Mole Concept Notes - Theory, Exercises, Answers & More

Topic Page No.
Theory 01 - 04
Exercise - 1 05 - 13
Answer Key 20 - 21
Contents
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Syllabus
MOLE CONCEPT
Concept of atoms and molecules; Dalton’s Atomic theory; Mole concept; Chemi-
cal formulae; Balanced chemical equations; Calculations (based on mole con-
cept) involving common oxidation-reduction, neutralisation and displacement
reactions; Concentration in terms of mole fraction, molarity & molality.
Name:____________________________ Contact No. __________________

MOLE CONCEPT # 1
F-106, Road No.2 Indraprastha Industrial Area, End of Evergreen Motor,
BSNL Lane, Jhalawar Road, Kota, Rajasthan (324005) Tel. : +91-744-242-5022, 92-14-233303
SIGNIFICANT FIGURES :
(A) Every scientific observation involves some degree of uncertainity depending upon the limitation of
instrument. To represent scientific data, role of significant figures has its own importance.
(B) Significant figures are equal to the number of digits in numbers with last digit uncertain and rest all
are certain digits i.e. all the digits of datum including the uncertain one , are called significant figures.
(C) Rules for determination significant figure:
(i) All non zero digits are significant.
Example : 3.14 has three significant figures
(ii) The zeros to the right of the decimal point are significant.
Example : 3.0 has two significant figures.
(iii) The zeros to the left of the first non zero digit in a number are not significant.
Example : 0.02 has one significant figure.
(iv) The zeros between two non zero digits are also significant.
Example : 6.01 has three significant figures.
(v) Exponential form : N × 10n. Where N show the significant figure.
Example : 1.86 × 104 has three significant figure.
(vi) Rounding off the uncertain digit :
(a) If the left most digit to be rounded off is more than 5 , the preceding number is increased by one.
Example : 2.16 is rounded to 2.2
(b) If the left most digit to be rounded off is less than 5 , the preceding number is retained.
Example : 2.14 is rounded off to 2.1
(c) If the left most digit to be rounded off is equal to 5 , the preceding number is not changed if it is even
and increased by one if it is odd.
Example : 3.25 is rounded off to 3.2 2.35 is round off to 2.4
LAWS OF CHEMICAL COMBINATION :
Law of definite proportion [Proust, 1799]
(a) According to the law , the composition of a compound always remains a constant i.e. the ratio of
weights of different elements in a compound ; no matter by whatever method , it is prepared or obtained
from different sources, remains always a constant
Example : In H2O ratio of weight = 1 : 8
In CO2 ratio of weight = 3 : 8
Law of multiple proportion [John Dalton, 1804]
According to this law, when two elements A and B combine to form more than one chemical compound
then different weights of A , which combine with a fixed weight of B , are in a proportion of simple whole
number
MOLE CONCEPT

MOLE CONCEPT # 2
Example : CO & CO2
12 : 16 & 12 : 32
ratio = 16 : 32
= 1 : 2
Law of reciprocal proportions [Ritche, 1792-94]
When two elements combines separately with third element and form different types of molecules, their
combining ratio is directly reciprocated if they combine directly
Example :
C with H form methane and with O form CO2. In CH4 , 12 grams of C reacts with 4 grams of H whereas
in CO2 12 gram of C reacts with 32 grams of O. Therefore when H combines with O they should combine
in the ratio of 4 : 32 (i.e. = 1 : 8) or in simple multiple of it. The same is found to be true in H2O molecule.
The ratio of weights of H and O in Water is 1 : 8
Gay-Lussac’s [1808] law of combining volumes :
This law states that under similar conditions of pressure and temperature, volume ratio of gases is always
in terms of simple integers.
Ex.
N2(g) + 3H2(g)  2NH3(g)
vol. ratio 1 : 3 : 2
AVOGADRO'S HYPOTHESIS :
or Vapour density =
Molecular weight
2
On hydrogen scale :
Relative atomic mass (R.A.M) =
atomhydrogenoneofmass
elementanofatomoneofMass
Relative atomic mass (R.A.M) =
atom12Coneofmass
12
1
elementanofatomoneofMass

Atomic mass unit (or amu) :
The atomic mass unit (amu) is equal to
th
12
1






mass of one atom of carbon-12 isotope.
 1 amu =
12
1
× mass of one C-12 atom
~ mass of one nucleon in C-12 atom.
= 1.66 × 10–24 gm or 1.66 × 10–27 kg
 Atomic mass = R.A.M × 1 amu

MOLE CONCEPT # 3
Y-map : Interconversion of mole - volume, mass and number of particles :
Mole
× 22.4 lt
 22.4 lt
Volume at STP

N
A
× N
A
Number
× mol. wt.
× At. wt.

 At. wt.
mol. wt.
Mass
Relative density or Vapour density :
Vapour density is defined as the density of the gas with respect to hydrogen gas at the same temperature
and pressure.
Vapour density =
2H
gas
d
d
V.D. =
2H
gas
M
M
=
2
Mgas
Mgas = 2 V.D.
Relative density can be calculated w.r.t. to other gases also.
 Mole-mole analysis :
This analysis is very much important for quantitative analysis point of view. Students are advised to
clearly understand this analysis.
Now consider again the decomposition of KClO3 .
2KClO3
 2KCl + 3O2
In very first step of mole-mole analysis you should read the balanced chemical equation like
2 moles KClO3 on decomposition gives you 2 moles KCl and 3 moles O2. and from the stoichi-
ometry of reaction we can write
2
KClOofMoles 3
=
2
KClofMoles
=
3
OofMoles 2
Now for any general balance chemical equation like
a A + b B  c C + d D
you can write.
a
reactedAofMoles
=
b
reactedBofmoles
=
c
producedCofmoles
=
d
producedDofmoles
Molarity (M) :
The number of moles of a solute dissolved in 1 L (1000 ml) of the solution is known as the
molarity of the solution.
i.e., Molarity of solution =
litreinsolutionofvolume
soluteofmolesofnumber

MOLE CONCEPT # 4
Molality (m) :
The number of moles of solute dissolved in1000 gm (1 kg) of a solvent is known as the molality of
the solution.
i.e., molality = 1000
graminsolventofmass
soluteofmolesofnumber

Mole fraction (x) :
The ratio of number of moles of the solute or solvent present in the solution and the total number
of moles present in the solution is known as the mole fraction of substances concerned.
% calculation :
The concentration of a solution may also expressed in terms of percentage in the following way.
 % weight by weight (w/w) : It is given as mass of solute present in per 100 gm of solution.
i.e. % w/w = 100
gminsolutionofmass
gminsoluteofmass

 % weight by volume (w/v) : It is given as mass of solute present in per 100 ml of solution.
i.e., % w/v = 100
mlinsolutionofolumev
gminsoluteofmass

 % volume by volume (v/v) : It is given as volume of solute present in per 100 ml solution.
i.e., % v/v = mlinsolutionofvolume
mlinsoluteofvolume
× 100
Miscellaneous:
AVERAGE/ MEAN ATOMIC MASS :
The weighted average of the isotopic masses of the element’s naturally occuring isotopes.
Mathematically, average atomic mass of X (Ax) =
100
xa.....xaxa nn2211 
Where :
a1, a2, a3 ........... atomic mass of isotopes.
and x1, x2, x3 ........... mole % of isotopes.

