NCERT Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Chemistry
Chapter : 1
Chapter Name : Some Basic Concepts Of Chemistry

Q1.1 Calculate the molar mass of the following:
(i) H O
2

(ii) CO 2

(iii) CH 4

Answer. (i) H O:
2

The molecular mass of water, H O 2

= (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen )

= [2(1.0084) + 1(16.00u)]

= 2.016u + 16.00u

= 18.016

= 18.02u

(ii) CO 2

The molecular mass of carbon dioxide, CO 2

= (1 × Atomic mass of carbon ) + (2 × Atomic mass of oxygen )

= [1(12.011u) + 2(16.00u)]

= 12.011u + 32.00u

= 44.01u

(iii) CH 4:

The molecular mass of methane, CH 4

= (1 × Atomic mass of carbon ) + (4 × Atomic mass of hydrogen )

= [1(12.011u) + 4(1.008u)]

= 12.011u + 4.032u

= 16.043u

Page : 25 , Block Name : Exercise

Q1.2 Calculate the mass percent of different elements present in sodium sulphate (Na SO ).
2 4

Answer. The molecular formula of sodium sulphate is Na SO 2 4

Molar mass of Na SO = [(2 × 23.0) + (32.066) + 4(16.00)]
2 4

= 142.066g
Mass of that element in the compound
Mass percent of an element = × 100
Molar mass of the compound

∴ Mass percent of sodium:

46.0g
= × 100
142.066g

= 32.379

= 32.4%

Mass percent of sulphur:
32.066g
= × 100
142.066g

= 22.57

= 22.6%

Mass percent of oxygen:
64.0g
= × 100
142.066g

= 45.049

= 45.05%

Page : 25 , Block Name : Exercise

Q1.3 Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1%
dioxygen by mass.

Answer. % of iron by mass = 69.9%[ Given ]
% of oxygen by mass = 30.1%[ Given ]

Relative moles of iron in iron oxide:
% of iron by mass
=
Atomic mass of iron
69.9
=
55.85

= 1.25

Relative moles of oxygen in iron oxide:
% of oxygen by mass
=
Atomic mass of oxygen

30.1
=
16.00

= 1.88

Simplest molar ratio of iron to oxygen:
= 1.25 : 1.88

= 1 : 1.5

= 2 : 3

∴ The empirical formula of the iron oxide is Fe2 O3

Page : 25 , Block Name : Exercise

Q1.4 Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.

Answer. The balanced reaction of combustion of carbon can be written as:
(i) As per the balanced equation, 1 mole of carbon burns in 1 mole of dioxygen (air) to Produced 1
mole of carbon dioxide.

(ii) According to the question, only 16 g of dioxygen is available. Hence, it will react with 0.5 mole
of carbon to give 22 g of carbon dioxide. Hence, it is a limiting reactant.
(iii) According to the question, only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16
g of dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.

Page : 25 , Block Name : Exercise

Q1.5 Calculate the mass of sodium acetate (CH 3 COONa) required to make 500 mL of 0.375
molar aqueous solution. Molar mass of sodium acetate is 82.0245 gmol .
−1

Answer. 0.375M aqueous solution of sodium acetate.
≡ 1000mL of solution containing 0.375 moles of sodium acetate.

∴ Number of moles of sodium acetate in 500 mL.
0.375
= × 500
1000

= 0.1875 mole
Molar mass of sodium acetate = 82.0245gmole
−1
( Given )

−1
∴ Required mass of sodium acetate = (82.0245gmol ) (0.1875 mole)

= 15.38g

Page : 25 , Block Name : Exercise

Q1.6 Calculate the concentration of nitric acid in moles per litre in a sample which has a density,
1.41gmL and the mass percent of nitric acid in it being 69%.
−1

Answer. Mass percent of nitric acid in the sample = 69%[ Given ]
Thus, 100g of nitric acid contains 69g of nitric acid by mass. Molar mass of nitric acid (HNO ) 3

