanol are present in 100 g of
the spirit.
 If the weight of ethanol in the spirit = 95 g
Then the weight of water in the spirit = 5g
95 g ethanol (C2H5OH) =
95
mole of ethanol
46
= 2.07 moles
 
2 5
Mol. mass C H OH 46

5 g of water (H2O) =
5
mole of water
18
 
2
Mol. mass H O 18

= 0.28 mole
 Mole fraction of ethanol
2.07 2.07
0.88
2.07 0.28 2.35
  

 Mole fraction of water = 1 – 0.88 = 0.12.
Prob 14. Calculate the molality and mole fraction of the solute in aqueous solution containing
3.0 g of urea (mol. mass = 60) per 250 g of water.
Sol. Mass of solute (urea) dissolved = 3.0 g
Mass of the solvent (water) = 250 g
Molecular mass of the solute = 60
Calculation of molality:
3.0 g of the solute =
3.0
mole 0.05mole
60

 
Mol. mass of urea 60

Thus 250 g of the solvent contain solute
0.05
1000 0.2 mole
250
  
Hence molality of the solution = 0.2 m
Calculation of mole fraction:
3.0g of solute =
3
moles 0.05 mole
60

250 g of water =
250
moles 13.89 moles
18


9.  
2
Mol. mass of H O 18

 Mole fraction of the solute
0.05 0.05
0.00359
0.05 13.89 13.94
  

Prob 15. The mole fraction of benzene in a solution in toluene is 0.50. Calculate the weight percent of benzene in the
solution.
Sol. Weight of benzene = xg
Weight of toluene = (100 – x)
Mol. wt. of benzene = (C6H5CH3) = 78
Mol. wt. of toluene (C6H5CH3) = 92
 
B
B T
n
0.50
n n
 

 
x /78
i.e. 0.50
x /78 100 x /92

 
 
x 78 92
0.50
78 92x 78 100 x

 
 
or 92x = 46x + 3900 – 39x
or 85 x = 3900 or x = 45.9%
Alternatively, B
B T
n
0.50
n n


B
B T
w /78
i.e. 0.5
w /78 w /92


B B T
w w w
or 0.5
78 78 92
 
 
 
 
B B T
2w w w
78 78 92
 
or B B T
B
w w w 92
or
78 92 w 78
 
or T B T
B B
w w w
92 170
1 1 or
w 78 w 78

   
or B
B T
w 78
0.459
w w 170
 

Hence wt% = 45.9
Objective:
Prob 1. A gas mixture of 3 litres of propane 3 8
(C H ) and butane 4 10
(C H )on complete combustion at o
25 C produced 10
litres 2
CO . Find the ratio of 3 8 4 10
C H : C H
(A) 1:2 (B) 1:1
(C) 2:3 (D) 2:1
Sol. 3 8 2 2 2 ( )
C H + 5O 3CO + 4H O


a 3a
4 10 2 2 2 ( )
13
C H + O 4CO + 5H O
2


b 4b

10. a + b = 3
3a + 4b = 10
Solving b = 1, a = 2 litre
 (D)
Prob 2. Calculate molarity of HCl of density 1.17g/ml
(A) 32 M (B) 34M
(C) 16M (D) 8 M
Sol. B
( )
n
M
V
 no. of moles of HCl in 1 litre HCl
1.17 1000
32
36.5

 
 (A)
Prob 3. 112 ml of a gas is produced at STP by the action of 0.412 gm of ROH alcohol with 3
CH MgI . Molecular mass of
alcohol is:
(A) 32g (B) 41.2g
(C) 82.4g (D) 156g
Sol.  
3 4
ROH + CH MgI CH + MgI OR


1 mole 1 mole
So gas produced is 4
CH
4
ROH
CH ROH
KOH
w
112
n n
22400 MW
  
22400
MW 0.412 82.4g
112
  
 (C)
Prob 4. 10 ml of liquid carbon disulphide (specific gravity = 2.63) is burnt in oxygen. Find the volume of the resulting
gases measured at STP
(A) 11.2 litre (B) 22.4 litre
(C) 44.8 litre (D) 67.2 litre
Sol. 10 ml of CS2 = 10  2.63 g = 26.3 g
2 2 2 2
CS 3O CO 2SO
 