MOLE CONCEPT # 5
PART - I : OBJECTIVE QUESTIONS
* Marked Questions are having more than one correct option.
Section (A) : Atoms, molecules, moles, avogadro's hypothesis
A-1. The charge on 1 gram ions of Al3+
is : (NA = Avogadro number, e = charge on one electron)
(A)
27
1
NAe coulomb (B)
3
1
× NAe coulomb (C)
9
1
× NAe coulomb (D) 3 × NAe coulomb
A-2. What is correct for 10 g of CaCO3 -
(A) It contains 1g-atom of carbon (B) It contains 0.3 g-atoms of oxygen
(C) It contains 12 g of calcium (D) None of these
A-3. Which of the following contains the largest number of atoms -
(A) 11g of CO2 (B) 4g of H2 (C) 5g of NH3 (D) 8g of SO2
A-4. If the atomic mass of Sodium is 23, the number of moles in 46 g of sodium is :
(A) 1 (B) 2 (C) 2.3 (D) 4.6
A-5. Out of 1.0 g dioxygen, 1.0 g (atomic) oxygen and 1.0 g of ozone, the maximum number of oxygen
atoms are contained in -
(A) 1.0 g of atomic oxygen. (B) 1.0 g of ozone.
(C) 1.0 g of oxygen gas. (D) All contain same number of atoms.
A-6. One mole of P4 molecules contains -
(A) 1 molecule (B) 4 molecules
(C) 1/4 × 6.022 × 1023
atoms (D) 24.088 × 1023
atoms
A-7. The number of sodium atoms in 2 moles of sodium ferrocyanide Na4[Fe(CN)6], is-
(A) 2 (B) 6.023 × 1023
(C) 8 × 6.02 × 1023
(D) 4 × 6.02 × 1023
A-8. Which of the following will contain same number of atoms as 20g of calcium?
(A) 24g magnesium (B) 12g carbon (C) 8g oxygen gas (D) 16 g oxygen atom
A-9. The percentage by mole of NO2 in a mixture of NO2(g) and NO(g) having average molecular mass 34 is :
(A) 25% (B) 20% (C) 40% (D) 75%
A-10*. In which of the following pairs do 1 g of each have an equal number of molecules?
(A) N2O and CO (B) N2 and C3O2 (C) N2 and CO (D) N2O and CO2
Section (B) : Density and vapour density
B-1. 5.6 litre of a gas at N.T.P. weighs equal to 8 gm the vapour density of gas is -
(A) 32 (B) 16 (C) 8 (D) 40.
B-2. 2 moles of H2 at NTP occupy a volume of
(A) 11.2 litre (B) 44.8 litre (C) 2 litre (D) 22.4 litre

MOLE CONCEPT # 6
B-3. 4.48 litres of methane at N.T.P. correspond to-
(A) 1.2 x 1022
molecules of methane (B) 0.5 mole of methane
(C) 3.2 gm of methane (D) 0.1 mole of methane
B-4. Mol. wt. = vapour density × 2, is valid for -
(A) metals (B) non metals (C) solids (D) gases
B-5. Number of moles of water in 488 gm of BaCl2.2H2O are - (Ba = 137)
(A) 2 moles (B* ) 4 moles (C) 3 moles (D) 5 moles
B-6. Density of ozone relative to methane under the same temperature & pressure conditions is :
(A) 1 (B) 3 (C) 1.5 (D) 2.5
B-7. Vapour density of a gas if its density is 0.178 g/L at NTP is :
(A) 0.178 (B) 2 (C) 4 (D) 0.089
B-8. A nugget of gold and quartz was found to contain x g of gold and y g of quartz and has density d. If the
densities of gold and quartz are d1 and d2 respectively then the correct relation is :
(A)
1d
x
+
2d
y
=
d
yx 
(B) xd1 + yd2 = (x + y) d
(C)
2d
x
+
1d
y
=
d
yx 
(D)
d
yx 
+
1d
x
+
2d
x
= 0
B-9. The vapour density of a gas A is twice that of a gas B. If the molecular weight of B is M, the molecular
weight of A will be :
(A) M (B) 2M (C) 3M (D)
2
M
B-10. Mass of H2O in 1000 kg CuSO4.5H2O is - (Cu = 63.5)
(A) 360.5 kg (B) 36.05 kg (C) 3605 kg (D) 3.605 kg
B-11. The relative density of a gas A with respect to another gas B is 2. The vapour density of the gas B is 20, the
vapour density of the gas A is :
(A) 30 (B) 40 (C) 50 (D) 60
B-12. The ratio of the weight of one litre of a gas to the weight of 1.0 L oxygen gas both measured at S.T.P. is
2.22. The molecular weight of the gas would be :
(A) 14.002 (B) 35.52 (C) 71.04 (D) 55.56
B-13. The molar mass of normal water is .......... as compared to heavy water.
(A) 10% less (B) 10% high (C) 2% less (D) zero% less
Section (C) : Percentage composition and molecular formula
C-1. The mass of carbon present in 0.5 mole of K4[Fe(CN)6] is -
(A) 1.8 gm (B) 18 gm (C) 3.6 gm (D) 36 gm
C-2. The percentage of nitrogen in urea is about-
(A) 38.4 (B) 46.6 (C) 59.1 (D) 61.3
C-3. The empirical formula of a compound of molecular mass 120 is CH2O. The molecular formula of the
compound is :
(A) C2H4O2 (B) C4H8O4 (C) C3H6O3 (D) all of these