−1
= {1 + 14 + 3(16)}gmol

= 1 + 14 + 48

−1
= 63gmol

∴ Number of moles in 69 g of HNO 3
69g
= −1
63gmol

= 1.095mol

Volume of 100g nitric acid solution
Mass of solution
=
density of solution

100g
= −1
1.41gmL

100g
=
−1
1.4 lg mL

−3
= 70.92mL ≡ 70.92 × 10 L

Concentration of nitric acid
1.095mole
=
−3
70.92×10 L

= 15.44mol/L

∴ Concentration of nitric acid = 15.44mol/L

Page : 26 , Block Name : Exercise

Q1.7 How much copper can be obtained from 100 g of copper sulphate (CuSO ) ? 4

Answer. 1 mole of CuSO contains 1 mole of copper.
4

Molar mass of cus O = (63.5) + (32.00) + 4(16.00)
4

= 63.5 + 32.00 + 64.00

= 159.5g

159.5g of CuSO contains 53.5 g of copper.
4

of cusO will contain of copper.
63.5×100g
⇒ 100g 4
159.5

AMount of copper that can be obtained from 100 g CuSO =
63.5×100
∴ 4
159.5

= 39.81g

Page : 26 , Block Name : Exercise

Q1.8 Determine the molecular formula of an oxide of iron, in which the mass percent of iron and
oxygen are 69.9 and 30.1, respectively.

Answer. Mass percent of iron (Fe) = = 69.9% (Given)
Mass percent of oxygen (O) = 30.1%( Given )
Number of moles of iron present in the oxide = 69.90

55.85

= 1.25

Number of moles of oxygen present in the oxide =
30.1

16.0

= 1.88

Ratioof irontooxygenintheoxide,

= 1.25 : 1.88
1.25 1.88
= :
1.25 1.25

= 1 : 1.5

= 2 : 3

∴ The empirical formula of the oxide is Fe2 O3

Empirical formula mass of Fe O = [2(55.85) + 3(16.00)]g
2 3

Molar mass of Fe O = 159.69g
2 3

Molar mass 159.69g
∴ n = =
Empirical formula mass 159.7g

= 0.999

= 1( approx )

Molecular formula of a compound is obtained by multiplying the empirical formula with n.

Thus, the empirical formula of the given oxide is Fe2 O3 and n is 1 .

Hence, the molecular formula of the oxide is Fe2 O3.

Page : 26 , Block Name : Exercise

Q1.9 Calculate the atomic mass (average) of chlorine using the following data:

Answer. The average atomic mass of chlorine.

= 26.4959 + 8.9568

= 35.4527u

∴ The average atomic mass of chlorine = 35.4527u

Page : 26 , Block Name : Exercise

Q1.10 In three moles of ethane (C H ), calculate the following:
2 6

(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.

Answer.
(i) 1 mole of C2 H6 contains 2 moles of carbon atoms.

∴ Number of moles of carbon atoms in 3 moles of C2 H6

= 2 × 3 = 6

(ii) 1 mole of C2 H6 contains 6 moles of hydrogen atoms.

∴ Number of moles of carbon atoms in 3 moles of C2 H6

= 3 × 6 = 18
23
(iii) 1 mole of C2 H6 contains 6.023 × 10 molecules of ethane.

∴ Number of molecules in 3 moles of C2 H6

23 23
= 3 × 6.023 × 10 = 18.069 × 10

Page : 26 , Block Name : Exercise

Q1.11 What is the concentration of sugar (C 12 H22 O11 ) in molL
−1
if its 20g are dissolved in
enough water to make a nal volume up to 2L?

Answer. Molarity (M) of a solution is given by,
Number of moles of solute
=
Volume of solution in Litres
Mass of sugar/molar mass of sugar
=
2L
20g/[(12×12)+(1×22)+(11×16)]g
=
2L
20g/342g
=
2L

0.0585mol
=
2L
−1
= 0.02925molL
−1
∴ Molar concentration of sugar = 0.02925molL

Page : 26 , Block Name : Exercise

Q1.12 If the density of methanol is 0.793kgL −1
. what is its volume needed for making 2.5 L of its
0.25 M solution?

Answer. Molar mass of methanol (CH OH) = (1 × 12) + (4 × 1) + (1 × 16)
3

−1
= 32gmol
−1
= 0.032kgmol
−1

Molarity of methanol solution =
0.793kgL

−1
0.032kgmol

−1
= 24.78molL

Since density is mass per unit volume
Applying,
M 1 V1 = M 2 V2

Given solution) (Solution to be prepared
−1 −1
(24.78molL ) V1 = (2.5L) (0.25molL )

V1 = 0.0252L

V1 = 25.22mL

Page : 26 , Block Name : Exercise

Q1.13 Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is
as shown below:
−2
1Pa = 1Nm

If mass of air at sea level is 1034 g cm −2
. calculate the pressure in pascal.