 
76 g 3 moles
= 3  22.4 L at STP
(D)
Prob 5. Which of the conc. Terms are temperature independent
(A) Molality, Mass %, Mole fraction (B) Mass%, Molarity, Normality
(C) Molality, Normality, Volume % (D) ppm, molaity, m/v%
Sol. (A)
Prob 6. Calculate the molarity of water in pure water.
(A) 5.55 (B) 55.5
(C) 0.55 (D) cant be calculated
Sol. 1 litre of pure water = 1000 cm3 = 1000 g (density of water = 1 gm/cc)
 Number of mole in 1 litre of pure water =
1000
55.55
18

(B)

11. Prob 7. 12 gm of Mg (At mass = 24) will react with an acid to give
(A) one mole of H2 (B) one mole of O2
(C) half moles of H2 (D) half mole of O2
Sol. 2
Mg 2H Mg H
 
 
 
24 g 1 mole
 12 g of Mg gives half mole of H2.
Hence (C) is correct.
Prob 8. Which of the following samples contains the largest number of atoms?
(A) 1 g of Ni(s) (B) 1 g of Ca(s)
(C) 1 g of N2(g) (D) 1 g of B(s)
Sol. 1 gram Nickel =
1
mole Niatom
59
; 1 gram Ca =
1
mole Ca atom
40
, 1 gram N2 has
1
mole of Natom
14
; 1 gram of B
=
1
mole B atom
11
.
So now it is clear, choice (D) is correct for larger number of atom.
Prob 9. How many gm of NaOH will be needed to prepare 250 ml 0.1 M solution?
(A) 1 gm (B) 10 gm
(C) 4 gm (D) 6 gm
Sol.
xgm
1000mL
40gm
Molarity 0.1
250mL

 
 x = 1 gm of NaOH
Hence (A) is correct
Fill in the blanks
Prob 10. The idea of equivalent masses of elements follows from the law of ………………….. proportions.
Sol. Reciprocal
Prob 11. In a chemical change mass is neither created nor destroyed. This is the law of …………………..
Sol. Conservation of mass
Prob 12. Molecular mass of oxygen is 32 amu and gram molecular mass of oxygen is ……………………….
Sol. 32 g
Prob 13. The number of moles of barium carbonate which contains 1.5 moles of oxygen atoms is ………………………..
Sol. 0.5
Prob 14. A mixture having 2 g of H2 and 32 g of oxygen occupies a volume of ……………………. at NTP.
Sol. 44.8 litre
Match the following
Prob 15. Match the following choosing one item from column X and the appropriate related item from column Y.
Column X Column Y
(a) Equal volumes of all gases contain equal
number of molecules at NTP.
(i) Dalton’s atomic theory
(b) The atom is indestructible. (ii) Law of conservation of mass
(c) All pure samples of the same compound
contain the same elements combined in
the same proportion by mass.
(iii) Avogadro’s law
(d) Total mass before and after the chemical
reaction is same.
(iv) Dulong and Petit’s law
(e) 6.4
Atomic mass
Specific heat

(v) Gay-Lussac law
(f) Gases react in simple ratio of their
volumes.
(vi) Law of constant proportions
Sol. (a) – (iii), (b) – (i), (c) – (vi), (d) – (ii), (e) – (iv), (f) – (v)