MOLE CONCEPT # 7
C-4. Insulin contains 3.4% sulphur by mass. What will be the minimum molecular weight of insulin -
(A) 94.117 (B) 1884 (C) 941 (D) 976
C-5. The percent of N in 66% pure (NH4)2 SO4 sample is -
(A) 32 (B) 28 (C) 14 (D) None of these
C-6. A compound was found to contain 5.37% nitrogen by mass. What is the minimum molecular weight
of compound-
(A) 26.07 (B) 2.607 (C) 260.7 (D) None
C-7. Calculate the molecular formula of compound which contains 20% Ca and 80% Br (by wt.) if molecular
weight of compound is 200. (Atomic wt. Ca = 40, Br = 80)
(A) Ca1/2Br (B) CaBr2 (C) CaBr (D) Ca2Br
C-8. The empirical formula of a compound is CH. Its molecular weight is 78. The molecular formula of the
compound will be -
(A) C2H2 (B) C3H3 (C) C4H4 (D) C6H6
C-9. An oxide of a metal (M) contains 40% by mass of oxygen. Metal (M) has atomic mass of 24. The
empirical formula of the oxide is-
(A) M2O (B) MO (C) M2O3 (D) M3O4
C-10. Two oxides of Metal contain 27.6% and 30% oxygen respectively. If the formula of first oxide is M3O4
then formula of second oxide is -
(A) MO (B) M2O (C) M2O3 (D) MO2
C-11. The simplest formula of a compound containing 50% of element X (atomic mass = 10) and 50% of the
element Y (atomic mass = 20) by weight is
(A) XY (B) X2Y (C) XY2 (D) X2Y3.
Section (D) : Balanced chemical equation analysis & Limiting Reagent
D-1. The mass of oxygen that would be required to produce enough CO, which completely reduces
1.6 kg Fe2O3 (at. mass Fe = 56) is :
(A) 240 gm (B) 480 gm (C) 720 gm (D) 960 gm
D-2. The moles of O2 required for reacting with 6.8 gm of ammonia.
(..... NH3 +..... O2  ..... NO + ..... H2O) is
(A) 5 (B) 2.5 (C) 1 (D) 0.5
D-3. What weight of CaCO3 must be decomposed to produce the sufficient quantity of carbon dioxide to convert
21.2 kg of Na2CO3 completely into NaHCO3. [Atomic mass Na = 23, Ca = 40]
CaCO3
 CaO + CO2
Na2 CO3 + CO2 + H2O  2NaHCO3
(A) 100 Kg (B) 20 Kg (C) 120 Kg (D) 30 Kg
D-4. 12 g of alkaline earth metal gives 14.8 g of its nitride. Atomic weight of metal is -
(A) 12 (B) 20 (C) 40 (D) 14.8
D-5. NX is produced by the following step of reactions
M + X2  M X2
3MX2 + X2  M3X8
M3 X8 + N2CO3  NX + CO2 + M3O4
How much M (metal) is consumed to produce 206 gm of NX. (Take at wt of M = 56, N=23, X = 80)
(A) 42 gm (B) 56 gm (C)
3
14
gm (D)
4
7
gm

MOLE CONCEPT # 8
D-6. How many mol Fe2+
ions are formed, when excess of iron is treated with 50mL of 4.0M HCl under
inert atmosphere ? Assume no change in volume -
(A) 0.4 (B) 0.1 (C) 0.2 (D) 0.8
D-7. A mixture containing 100 gm H2 and 100 gm O2 is ignited so that water is formed according to the reaction,
2H2 + O2  2H2O; How much water will be formed -
(A) 113 gm (B) 50 gm (C) 25 gm (D) 200 gm
D-8. 0.5 mole of H2SO4 is mixed with 0.2 mole of Ca (OH)2. The maximum number of moles of CaSO4
formed is -
(A) 0.2 (B) 0.5 (C) 0.4 (D) 1.5
D-9. For the reaction : A + 2B  C
5 mole of A and 8 mole of B will produce -
(A) 5 mole of C (B) 4 mole of C (C) 8 mole of C (D) 13 mole of C
D-10. 12 litre of H2 and 11.2 litre of Cl2 are mixed and exploded. The composition by volume of mixture
is -
(A) 24 litre of HCl (B) 0.8 litre Cl2 and 20.8 lit HCl.
(C) 0.8 litre H2 & 22.4 litre HCl (D) 22.4 litre HCl
D-11. How many mole of Zn(FeS2) can be made from 2 mole zinc, 3 mole iron and 5 mole sulphur.
(A) 2 mole (B) 3 mole (C) 4 mole (D) 5 mole
D-12. For the reaction 2P + Q  R, 8 mol of P and 5 mol of Q will produce
(A) 8 mol of R (B) 5 mol of R (C) 4 mol of R (D) 13 mol of R
D-13. Equal weight of 'X' (At. wt. = 36) and 'Y' (At. wt. = 24) are reacted to form the compound X2Y3. Then :
(A) X is the limiting reagent
(B) Y is the limiting reagent
(C) No reactant is left over and mass of X2Y3 formed is double the mass of ‘X’ taken
(D) none of these
D-14*. A + B  A3B2 (unbalanced)
A3B2 + C  A3B2C2 (unbalanced)
Above two reactions are carried out by taking 3 moles each of A and B and one mole of C. Then which
option is/are correct ?
(A) 1 mole of A3B2C2 is formed (B) 21 mole of AA3B2C2 is formed
(C) 1/2 mole of A3B2 is formed (D) 21 mole of AA3B2 is left finally
D.15* If 27 g of Carbon is mixed with 88 g of Oxygen and is allowed to burn to produce CO2 , then :
(A) Oxygen is the limiting reagent. (B) Volume of CO2 gas produced at NTP is 50.4 L.
(C) C and O combine in mass ratio 3 : 8. (D) Volume of unreacted O2 at STP is 11.2 L.
D-16. Phosphine (PH3) decomposes to produce vapours of phosphorus and H2 gas. What will be the
change in volume when 100 mL of phosphine is decomposed ? PH3  P4 + H2
(A) + 50 mL (B) 500 mL (C) + 75 mL (D) – 500 mL
D-17. How many moles of potassium chlorate need to be heated to produce 11.2 litre oxygen at N.T.P.
(A)
2
1
mol (B)
3
1
mol (C)
4
1
mol (D)
3
2
mol