Answer. Pressure is de ned as force acting per unit area of the surface.
F
P =
A
−2 2 2
1034g×9.8ms 1kg (100) cm
= × ×
2
cm 1000g 1m2

5 −1 −2
= 1.01332 × 10 kgm s

We know,
−2
1N = 1kgms

Then,
−2 −2 −2
1Pa = 1Nm = 1kgm s

−1 −2
1Pa = 1kgm s
5
∴ Pressure = 1.01332 × 10 Pa

Page : 26 , Block Name : Exercise

Q1.14 What is the SI unit of mass? How is it de ned?

Answer. The SI unit of mass is kilogram (kg). 1 kilogram is de ned as the mass equal to the mass of
the international prototype of kilogram.

Page : 26 , Block Name : Exercise

Q1.15 Match the following pre xes with their multiples:

Answer.

Page : 26 , Block Name : Exercise

Q1.16 What do you mean by signi cant gures?

Answer. Signi cant gures are those meaningful digits that are known with certainty. They
indicate uncertainty in an experiment or calculated value. For example, if 15.6 mL is the result of
an experiment, then 15 is certain while 6 is uncertain, and the total number of signi cant gures
are 3. Hence, signi cant gures are de ned as the total number of digits in a number including the
last digit that represents the uncertainty of the result.

Page : 26 , Block Name : Exercise

Q1.17 A sample of drinking water was found to be severely contaminated with chloroform CHCl 3

supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in the water sample.

Answer. (i) 1 ppm is equivalent to 1 part out of 1 million (10 ) parts. 6

∴ Mass percent of 15ppm chloroform in water.
15
= 6
× 100
10
−3
≈ 1.5 × 10 %

(ii) 100g of the sample contains 1.5 × 10 −3
g of CHCl3.
−2
⇒ 1000g of the sample contains 1.5 × 10 g of CHCl3

∴ Molality of chloroform in water.
−2
1.5×10 g
=
Molar mass of CHCl3

Molar mass of CHCl 3 = 12.00 + 1.00 + 3(35.5)
−1
= 119.5gmol
−2
∴ Molality of chloroform in water = 0.0125 × 10 m
−4
= 1.25 × 10 m

Page : 26 , Block Name : Exercise

Q1.18 Express the following in the scienti c notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012

Answer. (i) 0.0048 = 4.8 × 10 −3

(ii) 234, 000 = 2.34 × 10 5

(iii) 8008 = 8.008 × 10 3

(iv) 500.0 = 5.000 × 10 2

(v) 6.0012 = 6.0012

Page : 26 , Block Name : Exercise

Q1.19 How many signi cant gures are present in the following?
(i) 0.0025
(ii) 208
(iii) 5005
(iv) 126,000
(v) 500.0
(vi) 2.0034

Answer. (i) There are 2 signi cant gures.
(ii) There are 3 signi cant gures.
(iii) There are 4 signi cant gures.
(iv) There are 3 signi cant gures.
(v) There are 4 signi cant gures.

(vi) There are 5 signi cant gures.

Page : 26 , Block Name : Exercise

Q1.20 Round up the following upto three signi cant gures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808

Answer. (i) 34.2
(ii) 10.4
(iii) 0.0460
(iv) 2810

Page : 27 , Block Name : Exercise

Q1.21 The following data are obtained when dinitrogen and dioxygen react together to form
different compounds:

(a) Which law of chemical combination is obeyed by the above experimental data? Give its
statement.
(b) Fill in the blanks in the following conversions:
(i) 1 km = ...................... mm = ...................... pm
(ii) 1 mg = ...................... kg = ...................... ng
(iii) 1 mL = ...................... L = ...................... dm3

Answer. (a) If we x the mass of dinitrogen at 28 g, then the masses of dioxygen that will combine
with the xed mass of dinitrogen are 32 g, 64 g, 32 g, and 80 g. The masses of dioxygen bear a
whole number ratio of 1:2:2:5. Hence, the given experimental data obeys the law of multiple
proportions. The law states that if two elements combine to form more than one compound, then
the masses of one element that combines with the xed mass of another element are in the ratio of
small whole numbers.
(b) (i) 1km = 1km × 1000m
×
1km
×
100cm