12. Assertion–Reason type questions
The following questions consist of two statements each printed as Assertion and Reason. While answering these
questions, you are required to choose any one of the following five responses.
(A) If both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.
(B) If both Assertion and Reason are true but Reason is not a correct explanation of the Assertion.
(C) If Assertion is true but the Reason is false.
(D) If the Assertion is false but the Reason is true.
(E) If both Assertion and Reason are false.
Prob 16. Assertion: Cinnabar is a chemical compound whereas brass is a mixture.
Reason: Cinnabar always contains 6.25 times as much mercury as sulphur by weight. Brass can be made
with widely different ratios of copper and zinc.
Sol. A
Prob 17. Assertion: A single C12 atom has a mass exactly 12 amu and a mole of these atoms has a mass of exactly 12
grams.
Reason: A mole of atoms of any element has a mass in grams equal to its atomic weight.
Sol. A
Prob 18. Assertion: Pure water obtained from different sources such as river, well, spring, sea etc. always contains
hydrogen and oxygen in the ratio of 1 : 8 by mass.
Reason: Mass of reactants and products during chemical or physical change is always the same.
Sol. B
Prob 19. Assertion: In a gaseous reaction, the ratio of volumes of reactants and gaseous products is in agreement with
their molar ratio.
Reason: Volume of gas is inversely proportional to its moles at particular temperature and pressure.
Sol. C
Prob 20. Assertion: The standard unit for expressing the mass of atoms is amu.
Reason: amu is also called as avogram.
Sol. B
ASSIGNMENT PROBLEMS
Subjective:
Level - O
1. Define Element, Compound and Mixture.
2. Give three main points of difference between a compound and a mixture.
3. 0.44 g of hydrocarbon on complete combustion with oxygen gave 1.8 g of water and 0.88 g of carbon dioxide. Show
that these results are in agreement with the law of conservation of mass.
4. Give one experiment involving a chemical reaction to prove that the law of conservation of mass is true.
5. N and O combine with H to form 3
NH and 2
H O and they combine with each other to form 2
NO . Which law is
illustrated? Explain?
6. 10 mL of hydrogen combine with 5 mL of oxygen to yield water. When 200 mL of hydrogen at NTP are passed over
heated CuO, the latter loses 0.144 g of its mass. Do these results agree with the law of constant composition?
7. Define Gay Lussac’s Law of gaseous volumes.

13. 8. Air contains 20% of oxygen by volume Calculate the theoretical volume of air which will be required for the burning of
200 cm3 of acetylene gas completely. All volumes are measured under the same conditions of temperature and
pressure.
9. What are the postulates of Dalton’s Atomic Theory? How do the laws of chemical combination follow from it?
10. How is mole related to?
(a) number of atoms / molecules
(b) mass of the substance
(c) volume of the gaseous substance?
11. How many molecules of the water and atoms of oxygen are present in 0.9 g of water?
12. What is the difference between
(a) Normality and Molarity?
(b) Molarity and Molality?
13. Define Empirical formula and Molecular formula. What is the relationship between them?
14. What is a limiting reactant? Explain with suitable example.
15. The vapour density of gaseous element is 5 times that of oxygen under similar conditions of temperature and
pressure. If the molecule of the element is triatomic, what will be its atomic mass?
Level – I
1. Calculate the volume of hydrogen liberated at NTP when 500 cm3 of 0.5 M sulphuric acid reacts with excess of zinc.
2. How many grams of barium chlorine (BaCl2) are needed to prepare 100 cm3 of 0.250 M BaCl2 solution?
3. Give one limitation of the law of constant composition.
4. Which law co-relates the mass and volume of a gas?
5. What is the difference between the mass of a molecule and molecular mass?
6. Why is the value Avogadro’s number 6.022  1023 and not any other value?
7. Where do we use the words mole and mol?
8. What is the basic difference between empirical and molecular formulae?
9. Why law of conservation of mass should better be called as law of conservation of mass and energy?
10. From 200 mg of carbon dioxide, 1021 molecules are removed. How many moles of the gas are left?
11. You are supplied with a gas containing 0.32 g of oxygen. Calculate the number of moles and number of molecules
present in it.
12. Determine the empirical formula of a compound having percentage composition as:
Fe = 20%; S = 11.5%; O = 23.1% and H2O molecules = 45.4%
13. 10 grams of a sample of potassium chlorate gave on complete combustion 2.24 litres of oxygen at NTP. What is the
percentage purity of the sample of potassium chlorate?
14. Find out the weight of calcium carbonate that must be decomposed to produce sufficient quantity of carbon dioxide
to convert 10.6 g of sodium carbonate completely into sodium bicarbonate.
15. 4.90 g of KClO3 when heated produced 1.92 g of oxygen and the residue (KCl) left behind weighs 2.96 g. Show that
these results illustrate the law of conservation of mass.