MOLE CONCEPT # 9
D-18. The molar ratio of Fe++
to Fe+++
in a mixture of FeSO4 and Fe2(SO4)3 having equal number of sulphate ion
in both ferrous and ferric sulphate is
(A) 1 : 2 (B) 3 : 2 (C) 2 : 3 (D) can't be determined
D-19. If 1
2
1
moles of oxygen combine with Al to form Al2O3, the weight of Al used in the reaction is : (Al = 27)
(A) 27g (B) 54g (C) 40.5g (D) 81 g
D-20. If 1 gm of HCl and 1 gm of MnO2 heated together the maximum weight of Cl2 gas evolved will be
[MnO2 + 4HCl  MnCl2 + Cl2 + 2H2O] [Atomic mass of Mn=55]
(A) 2gm (B) 0.975 gm (C) 0.486 gm (D) 0.972 gm
D-21. 0.5 mole of H2SO4 is mixed with 0.2 mole of Ca (OH)2. The maximum number of moles of CaSO4 formed
is
(A) 0.2 (B) 0.5 (C) 0.4 (D) 1.5
D-22. A sample of ammonium phosphate (NH4)3PO4 contains 3.18 mol of H atoms. The number of mol of
O atoms in the sample is :
(A) 0.265 (B) 0.795 (C) 1.06 (D) 3.18
Section (E) : Principle of Atom Conservation (POAC)
E-1. A sample of calcium carbonate is 80% pure. 25g of this sample is treated with excess of HCl. How much
volume of CO2 will be obtained at NTP.
(A) 4.48 litre (B) 5.6 litre (C) 11.2 litre (D) 2.24 litre
E-2. 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 1.00g of pure silver. (Ag) -
(A) 74.6g (B) 85.7g (C) 107.9g (D) 134.0g
E-3. 25.4 g of iodine and 14.2g of chlorine are made to react completely to yield a mixture of Cl and Cl3.
Calculate the number of moles of Cl and Cl3 formed.
(A) 0.1 mole, 0.1 mole (B) 0.1 mole, 0.2 mole (C) 0.5 mole, 0.5 mole (D) 0.2 mole, 0.2 mole
E-4. What weights of P4O6 and P4O10 will be produced by the combustion of 31g of P4 in 32g of oxygen leaving
no P4 and O2.
(A) 2.75g, 219.5g (B) 27.5g, 35.5g (C) 55g, 71g (D) 17.5g, 190.5g
E-5. 21.6 g of silver coin is dissolved in HNO3. When NaCl is added to this solution, all silver is precipitated as
AgCl. The weight of AgCl is found to be 14.35 g then % of silver in coin is :
(A) 50% (B) 75% (C) 100% (D) 15%
Section (F) : Concentration terms
F-1. What volume of a 0.8 M solution contains 100 milli moles of the solute?
(A) 100 mL (B) 125 mL (C) 500 mL (D) 62.5 mL
F-2. If 500 ml of 1 M solution of glucose is mixed with 500 mlof 1 M solution of glucose final molarity of solution
will be :
(A) 1 M (B) 0.5 M (C) 2 M (D) 1.5 M
F-3. The volume of water that must be added to a mixture of 250 ml of 0.6 M HCl and 750 ml of
0.2 M HCl to obtain 0.25 M solution of HCl is :
(A) 750 ml (B) 100 ml (C) 200 m (D) 300 m
F-4. What volume of 0.10 M H2SO4 must be added to 50 mL of a 0.10 M NaOH solution to make a solution in
which the molarity of the H2SO4 is 0.050 M ?
(A) 400 mL (B) 50 mL (C) 100 mL (D) 150 mL

MOLE CONCEPT # 10
F-5. 500 mL of a glucose solution contains 6.02 × 1022
molecules. The concentration of the solution is
(A) 0.1 M (B) 1.0 M (C) 0.2 M (D) 2.0 M
F-6. Mole fraction of A in H2O is 0.2. The molality of A in H2O is :
(A) 13.9 (B) 15.5 (C) 14.5 (D) 16.8
F-7. The molarity of the solution containing 2.8%( mass / volume) solution of KOH is : (Given atomic mass of
K = 39 ) is :
(A) 0.1 M (B) 0.5 M (C) 0.2 M (D) 1 M
F-8. What is the molarity of H2SO4 solution that has a density of 1.84 g/cc and contains 98% by mass of
H2SO4? (Given atomic mass of S = 32)
(A) 4.18 M (B) 8.14 M (C) 18.4 M (D) 18 M
F-9. The molality of a sulphuric acid solution is 0.2. Calculate the total weight of the solution having
1000 gm of solvent.
(A) 1000 g (B) 1098.6 g (C) 980.4 g (D) 1019.6g
F-10. Suppose you want an acidic solution to carry out a chemical reaction with 2 moles of NaOH. Which sample
of acid is the best choice for you. (At. wt. : S = 32, Cl = 35.5)
(A) 1 M H2SO4 (50 Rs per lt.) (B) 1 M H2SO4 (56 Rs per lt.)
(C) 1 M HCl (30 Rs per lt.) (D) 1 M HCl (27 Rs per lt.)
F-11. 2M of 100 ml Na2 SO4 is mixed with 3M of 100 ml NaCl solution and 1M of 200 ml CaCl2 solution. Then the
ratio of the concentration of cation and anion.
(A) 1/2 (B) 2 (C) 1.5 (D) 1
F-12. H3PO4 (98 g mol–1
) is 98% by mass of solution. If the density is 1.8 g/ml, the molarity is
(A) 18 M (B) 36 M (C) 54 M (D) 0.18 M
F-13. 50 mL solution of BaCl2 (20.8% w/v) and 100 mL solution of H2SO4 (9.8% w/v) are mixed
(Ba = 137, Cl = 35.5, S = 32)
BaCl2 + H2SO4  BaSO4  + 2HCl
BaSO4 formed is :
(A) 23.3 g (B) 46.6 g (C) 29.8 g (D) 11.65 g
F-14. A solution of glucose received from some research laboratory has been marked mole fraction x and molality
(m) at 10ºC. When you will calculate its molality and mole fraction in your laboratory at 24ºC you will find
(A) mole fraction (x) and molality (m) (B) mole fraction (2x) and molality (2m)
(C) mole fraction (x/2) and molality (m/2) (D) mole fraction (x) and (m ± dm) molality
F-15. The volume of water is required to make 0.20 M solution from 16 mL of 0.5 M solution is
(A) 40 ml (B) 16 ml (C) 50 ml (D) 24ml
F-16. A solution of FeCl3 is
30
M
its molarity for Cl–
ion will be
(A)
90
M
(B)
30
M
(C)
10
M
(D)
5
M

MOLE CONCEPT # 11
PART - II : MISCELLANEOUS QUESTIONS
Comprehensions Type
Comprehension # 1
Read the following comprehension carefully and answer the questions.
10 moles of SO2 and 4 moles of O2 are mixed in a closed vessel of volume 2 litres. The mixture
is heated in presence of Pt catalyst. Following reaction takes place :
2SO2(g) + O2(g)  2SO3(g)
Assuming the reaction proceeds to completion.
1. Select the correct statement -
(A) SO2 is the limiting reagent (B) O2 is the limiting reagent
(C) both SO2 and O2 are limiting reagent (D) cannot be predicted
2. Number of moles of SO3 formed in the reaction will be -
(A) 10 (B) 4 (C) 8 (D) 14
3. Number of moles of excess reactant remaining -
(A) 4 (B) 2 (C) 6 (D) 8
Comprehension # 2
The concentrations of solutions can be expressed in number of ways; viz : mass fraction of solute (or mass
percent), Molar concentration (Molarity) and Molal concentration (molality). These terms are known as
concentration terms and also they are related with each other i.e. knowing one concentration term for the
solution, we can find other concentration terms also. The definition of different concentration terms are
given below :
Molarity : It is number of moles of solute present in one litre of the solution.
Molality : It is the number of moles of solute present in one kg of the solvent
Mole Fraction = solventofmolessoluteofmoles
soluteofmoles