1m
10mm

1cm
6
∴ 1km = 10 mm
1000m 1pm
1km = 1km × × −12
1km 10 m
15
∴ 1km = 10 pm

Hence , 1km = 10 mm = 10
6 15
pm

(ii) 1mg = 1mg ×
1g 1kg
×
1000mg 1000g

−6
⇒ 1mg = 10 kg
1g 1ng
1mg = 1mg × ×
1000mg −9
10 g

6
⇒ 1mg = 10 ng
−6 6
∴ 1mg = 10 kg = 10 ng

(iii) 1mL = 1mL ×
1L

1000mL
−3
⇒ 1mL = 10 L
3 3 1dm×1dm×1dm
1mL = 1cm = 1cm
10cm×10cm×10cm
−3 3
⇒ 1mL = 10 dm
−3 −3 3
∴ 1mL = 10 L = 10 dm

Page : 27 , Block Name : Exercise

Q1.22 If the speed of light is 3.0 × 10 ms 8 −1
. calculate the distance covered by light in 2.00 ns.

Answer. According to the question:
Time taken to cover the distance = 2.00 ns
−9
= 2.00 × 10 s

Speed of light = 3.0 × 10 ms 8 −1

Distance travelled by light in 2.00 ns
= Speed of light x Time taken
8 −1 −9
= (3.0 × 10 ms ) (2.00 × 10 s)

−1
= 6.00 × 10 m

= 0.600m

Page : 27 , Block Name : Exercise

Q1.23 In a reaction
A + B2 → AB2

Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B

Answer. A reagent determines the extent of a reaction. It is the reactant which is the rst to get
consumed during a reaction, thereby causing the reaction to stop and limiting the amount of
products formed.
(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B. Thus, 200 molecules of
B will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the Iimiting
reagent.
(ii) According to the reaction, 1 mol of A reacts with 1 mol of 3. Thus, 2 mol of A will react with
only 2 mol of B. As a result, 1 mol of A will not be consumed. Hence, A is the limiting reagent.
(iii) According to the given reaction, 1 atom of A combines with 1 molecule of B. Thus, all 100

atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric where no
limiting reagent is present.
(iv) 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of B will combine with only
2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, 3 is the limiting reagent.
(v) According to the reaction, 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of
A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be left as such. Hence, A
is the limiting reagent.

Page : 27 , Block Name : Exercise

Q1.24 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 × 10 g dinitrogen reacts with 1.00 × 10 g of
3 3

dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?

Answer. (i) Balancing the given chemical equation,
N2(g) + 3H2(g) ⟶ 2NH3(g)

From equation, 1 mole (28 g) of nitrogen reacts with 3 mole (6 g) of dihydrogen to give 2 mole ( 34
g) of ammonia.
of di nitrogen will react with dihydrogen i.e., 2.00 × 10 g of
3 6g 3 3
⇒ 2.00 × 10 g × 2.00 × 10 g
28g

nitrogen will react with 428.6 of dihydrogen.
Given,
Amount of dihydrogen = 1.00 × 10 g 3

Hence, N is the limiting reagent.
2

∴ 28g of N2 produces 34g of NH3
34g
Hence, mass of ammonia produced by 2000 g of N = 2
28g
× 2000g

= 2428.57g

(ii) N is the limiting reagent and H is the excess reagent. Hence, H will remain unreacted.
2 2 2

(iii) Mass of dihydrogen left unreacted = 1.00 × 10 g − 428.6g 3

= 571.4g

Page : 27 , Block Name : Exercise

Q1.25 How are 0.50 mol Na CO and 0.50M Na CO different?
2 3 2 3

Answer. Molar mass of Na CO 2 3 = (2 × 23) + 12.00 + (3 × 16)
−1
= 106gmol

Now, 1 mole of Na CO2 3 means 106g of Na2 CO3
106g
∴ 0.5mol of Na2 CO3 = × 0.5molNa2 CO3
1mole

= 53gNa2 CO3

⇒ 0.50M of Na2 CO3 = 0.50mol/LNa2 CO3

Hence, 0.50mol of Na CO is present in 1 L of water or 53 g of Na CO is present in 1 L of water.
2 3 2 3

Page : 27 , Block Name : Exercise

Q1.26 If 10 volumes of dihydrogen gas reacts with ve volumes of dioxygen gas, how many volumes
of water vapour would be produced?