14. Level – II
1. The masses of equal volumes of a gas and hydrogen are 25.6 g of 0.8 g respectively under same conditions of
temperature and pressure. Find the molecular mass of the gas.
2. A typical analysis of pyrex glass showed 12.9% B2O3, 2.2% Al2O3, 3.8% Na2O, 0.4% K2O and 80.7% SiO2. What is
the ratio of silicon to boron atoms in glass?
3. 2.0 g of hydrated copper sulphite gave on heating, 1.2786 g of anhydrous salt. Calculate the number of molecules of
water of crystallization per molecule of hydrated salt. [Cu = 63.5,
S = 32]
4. The average molar mass of mixture of methane (CH4) and ethene (C2H4) present in the ratio of a:b is found to be
20.0 g mol1
. If the ratio were reversed, what would be the molar mass of the mixture?
5. Sodium chlorate, 3
NaClO , can be prepared by the following series of reactions
4 2 2 2
2KMnO + 16 HCl 2KCl + 2MnCl + 8H O + 5Cl


2 2 3 2 2 2
6Cl + 6Ca(OH) Ca(ClO ) + 5CaCl + 6H O


3 2 2 4 4 3
Ca(ClO ) + Na SO CaSO + 2NaClO


What mass of 3
NaClO can be prepared from 100 cc of concentrated HCl -3
(density = 1.18 g cm , 36 mass % HCl)?
Assume all other substances are present in excess amounts.
6. A mineral consists of an equimolar mixture of the carbonates of two bivalent metals. One metal is present to the
extent of 13.2% by weight. 2.58 g. of the mineral on heating lost 1.233 g. of 2
CO . Calculate the % by weight of the
other metal.
7. Chlorine gas can be produced in the laboratory by the reaction
2 2 7 3 2 2
K Cr O + 14HCl 2KCl + 2CrCl + 7H O + 3Cl (g)


If a 68.0 g sample of that is 96% 2 2 7
K Cr O is allowed to react with 320 cc of a HCl solution having density of
-3
1.15 g cm and containing 30 mass % HCl, what mass of 2
Cl is generated?
8. 200 gms of ‘marble chips’ are dropped into one kilogram of solution of HCl containing one-tenth of its weight of the
pure acid. How much of chips will remain undissolved. What weight of anhydrous calcium chloride and what weight
of 2
CO gas could be obtained from it?
9. 1.6 gm of pyrolusite was treated with 50 mL of 0.5 M oxalic acid and some sulphuric acid. The oxalic acid left
undecomposed was raised to 250 ml in a flask. 25 ml this solution when treated with 0.02 M 4
KMnO required 32 mL
of the solution. Find the % of 2
MnO in the sample and also the percentage of available oxygen.
10. Upon mixing 45.0 ml of 0.25 M lead nitrate solution with 25.0 ml of 0.10 M chromic sulphate solution, precipitation of
lead sulphate takes place. How many moles of lead sulphate are formed? Also calculate the molar concentration of
the species left behind in the final solution. Assume lead sulphate is completely insoluble (At. wt. of Pb = 207.2).
Objective:
Level – I
1. Potassium combines with two isotopes of chlorine (35Cl and 37Cl) respectively to form two samples of KCl. Their
formation follows the law of:
(A) constant proportions (B) multiple proportions
(C) reciprocal proportions (D) none of these
2. H2S contains 5.88% hydrogen, H2O contains 11.11% hydrogen while SO2 contains 50% sulphur. These figures
illustrate the law of:
(A) conservation of mass (B) constant proportions
(C) multiple proportions (D) reciprocal proportions
3. The best standard of atomic mass is:
(A) carbon-12 (B) oxygen-16
(C) hydrogen-1.008 (D) chlorine-35.5