If molality of the solution is givenas 'a' then mole fraction of the solute can be calculated by
Mole Fraction =
solventM
1000
a
a

; =
)1000Ma(
Ma
solvent
solvent


where a = molality and Msolvent = Molar mass of solvent
We can change : Mole fraction  Molality  Molarity
4. 60 gm of solution containing 40% by mass of NaCl are mixed with 100 gm of a solution containing 15% by
mass NaCl. Determine the mass percent of sodium chloride in the final solution.
(A) 24.4% (B) 78% (C) 48.8% (D) 19.68%
5. What is the molality of the above solution.
(A) 4.4 m (B) 5.5 m (C) 24.4 m (D) none
6. What is the molarity of solution if density of solution is 1.6 gm/ml
(A) 5.5 M (B) 6.67 M (C) 2.59 M (D) none

MOLE CONCEPT # 12
Comprehension # 3
According to the Avogadro’s law, equal number of moles of gases occupy the same volume at identical
condition of temperature and pressure. Even if we have a mixture of non-reacting gases then Avogadro’s
law is still obeyed by assuming mixture as a new gas.
Now let us assume air to consist of 80% by volume of Nitrogen (N2) and 20% by volume of oxygen (O2). If
air is taken at STP then its 1 mol would occupy 22.4 L. 1 mol of air would contain 0.8 mol of N2 and 0.2 mol
of O2 hence the mole fractions of N2 and O2 are given by 8.0X 2N  , 2.0X 2O 
7. Volume occupied by air at NTP containing exactly 11.2 gm of Nitrogen :
(A) 22.4 L (B) 8.96 L (C) 11.2 L (D) 2.24 L
8. If air is treated as a solution of O2 and N2 then % W/W of oxygen is :
(A)
9
10
(B)
9
200
(C)
9
700
(D)
9
350
9. Density of air at NTP is :
(A) 1 g/L (B)
7
9
g/L (C)
7
2
g/L (D) can’t be determined
Match the Column Type :
10. Column-I Column-II
(A) 100 ml of 0.2 M AlCl3 solution + 400 ml (p) Total concentration of cation(s) = 0.12 M
of 0.1 M HCl solution
(B) 50 ml of 0.4 M KCl + 50 ml H2O (q) [SO4
2–
] = 0.06 M
(C) 30 ml of 0.2 M K2SO4 + 70 ml H2O (r) [SO4
2–
] = 2.5 M
(D) 200 ml 24.5% (w/v) H2SO4 (s) [Cl¯] = 0.2 M
11. Column I Column II
(A) Zn(s) + 2HCl(aq)  ZnCl2(s) + H2(g) (p) 50% of excess reagent left
above reaction is carried out by taking
2 moles each of Zn and HCl
(B) AgNO3(aq) + HCl(aq)  AgCl(s) + HNO3(g) (q) 22.4 L of gas at STP is liberated
above reaction is carried out by taking
170 g AgNO3 and 18.25 g HCl (Ag = 108)
(C) CaCO3(s)  CaO(s) + CO2(g) (r) 1 moles of solid (product) obtained.
100 g CaCO3 is decomposed
(D) 2KClO3(s)  2KCl(s) + 3O2(g) (s) HCl is the limiting reagent
2/3 moles of KClO3 decomposed

MOLE CONCEPT # 13
Assertion / Reason Type
Direction :
Each question has 5 choices (A), (B), (C), (D) and (E) out of which ONLY ONE is correct.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(C) Statement-1 is True, Statement-2 is False.
(D) Statement-1 is False, Statement-2 is True.
(E) Statement-1 and Statement-2 both are False.
12. Statement-1 : A one molal solution prepared at 20°C will retain the same molality at 100°C, provided there
is no loss of solute or solvent on heating.
Statement-2 : Molality is independent of temperature.
13. Statement-1 : Molality and mole fraction concentration units do not change with temperature.
Statement-2 : These units are not defined in terms of any volume.
14. Statement-1 : The molality and molarity of very dilute aqueous solutions differ very little.
Statement-2 : The density of water is about 1.0 g cm–3
at room temperature.
15. Statement-1 : For calculating the molality or the mole fraction of solute, if the molarity is known, it is
necessary to know the density of the solution.
Statement-2 : Molality, molarity and the mole fraction of solute can be calculated from the weight percentage
and the density of the solution
16. Statement-1 : The ratio of the mass of 100 billion atoms of magnesium to the mass of 100 billion atoms of
lead can be expressed as
207
24
.
Statement-2 : Atomic weights are relative masses.
17. Statement-1 : The average mass of one Mg atom is 24.305 amu, which is not the actual mass of one Mg
atom.
Statement-2 : Three isotopes, 24
Mg, 25
Mg and 26
Mg, of Mg are found in nature.
18. Statement-1 : A molecule of butane, C4 H10 has a mass of 58.12 amu.
Statement-2 : One mole of butane contains 6.022 × 1023
molecules and has a mass of 58.12 g.
19. Statement-1 : Both 12 g. of carbon and 27 g. of aluminium will have 6.02 × 1023
atoms.
Statement-2 : Gram atomic mass of an element contains Avogadro’s number of atoms.

MOLE CONCEPT # 14
PART - I : MIXED OBJECTIVE
Single choice type
1. The % loss in weight after heating a pure sample of potassium chlorate KClO3 (M. wt. 122.5) will be:
(A) 12.25 (B) 24.50 (C) 39.17 (D) 49.00
2. When 100g of ethylene polymerises entirely to polyethene, the weight of polyethene formed as per the
equation n(C2H4)  (CH2CH2)n is.
(A) (n/2)g (B) 100g (C) (100/n)g (D) 100ng
3. Formation of polyethene from calcium carbide takes place as follows :
CaC2+H2O  Ca(OH)2 + C2H2  C2H4;
n(C2H4)  (CH2CH2)n.
The amount of polyethylene possibly obtainable from 64.0 kg CaC2 can be
(A) 28Kg (B) 14kg (C) 21kg (D) 42 kg
4. In an organic compound of molar mass greater than 100 containing only C, H and N, the percentage of C
is 6 times the percentage of H while the sum of the percentages of C and H is 1.5 times the percentage of
N. What is the least molar mass?
(A) 175 (B) 140 (C) 105 (D) 210
5. 1 mol of iron (Fe) reacts completely with 0.65 mol O2 to give a mixture of only FeO and Fe2O3. Mole ratio
of ferrous oxide to ferric oxide is :
(A) 3 : 2 (B) 4 : 3 (C) 20 : 13 (D) none of these
6. XeF6 fluorinates 2 to F7 and liberates Xenon(g). 210 mmol of XeF6 can yield a maximum of_____ mmol
of IF7
(A) 420 (B) 180 (C) 210 (D) 245
7. The sodium salt of methyl orange has 7% sodium. What is the minimum molecular weight of the compound?
(A) 420 (B) 375 (C) 329 (D) 295
8. In the preceding problem, if the compound contains 12.8% nitrogen and 9.8% sulphur how many nitrogen
and sulphur atoms are present per atom of sodium?
(A) 2 and 1 (B) 1 and 3 (C) 1 and 2 (D) 3 and 1
9. A certain mixed oxide of iron contains 2.5 grams of oxygen for every 7.0 grams of iron. If it is regarded as
a mixture of FeO and Fe2O3 in the weight ratio a : b, what is a : b, (atomic weight of iron = 56)
(A) 9 : 10 (B) 9 : 20 (C) 14 : 5 (D) 1 : 1
10. One mole of a mixture of N2, NO2 and N2O4 has a mean molar mass of 55.4. On heating to a temperature
at which all the N2O4 may be presumed to have dissociated : N2O4 2NO2, the mean molar mass
tends to the lower value of 39.6. What is the mole ratio of N2 : NO2 : N2O4 in the original mixture ?
(A) 0.5 : 0.1 : 0.4 (B) 0.6 : 0.1 : 0.3 (C) 0.5 : 0.2 : 0.3 (D) 0.6 : 0.2 : 0.2
11. One mole of potassium chlorate (KClO3) is thermally decomposed and excess of aluminium is burnt in the
gaseous product. How many mol of aluminium oxide (Al2O3) are formed ?
(A) 1 (B) 1.5 (C) 2 (D) 3