Answer. Reaction of dihydrogen with dioxygen can be written as:
2H2(s) + O2(x) ⟶ 2H2 O(x)

Now, two volumes of dihydrogen react with one volume of dihydrogen to produce two volumes of
water vapour. Hence, ten volumes of dihydrogen will react with ve volumes of dioxygen to
produce ten volumes of water vapour.

Page : 27 , Block Name : Exercise

Q1.27 Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg

Answer. (i) 28.7pm :
−12
1pm = 10 m
−12
∴ 28.7pm = 28.7 × 10 m

−11
= 2.87 × 10 m

(ii) 15.15pm :
−12
1pm = 10 m
−12
∴ 15.15pm = 15.15 × 10 m

−12
= 1.515 × 10 m

(iii) 25365mg :
−3
1mg = 10 g
4 −3
25365mg = 2.5365 × 10 × 10 g

Since,
−3
1g = 10 kg
1 −1 −3
2.5365 × 10 g = 2.5365 × 10 × 10 kg

−2
∴ 25365mg = 2.5365 × 10 kg

Page : 27 , Block Name : Exercise

Q1.28 Which one of the following will have the largest number of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1g of Cl (g)
2

Answer. 1g of Au(s) = 1

197
mol of Au(s)

23
6.022 × 10
= atoms of Au (s)
197
21
= 3.06 × 10 atoms of Au(s)
1
1g of Na(s) = mol of Na(s)
23
23
6.022×10
= atoms of Na(s)
23
23
= 0.262 × 10 atoms of Na(s)

21
= 26.2 × 10 atoms of Na(s)
1
1 g of L(s) = mol of Li(s)
7
33
6.022×10
= atomsof Li(s)
7
23
= 0.86 × 10 atoms of Li (s)

21
= 86.0 × 10 atoms of Li (s)
1
1g of Cl2 (g) = mol of Cl2 (g)
71

( Molar mass of Cl molecule = 35.5 × 2 = 71gmol
2
−1
)
33
6.022×10
= atoms of Cl2 (g)
71
23
= 0.0848 × 10 atoms of Cl2 (g)

21
= 8.48 × 10 atoms of Cl2 (g)

H ence, 1gof Li(s)willhavethelargestnumberof atoms.

Page : 28 , Block Name : Exercise

Q1.29 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).

Answer. Mole fraction of C H OH =
Number of moles of C2 H5 OH

2 5
Number of moles of solution

Number of moles present in 1 L water:
1000g
nH2 O = −1
18gmol

nH O = 55.55mol
2

Substituting the value of n H2 O in equation (1),
nC H OH
2 3
= 0.040
nC H OH +55.55
2 3

nC H OH = 0.040nC H OH
2 3 2 3
+ (0.040)(55.55)
0.96nC H OH = 2.222mol
2 3

2.222
hC H OH = mol
2 3 0.96

hC H OH ≡ 2.314mol
2 3

Molarity of solution =
2.314mol
∴
1L

= 2.314M

Page : 28 , Block Name : Exercise

Q1.30 What will be the mass of one 12
C atom in g?

Answer. 1 mole of carbon atoms = 6.023 × 10 23
atoms of carbon
= 12g of carbon

=
12 12g
∴ Mass of one C 23
6.022×10
−23
= 1.993 × 10 g

Page : 28 , Block Name : Exercise

Q1.31 How many signi cant gures 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

Answer. (i)
0.02856×298.15×0.112

0.5785

Least precise number of calculation = 0.112
∴ Number of signi cant gures in the answer.

= Number of signi cant gures in the least precise number.
=3
(ii) 5 × 5.364
Least precise number of calculation = 5.364
∴ Number of signi cant gures in the answer = Number of signi cant gures in 5. 364

=4
(iii) 0.0125 + 0.7864 + 0.0215
Since the least number of decimal places in each term is four, the number of signi cant gures in
the answer is also 4.

Page : 28 , Block Name : Exercise

Q1.32 Use the data given in the following table to calculate the molar mass of naturally occuring
argon isotopes:

Answer. Molar mass of argon
0.337 0.063 90.60 −1
= [(35.96755 × ) + (37.96272 × ) + (39.9624 × )] gmol
100 100 100

−1
= [0.121 + 0.024 + 39.802]gmol

−1
= 39.947gmol

Page : 28 , Block Name : Exercise

Q1.33 Calculate the number of atoms in each of the following (i) 52 moles of Ar (ii) 52 u of He (iii)
52 g of He.

