Chemistry Basics

NCERT Solution

Question 1: Calculate the molar mass of the following:

(i) H2O

Answer: Molar mass of H2O

= 2 × 1.008 + 16.00 = 2.02 + 16.00 = 18.02 g

(ii) CO2

Answer: Molar mass of CO2

= 12 + 2 × 16 = 12 + 32 = 44 g

(iii) CH4

Answer: Molar mass of CH4

= 12.00 + 4 × 1.008 = 12.00 + 4.03 = 16.03 g


Question 2: Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4).

Answer: Molar Mass of Na2SO4

= 2 × 23 + 32 + 4 × 16

= 46 + 32 + 64 = 142 g

Mass % of an element = Molar mass of element in compound ÷ Molar mass of compound × 100

So, mass % of sodium `=(46)/(142)× 100=32.39%`

Mass % of sulphur `=(32)/(142)×100=22.53%`

Mass % of oxygen `=(64)/(142)×=45.07%`

Question 3: Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% di-oxygen by mass.

Answer: Atomic mass of iron = 55.85 and atomic mass of oxygen = 16.00

Molar mass of iron in given compound `=(69.9)/(55.85)×100=125` g

Molar mass of oxygen in given compound `=(30.1)/(16)×100=188` g

Ratio of molar masses `=(188)/(125)=3/2`

So empirical formula of given compound is Fe2O3


Question 4: Calculate the amount of carbon dioxide that could be produced when

  1. 1 mole of carbon is burnt in air.
  2. 1 mole of carbon is burnt in 16 g of dioxygen.
  3. 2 moles of carbon are burnt in 16 g of dioxygen.

Answer: Balanced equation for combustion of carbon is as follows:

C + O2 → CO2

This equation shows 1 mole of carbon is burnt in 1 mole of oxygen to produce 1 mole of carbon dioxide.

  1. 1 mole carbon dioxide
  2. As molar mass of oxygen molecule is 32 g so combustion in 16 g oxygen will give 0.5 mole carbon dioxide
  3. As oxygen is the limiting reactant, so in this case also only 0.5 mole carbon dioxide will be produced.

Question 5: Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1.

Answer: Molar mass of sodium acetate = 82.0245 g per mole

So, this quantity will be present in 1 liter of aqueous solution of sodium acetate. The molarity of solution will be 1 M.

So, molar mass in 0.375 M 1 liter solution = 0.375 × 82.0245

So, molar mass in 0.375 M 500 mL solution = 0.375 × 82.0245 ÷ 2 = 15.380 g

Question 6: 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%.

Answer: Mass percent of nitric acid in given solution = 69%

So, the solution contains 69 g nitric acid per 100 g of solution

Molar Mass of HNO3 = 1 + 14 + 3 × 16 = 1 + 14 + 48 = 63 g

No. of moles in 69 g nitric acid = 69 g ÷ 63 g = 1.09

Density = 1.41 g per mL

So, volume of 100 g nitric acid solution = 100 g ÷ 1.41 g per mL = 70.92 mL = 0.07092 L

Molarity of nitric acid = 1.095 M ÷ 0.07092 L = 15.44 M L-1


Question 7: How much copper can be obtained from 100 g of copper sulphate (CuSO4)?

Answer: Molar mass of CuSO4 = 63.5 + 32 + 4 × 16

= 63.5 + 32 + 64 = 159.5 g

Amount of copper in 100 g copper sulphate `=(63.5)/(159.5)×100=39.81` g

Question 8: Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively.

Answer: Same as solution for question 3

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

% Natural AbundanceMolar Mass
35Cl75.7734.9689
37Cl24.2336.9659

Answer: Average atomic mass of chlorine

`=(75.77×34.9689+24.23×36.9659)/(100)`

`=(2649.59+895.68)/(100)=35.45`



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