15. 4. The chemical formula of a particular compound represents:
(A) the size of its molecule
(B) the shape of its molecule
(C) the total number of atoms in a molecule
(D) the number of different types of atoms in a molecule
5. Two containers P and Q of equal volume (1 litre each) contain 6 g of O2 and SO2 respectively at 300 K and 1
atmosphere. Then
(A) Number of molecules in P is less than that in Q
(B) Number of molecules in Q is less than that in P
(C) Number of molecules in P and Q are same
(D) Either (A) or (B)
6. The product of atomic mass and specific heat of any element is a constant, approximately 6.4. This is known as:
(A) Dalton’s law (B) Avogadro’s law
(C) Gay-Lussac law (D) Dulong Petit’s law
7. 250 ml of a sodium carbonate solution contains 2.65 grams of Na2CO3. If 10 ml of this solution is diluted to one litre,
what is the concentration of the resultant solution? (mol wt. of Na2CO3 = 106)
(A) 0.1 M (B) 0.001M
(C) 0.01 M (D) 104
M
8. 7.5 grams of a gas occupy 5.6 litres of volume at STP. The gas is
(A) NO (B) N2O
(C) CO (D) CO2
9. The weight of a molecule of the compound C60H122 is
(A) 1.4  1021
g (B) 1.09  1021
g
(C) 5.025  1023 g (D) 16.023  1023 g
10. 1.0 mole of CO2 contains:
(A) 6.02 × 1023 atoms of C (B) 6.02 × 1023 atoms of O
(C) 18.1 × 1023 molecules of CO2 (D) 3 g-atoms of CO2
11. The number of atoms in 1.4 g nitrogen gas is:
(A) 6.02 × 1022 (B) 3.01 × 1022
(C) 1.20 × 1023 (D) 6.02 × 1023
12. Which of the following has the smallest number of molecules?
(A) 22.4 × 103 ml of CO2 gas (B) 22 g of CO2 gas
(C) 11.2 litre of CO2 gas (D) 0.1 mole of CO2 gas
13. The number of grams of H2SO4 present in 0.25 mole of H2SO4 is
(A) 0.245 (B) 2.45
(C) 24.5 (D) 49.0
14. At NTP 1.0 g hydrogen has volume in litre:
(A) 1.12 (B) 22.4
(C) 2.24 (D) 11.2
15. 19.7 kg of gold was recovered from a smuggler. The atoms of gold recovered are: (Au = 197)
(A) 100 (B) 6.02 × 1023
(C) 6.02 × 1024 (D) 6.02 × 1025
16. The molecular mass of CO2 is 44 amu and Avogadro’s number is 6.02 × 1023. Therefore, the mass of one molecule
of CO2 is:
(A) 7.31 × 10–23 (B) 3.65 × 10–23
(C) 1.01 × 10–23 (D) 2.01 × 10–23
17. The number of moles of H2 in 0.224 litre of hydrogen gas at NTP is:
(A) 1 (B) 0.1
(C) 0.01 (D) 0.001
18. Irrespective of the source, pure sample of water always yields 88.89% mass of oxygen and 11.11% mass of
hydrogen. This is explained by the law of:
(A) conservation of mass (B) constant composition
(C) multiple proportion (D) constant volume

16. 19. Given the numbers: 161 cm; 0.161 cm; 0.0161 cm. The number of significant figure for three numbers is:
(A) 3, 4, 5 (B) 3, 3, 3
(C) 3, 3, 4 (D) 3, 4, 4
20. Which one of the following law directly explains the law of conservation of mass?
(A) Hund’s rule (B) Dalton’s law
(C) Avogadro’s law (D) Berzelius hypothesis
Level – II
1. 6.023 ×1023 molecules of Ca (OH)2 react with 3.01×1022 molecules of HCl, number of moles of CaCl2 obtained are
(A) 0.05 (B) 0.10
(C) 0.025 (D) 3.01
2. A copper sulphate solution contains 1.595% of CuSO4 by weight. Its density is 1.2 g / ml, Its molarity will be
(A) 0.12 (B) 0.06
(C) 1.20 (D) 1.595
3. Which of the following samples contains 2.0  1023 atoms?
(A) 8.0 g O2 (B) 3.0 g Be
(C) 8.0 g C (D) 19.0 g F2
4. Choose the wrong statement:-
(A) 1 Mole means 6.02  1023 particles
(B) Molar mass is mass of one molecule
(C) Molar mass is mass of one mole of a substance
(D) Molar mass is molecular mass expressed in grams
5. What quantity of limestone (CaCO3) on heating will give 56 Kg of CaO?
(A) 1000 Kg (B) 44 Kg
(C) 56 Kg (D) 100 Kg
6. Simplest formulae of a compound containing 50% of element X (atomic weight 10) and 50% of element Y (atomic
weight 20) is
(A) XY (B) X2Y
(C) XY3 (D) X2Y3
7. At room temperature and pressure two flask of equal volumes are filled with H2 and SO2 respectively. Particles which
are equal in number in two flasks are
(A) Atoms (B) Electrons
(C) Molecules (D) Neutrons
8. Chlorophyll contains 2.68% of magnesium by mass. Calculate the number of magnesium atoms in 3.00 gms of
chlorophyll.
(A) 2.01  1021 atoms (B) 6.023  1023 atoms
(C) 1.7  1020 atoms (D) 2.8  1022 atoms
9. What is the total number of atoms present in 25.0 mg of camphor C10H16O?
(A) 9.89  1019 (B) 6.02  1020
(C) 9.89  1020 (D) 2.67  1021
10. 2 mol of H2S and 11.2 L SO2 at N.T.P. reacts to form x mol of sulphur; x is
SO2 + 2H2S  3S + 2H2O
(A) 1.5 (B) 3
(C) 11.2 (D) 6
11. How many grams of phosphoric acid (H3PO4) would be needed to neutralise 100 g of magnesium hydroxide
(Mg(OH)2).
(A) 66.7 g (B) 252
(C) 112.6 g (D) 168 g
12. If 0.5 mol of BaCl2 is mixed with 0.2 mol of Na3PO4, the maximum number of mol of Ba3(PO4)2 that can be formed is
(A) 0.7 (B) 0.5
(C) 0.2 (D) 0.1

