Class 11 Chemistry

# 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 Abundance Molar Mass 35Cl 75.77 34.9689 37Cl 24.23 36.9659

Answer: Average atomic mass of chlorine

=(75.77×34.9689+24.23×36.9659)/(100)

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