Answer. (i) 1 mole of Ar = 6.022 × 10 23
atoms of Ar
23
∴ 52 mol of Ar = 52 × 6.022 × 10 atoms of Ar
25
= 3.131 × 10 atoms of Ar

(ii) 1 atom of He = 4 u of He
Or,
4u of He = 1 atom of He

1u of He = atom of He
1

4

52u of He = 52
atom of He
4

= 13 atoms of He
(iii) 4g of He = 6.022 × 10 23
atoms of He
23

52 g of He = atoms of He
6.022×10 ×52
∴
4

atoms of He
24
= 7.8286 × 10

Page : 28 , Block Name : Exercise

Q1.34 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 weigh 11.6 g. Calculate
(i) empirical formula,
(ii) molar mass of the gas, and (iii) molecular formula.

Answer. (i) 1 mole (44 g) of CO 2 contains 12g of carbon.

will contain carbon =
12g
∴ 3.38g of CO2 × 3.38g
44g

= 0.9217g

18 g of water contains 2 g of hydrogen.
g of water will contain hydrogen =
2g
∴ 0.690 × 0.690
18g

= 0.0767g

Since carbon and hydrogen are the only constituents of the compound, the total mass of the
compound is:
= 0.9217g + 0.0767g

= 0.9984g
0.9217g
∴ Percent of C in the compound = 0.9984g
× 100

= 92.32%
0.0767g
Percent of H in the compound = 0.9984g
× 100

= 7.68%

Moles of carbon in the compound = 92.32

12.00

= 7.69

Moles of hydrogen in the compound = 7.68

1

= 7.68

∴ Ratio of carbon to hydrogen in the compound = 7.69 : 7.68

= 1 : 1

Hence, the empirical formula of the gas is CH.

(ii) Given,
Weight of 10.0L of the gas (at S.T.P) = 11.6g
L of gas at STP =
11.6g
∴ Weight of 22.4 × 22.4L
10.0L

= 25.984g

≈ 26g

Hence, the molar mass of the gas is 26 g.

(iii) Empirical formula mass of CH = 12 + 1 = 13 g
Molar mass of gas
n =
Empirical formula mass of gas

26g
=
13g

n = 2

∴ Molecular formula of gas = (CH)n

= C 2 H2

Page : 28 , Block Name : Exercise

Q1.35 Calcium carbonate reacts with aqueous HCl to give CaCl and CO according to the
2 2

reaction, CaCO (s) + 2HCl(aq) → CaCl (aq) + CO (g) + H O(l).
3 2 2 2

What mass of CaCO is required to react completely with 25 mL of 0.75 M HCl?
3

Answer. 0.75M of HCl ≡ 0.75mol of HCl are present in 1L of water
−1
≡ [(0.75 mol) × (36.5gmol )] HCl is present in 1L of water

≡ 27.375g of HCl is present in 1L of water
Thus, 1000 mL of solution contains 27.375 g of HCL.
∴ Amount of HCl present in 25mL
27.375g
= × 25mL
1000mL

= 0.6844g

From the given chemical equation,
CaCO3(s) + 2HCl(aq) ⟶ CaCl2(aq) + CO2(g) + H2 O(g)

2 mol of HCl(2 × 36.5 = 71g) react with 1 mol of CaCO (100g)3

that will react with 0.6844g =
100
∴ Amount of CaCO3 × 0.6844g
71

= 0.9639g

Page : 28 , Block Name : Exercise

Q1.36 Chlorine is prepared in the laboratory by treating manganese dioxide (MnO ) with aqueous
2

hydrochloric acid according to the reaction
4HCl(aq) + MnO2 (s) → 2H2 O(l) + MnCl2 (aq) + Cl2 (g)

How many grams of HCl react with 5.0 g of manganese dioxide?

Answer. 1 mol [55 + 2 × 16 = 87g]MnO reacts completely with 4 mol [4 × 36.5 = 146g] of
2

HCL.
∴ 5.0g of MnO 2 will react with
146g
=
87g
× 5.0g of HCL
= 8.4g of HCl

Hence, 8.4g of HCl will react completely with 5.0g of manganese dioxide.

Page : 28 , Block Name : Exercise