MOLE CONCEPT # 15
12. A certain organic substance used as a solvent in many reactions contains carbon, hydrogen, oxygen and
sulphur. Weight % of hydrogen in the compound is 7.7. The weight ratio C : O : S = 3 : 2 : 4. What is the
least possible molar mass of the compound ?
(A) 86 (B) 63 (C) 94 (D) 78
13. When x grams of carbon are heated with y grams of oxygen in a closed vessel, no solid residue is left
behind. Which of the following statements is correct ?
(A) y/x must lie between 1.33 and 2.67 (B) y/x must be greater than or equal 2.67.
(C) y/x must be less than or equal 1.33 (D) y/x must be greater than or equal 1.33.
14. A certain compound has the molecular formula X4O6. If 10g of X4O6 has 5.72g X, atomic mass of X is :
(A) 32 amu (B) 37 amu (C) 42 amu (D) 98 amu
15. The molarity of Cl¯ in an aqueous solution which was (w/V) 2% NaCl, 4% CaCI2 and 6% NH4Cl will be
(A) 0.342 (B) 0.721 (C) 1.12 (D) 2.18
16. Mole fraction of ethyl alcohol in aqueous ethyl alcohol (C2H5OH) solution is 0.25. Hence percentage of
ethyl alcohol by weight is
(A) 54% (B) 25% (C) 75% (D) 46%
17. If a piece of iron gains 10% of its weight due to partial rusting into Fe2O3 the percentage of total iron that
has rusted is :
(A) 23 (B) 13 (C) 23.3 (D) 25.67
18. Molarity of H2SO4 is 18 M. Its density is 1.8 g/cm3
, hence molality is
(A) 18 (B) 100 (C) 36 (D) 500
19. Cortisone is a molecular substance containing 21 atoms of carbon per molecule. The mass percentage of
carbon in cortisone is 69.98%. Its molar mass is :
(A) 176.5 (B) 252.2 (C) 287.6 (D) 360.1
20. A mineral water sample was analysed and found to contain 1 × 10–3
% ammonia (w/w). The mole of
dissolved ammonia gas in one litre water bottle is (dwater  1 gm/ml)
(A) 5.8 ×10–4
mol (B) 1 × 10–2
mol (C) 0.58 ×10–2
mol (D) same as w/w
21. 2 litres of a mixture of nitrous and nitric oxides at STP have a mean molecular weight of 39.8. What volume
of nitrogen measured at STP could be obtained when the mixture has been passed over red hot copper?
(A) 1.7 litres (B) 1.9 litres (C) 1.5 litres (D) 1.85 litres
22. 120 g Mg was burnt in air to give a mixture of MgO and Mg3N2. The mixture is now dissolved in HCl to form
MgCl2 and NH4Cl, if 107 grams NH4Cl is produced. Then the moles of MgCl2 formed is :
(At. wt. Mg = 24, N = 14, Cl = 35.5)
(A) 3 moles (B) 6 moles (C) 5 moles (D) 10 moles
23. On mixing 15.0 ml of ethyl alcohol of density 0.792 g ml 1
with 15 ml of pure water at 40
C, the resulting
solution is found to have a density of 0.924 g ml 1
. The percentage contraction in volume is :
(A) 8 % (B) 2 % (C) 3 % (D) 4 %
24. The mass of CaCO3 produced when carbon dioxide is passed in excess through 500 ml of 0.5 M
Ca(OH)2 will be-
(A) 10 gm (B) 20 gm (C) 50 gm (D) 25 gm.
25. What approximate volume of 0.40 M Ba(OH)2 must be added to 50.0 mL of 0.30 M NaOH to get a solution
in which the molarity of the OH–
ions is 0.50 M?
(A) 33 mL (B) 66 mL (C) 133 mL (D) 100 mL