17. 13. The number of electron in the telluride ion. Te2– is
(A) 50 (B) 52
(C) 53 (D) 54
14. An ore contains 1.34% of the mineral argentite, Ag2S, by weight. How many grams of this ore would have to be
processed in order to obtain 1.00 g of pure solid silver, Ag?
(A) 74.6 g (B) 85.7 g
(C) 134.0 g (D) 171.4 g
15. Hydrogen evolved at NTP on complete reaction of 27 gm of Al with excess of aq NaOH would be
(Chemical reaction: 2Al + 2NaOH + 2H2O  2NaAlO2 + 3H2)
(A) 22.4 lit (B) 44.8 lit
(C) 67.2 lit (D) 33.6 lit
ANSWERS TO ASSIGNMENT PROBLEMS
Subjective:
Level – O
6. Yes 8. 2500 cm3 of air
11. 3.011  1022 15. 53.33
Level – I
1. 5600 cm3 2. 5.20 g
3. When isotopes of an element take part in the formation of a compound, then the same compound has different ratios
of the elements.
4. It is Avogadro’s law and states that equal volumes of all gases under similar conditions of temperature and pressure
contain equal number of moles (or molecules).
5. Mass of a molecule is that of a single molecule also known as its actual mass. But molecular mass is the mass of
Avogadro’s number (6.022  1023) of molecules.
6. It represents the number of atoms in one gram atom of an element or the number of molecules in one mole of a
compound. If we divide the atomic mass of an element by actual mass its atom, the value is 6.022  1023. Similarly,
by dividing the molecular mass of a compound by the actual mass of its molecules, the same result is obtained.
7. In the part, we use the word mole while as a unit, we call it mol.
8. Empirical formula gives the simplest ratio of the atoms of different elements in the molecules a substance but the
molecules formula gives their actual ratio.
9. In nuclear reaction, it is observed that the mass of the product is less than the mass of reactants. The difference of
mass, called the mass defect, is converted into energy according to Einstein equation, E = mc2. Hence we better
call it as law of conservation of mass and energy.
10. 2.68  103
11. 0.01 mole, 6.022  1021 molecules
12. FeSO4.7H2O 13. 81.7%
14. 10 g
15. Mass of KClO3 taken = 4.90 g
Total mass of the products (KCl + O2) = 2.96 + 1.92 = 4.88 g
Level –II
1. 64.0 4. 24 g mol1
5. 12.91g 6. 21.68%
7. 47.31g 8. 63.01g, 152.06g; 60.27g
9. 9% 10. 0.0075 mol

18. Objective:
Level – I
1. D 2. D 3. A
4. D 5. B 6. D
7. B 8. A 9. A
10. A 11. A 12. D
13. C 14. D 15. D
16. A 17. C 18. B
19. B 20. C
Level – II
1. C 2. A 3. B
4. B 5. D 6. B
7. C 8. A 9. A
10. A 11. C 12. D
13. B 14. B 15. D