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 mole of carbon is burnt in air.
- 1 mole of carbon is burnt in 16 g of dioxygen.
- 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 mole carbon dioxide
- As molar mass of oxygen molecule is 32 g so combustion in 16 g oxygen will give 0.5 mole carbon dioxide
- 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`