MOLE CONCEPT # 16
More than one choice type
26.* If 100 ml of 1M H2SO4 solution is mixed with 100 ml of 98%(w/w) H2SO4 solution (d = 0.1 gm/ml) then :
(A) concentration of solution remains same (B) volume of solution become 200 ml
(C) mass of H2SO4 in the solution is 98 gm (D) mass of H2SO4 in the solution is 19.6 gm
27.* A sample of a mixture of CaCl2 and NaCl weighing 4.44 gm was treated to precipitate all the Ca as CaCO3,
which was then heated and quantitatively converted to 1.12g of CaO. Calculate the % of CaCl2 in the
mixture.(At . wt. Ca = 40, Na = 23, Cl = 35.5)
(A) Mixture contains 50% NaCl (B) Mixture contains 60% CaCl2
(C) Mass of CaCl2 is 2.22 g (D) Mass of CaCl2 1.11 g
28.* 3 moles of the gas C2 H6 is mixed with 60 gm of this gas and 2.4 × 1024
molecules of the gas is removed.
The left over gas is combusted in the presence of excess oxygen then
(NA = 6 × 1023
) (Density of water = 1gm/ml)
(A) 2 Moles of C2H6 left for combustion
(B) Volume of CO2 at S.T.P. produced after combustion 44.8 litre.
(C) Volume of liquid water produced is 54 ml
(D) None
PART - II : SUBJECTIVE QUESTIONS
1. g of green algae absorbs 6 × 10–3
mol CO2 per hour by photosynthesis. If the fixed C atoms are all stored
after photosynthesis as starch (C6H10O5), how long will it take for the algae to double their own weight?
[Mw of (C6H10O5)n = 162].
2. 3 g of ethane C2H6 on complete combustion gave 8.8 g of CO2 and 5.4 g of water. Show that the results are
in accordance with the law of conservation of mass.
3. 500 mL of 0.2 M NaCl sol. is added to 100 mL of 0.5 AgNO3 solution resulting in the formation of white
precipitate of AgCl. How many moles and how many grams of AgCl are formed? Which is the limiting
reagent?
4. Asubstance used as a water softener has the following mass percentage composition : 42.07% Na, 18.9%
P, and 39.04% of oxygen. Determine the empirical formula of the compound. (Na = 23, P = 31, 0 = 16)
5. A plystyrene, having formula Br3C6H2(C8H8)n, was prepared by heating styrene with to contain 10.46%
bromine by weight, find the value of n.
6. A sample containing only CaCO3 and MgCO3 is ignited to CaO and MgO. The mixture of oxides produced
weight exactly half as much as the original sample. Calculate the percentages of CaCO3 and MgCO3 in the
sample.
7. When 4 gm of a mixture of NaHCO3 and NaCl is heated, 0.66 gm CO2 gas is evolved. Determine the
percentage composition of the original mixture.
8. 2PbS + 3O2  2PbO + 2SO2
3SO2 + 2HNO3 + 2H2O  3H2SO4 + 2NO
According to the above sequence of reactions, how much H2SO4 will 1146 gm of PbS produce?
9. A 10 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate
calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass
of CaO is 1.12gm. Calculate % by mass of NaCl in the original mixture.

MOLE CONCEPT # 17
10. Equal weights of mercury and iodine are allowed to react completely to form a mixture of mercurous and
mercuric iodide leaving none of the reactants. Calculate the ratio by weight of Hg2I2 and HgI2 formed
(Hg = 200 , I = 127)
11. What volume of water is required to make 0.20 M solution from 1600 mL of 0.2050 M solution?
12. 20 mL of 0.2 M Al2(SO4)3 is mixed with 20 mL of 0.6 M BaCl2. Calculate the concentrated of each ion in
solution.
13. Calculate molality of 1 litre solution of 93% H2SO4 by volume. The density of solution is 1.84 g mL–1
.
14. What would be the molality of a solution obtained by mixing equal volumes of 30% by weight H2SO4
(d = 1.218 g mL–1
) and 70% by weight H2SO4 (d = 1.610 g mL–1
)? If the resulting solution has density
1.425 g mL–1
, calculate its molality.
15. What volume of 95% H2SO4 by weight (d = 1.85 g mL–1
) and what mass of water must be taken to prepare
100 mL of 15% solution of H2SO4 (d = 1.10 g mL–1
)
16. A compound of carbon, hydrogen, and nitrogen contains the three elements in the respective ratio of
9 : 1 : 3 : 5. Calculate the empirical formula. If the molecular weight of the compound is 108, what is its
molecular formula?
17. 0.45 g of an organic compound containing only C, H, and N on combustion gave 1.1 g of CO2 and 0.3 g of
H2O. What is the percentage of C, H and N in the organic compound.
18. What weight of Na2CO3 of 95% purity would be required to neutralise 45.6 mL of 0.23 N acid?
PART - I : IIT-JEE PROBLEMS (PREVIOUS YEARS)
* Marked Questions are having more than one correct option.
1. How many moles of electron weigh one kilogram : [JEE-2002, 3/150]
(A) 6.023 × 1023
(B)
108.9
1
× 1031
(C)
108.9
023.6
× 1054
(D)
023.6108.9
1

× 108
2. Which has maximum number of atoms : [JEE-2003, 3/144]
(A) 24 g of C (12) (B) 56 g of Fe (56) (C) 27 g of Al (27) (D) 108 g Ag (108)
Paragraph for Question Nos. (3) to (5)
Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules
(approximately 6.023 × 1023
) are present in a few grams of any chemical compound varying with their
atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced.
This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry
and radiochemistry. The following example illustrates a typical case, involving chemical / electrochemical
reaction, which requires a clear understanding of the mole concept.
A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to
the evolution of chlorine gas at one of the electrodes (atomic mass : Na = 23, Hg = 200 ; 1 Faraday = 96500
coulombs).
**[At the anode :
2Cl–
 Cl2 + 2e –
At the cathode :
Na+
+ e–
 Na
Na + Hg  NaHg (sodium amalgam)]
** (These reactions were not present in IIT-JEE paper)

MOLE CONCEPT # 18
3. The total number of moles of chlorine gas evolved is : [JEE-2007, 4/162]
(A) 0.5 (B) 1.0 (C) 2.0 (D) 3.0
4. If the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution is :
[JEE-2007, 4/162]
(A) 200 (B) 225 (C) 400 (D) 446
5. The total charge (coulombs) required for complete electrolysis is : [JEE-2007, 4/162]
(A) 24125 (B) 48250 (C) 96500 (D) 193000
6. Given that the abundances of isotopes 54
Fe, 56
Fe and 57
Fe are 5%, 90% and 5%, respectively, the atomic
mass of Fe is : [JEE-2009, 3/160]
(A) 55.85 (B) 55.95 (C) 55.75 (D) 56.05
PART - II : AIEEE PROBLEMS (PREVIOUS YEARS)
1. Which of the following concentration factor is affected by change in temperature ? [AIEEE 2002]
(1) Molarity (2) Molality (3) Mole fraction (4) Weight fraction
2. Number of atoms in 560g of Fe (atomic mass 56g mol–1
) is : [AIEEE 2002]
(1) Twice that of 70g N (2) Half that of 20g H (3) Both (1) and (2) (4) None of these
3. In an organic compound of molar mass 108 g mol–1
C,H and N atoms are present in 9 : 1 : 3.5 by weight.
Molecular formula can be : [AIEEE 2002]
(1) C6H8N2 (2) C7H10N (3) C5H6N3 (4) C4H18N3
4. What volume of hydrogen gas at 273 K and 1 atm pressure will be consumed in obtaining 21.6 gm of
elemental boron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen- [AIEEE 2003]
(1) 44.8 lit. (2) 22.4 lit. (3) 89.6 lit. (4) 67.2 lit.
5. 6.02 × 1020
molecules of urea are present in 100 ml of its solution. The concentration of urea solution is -
[AIEEE 2004]
(A) 0.001 M (B) 0.01 M (C) 0.02 M (D) 0.1 M
6. If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit,
the mass of one mole of a substance will [AIEEE 2005]
(1) decrease twice (2) increase two fold
(3) remain unchanged (4) be a function of the molecular mass of the substance
7. Two solution of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5M first
solution + 250 ml of 1.2M second solution.
What is the molarity of the final mixture ? [AIEEE 2005]
(1) 2.70M (2) 1.344M (3) 1.50M (4) 1.20M
8. Density of a 2.05M solution of acetic acid in water is 1.02 g/ml. The molality of the solution is :
[AIEEE-2006]
(1) 1.14 mol kg–1
(2) 3.28 mol kg–1
(3) 2.28 mol kg–1
(4) 0.44 mol kg–1
9. How many moles of magnesium phosphate, Mg3(PO4)2 will contain 0.25 mole of oxygen atoms ?
[AIEEE-2006]
(1) 0.02 (2) 3.125 × 10–2
(3) 1.25 × 10–2
(4) 2.5 × 10–2
10. The density (in g mL–1
) of a 3.60 M sulphuric acid solution that is 29% (H2SO4 molar mass = 98 g mol–1
) by
mass will be [AIEEE-2007, 3/120]
(1) 1.22 (2) 1.45 (3) 1.64 (4) 1.88
11. In the reaction [AIEEE-2007, 3/120]
2AI(s) + 6HCl(aq)  2Al3+
(aq) + 6Cl–
(aq) + 3H2 (g)
(1) 6L HCl(aq) is consumed for every 3L H2 produced.
(2) 33.6 L H2(g) is produced regardless temperature and pressure for every moles that reacts.
(3) 67.2 L H2(g) at STP is produced for every mole of Al that reacts .
(4) 11.2 L H2(g) at STP is produced for every mole of HCl(aq) consumed.

MOLE CONCEPT # 19
NCERT QUESTIONS
1. Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
2. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL–1
and
the mass per cent of nitric acid in it being 69%.
3. Calculate the atomic mass (average) of chlorine using the following data :
% Natural Abudance Molar Mass
35
Cl 75.77 34.9689
37
Cl 24.23 36.9659
4. If the density of methanol is 0.793 kg L–1
, what is its volume needed for making 2.5 L of its 0.25 M solution?
5. How many significant figures are present in the following?
(i) 0.0025 (ii) 208 (iii) 5005
(iv) 126,000 (v) 500.0 (vi) 2.0034
6. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical
equation :
N2(g) + H2(g)  2NH3(g)
(i) Calculate the mass of ammonia produced if 2.00 × 103
g dinitrogen reacts with 1.00 × 103
g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
7. Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040
(assume the density of water to be one).
8. How many significant figures should be present in the answer of the following calculations ?
(i)
0.02856 298.15 0.112
0.5785
 
(ii) 5 × 5.364 (iii) 0.0125 + 0.7864 + 0.0215
8. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives
3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of
this welding gas is found to weight 11.6 g. Calculate :
(i) empirical formula,
(ii) molar mass of the gas, and
(iii) molecular formula.
9. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction,
CaCO3(s) + 2 HCl (aq)  CaCl2 (aq) + CO2(g) + H2O()
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?
10. Chlorine is prepared in the laboratory by treating manganese dioixde (MnO2) with aqueous hydrochloric
acid according to the reaction.
4 HCl (aq) + MnO2(s)  2H2O () + MnCl(aq) + Cl2(g)
How many grams of HCl react with 5.0 g of manganese dioxide?

MOLE CONCEPT # 20
Exercise # 1
PART - I
A-1. (D) A-2. (B) A-3. (B) A-4. (B) A-5. (D) A-6. (D) A-7. (C)
A-8. (C) A-9. (A) A-10*. (CD) B-1. (B) B-2. (B) B-3. (C) B-4. (D)
B-5. (B) B-6. (B) B-7. (B) B-8. (A) B-9. (B) B-10. (B) B-11. (B)
B-12. (C) B-13. (A) C-1. (D) C-2. (B) C-3. (B) C-4. (C) C-5. (C)
C-6. (C) C-7. (B) C-8. (D) C-9. (B) C-10. (C) C-11. (B) D-1. (B)
D-2. (D) D-3. (B) D-4. (C) D-5. (A) D-6. (B) D-7. (A) D-8. (A)
D-9. (B) D-10. (C) D-11. (A) D-12. (C) D-13. (C) D-14*. (BD)
D.15* (BCD) D-16. (C) D-17. (B) D-18. (B) D-19. (B) D-20. (C) D-21. (A)
D-22. (C) E-1. (A) E-2. (B) E-3. (A) E-4. (B) E-5. (A) F-1. (B)
F-2. (A) F-3. (C) F-4. (C) F-5. (C) F-6. (A) F-7. (B) F-8. (C)
F-9. (D) F-10. (A) F-11. (D) F-12. (A) F-13. (D) F-14. (A) F-15. (D)
F-16. (C)
PART - II
1. (B) 2. (C) 3. (B) 4. (A) 5. (B) 6. (B) 7. (C)
8. (B) 9. (B) 10. (A) p, s ; (B) s ; (C) p, q ; (D) r
11. (A) p, q, r, s ; (B) p, s ; (C) q, r ; (D) q
12. (A) 13. (A) 14. (A) 15. (B) 16. (A) 17. (A) 18. (A)
19. (A)
Exercise # 2
PART - I
1. (D) 2. (B) 3. (A) 4. (B) 5. (B) 6. (B) 7. (C)
8. (D) 9. (A) 10. (A) 11. (A) 12. (D) 13. (D) 14. (A)
15. (D) 16. (D) 17. (C) 18. (D) 19. (D) 20. (A) 21. (A)
22. (C) 23. (C) 24. (D) 25. (A) 26. (ABD) 27. (AC) 28. (BC)

MOLE CONCEPT # 21
PART - II
1. 10 3. 0.05 mole, AgNO3
4. Na3
PO4
5. n = 19
6. CaCO3 = 28.4%; MgCO3 = 71.6% 7. 63 % , 37% 8. 470.4 gm
9. %NaCl = 77.8% 10. 0.532 : 1.00 11. 40 mL 12.
24
[Cl ] 0.6M
40

 
13. 10.42 14. 11.22 15. 90.6 g 16. C6
H8
N2
17. % of C = 66.66%, % of N = 25.93% 18. 0.5978 g
Exercise # 3
PART - I
1. (D) 2. (A) 3. (B) 4. (D) 5. (D) 6. (B)
PART - II
1. (1) 2. (3) 3. (1) 4. (4) 5. (B) 6. (3) 7. (2)
8. (3) 9. (2) 10. (1) 11. (4)
Exercise # 4
5. (i) 2 (ii) 3 (iii) 4 (iv) 3 (v) 4 (vi) 5
6. (i) 2.43 × 103
g (ii) yes
(iii) Hydrogen will remain unreacted ; 5.72 × 102
g
8. (i) 3 (ii) 4 (iii) 4
8. Empirical formula CH, molar mass 26.0 g mol–1
, molecular formula C2
H2
9. 0.94 g CaCO3
10. 8.40 g HCl

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