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Learn NCERT Exercise 3 with Free Lessons & Tips

A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

Mass of boron = 0.096 g (Given)

Mass of oxygen = 0.144 g (Given)

Mass of sample = 0.24 g (Given)

Thus, percentage of boron by weight in the compound =

= 40%

And, percentage of oxygen by weight in the compound =

= 60%

Comments

When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combinations will govern your answer?

Carbon + OxygenCarbon dioxide

3 g of carbon reacts with 8 g of oxygen to produce 11 g of carbon dioxide.

If 3 g of carbon is burnt in 50 g of oxygen, then 3 g of carbon will react with 8 g of oxygen. The remaining 42 g of oxygen will be left un-reactive.

In this case also, only 11 g of carbon dioxide will be formed.

The above answer is governed by the law of constant proportions.

Comments

What are polyatomic ions? Give examples?

A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example, ammonium ion, hydroxide ion (OH), carbonate ion, sulphate ion.

Comments

Write the chemical formulae of the following:

(a) Magnesium chloride

(b) Calcium oxide

(c) Copper nitrate

(d) Aluminium chloride

(e) Calcium carbonate

(a) Magnesium chloride → MgCl2

(b) Calcium oxide → CaO

(c) Copper nitrate → Cu (NO3)2

(d) Aluminium chloride → AlCl3

(e) Calcium carbonate → CaCO3

Comments

Calculate the molar mass of the following substances:

(a) Ethyne, C2H2

(b) Sulphur molecule, S8

(c) Phosphorus molecule, P4 (atomic mass of phosphorus = 31)

(d) Hydrochloric acid, HCl

(e) Nitric acid, HNO3

(a) Molar mass of ethyne, C2H2 = 2 × 12 + 2 × 1 = 26 g

(b) Molar mass of sulphur molecule, S8 = 8 × 32 = 256 g

(c) Molar mass of phosphorus molecule, P4 = 4 × 31 = 124 g

(d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5 g

(e) Molar mass of nitric acid, HNO3 = 1 + 14 + 3 × 16 = 63 g

Comments

What is the mass of−−

(a) 1 mole of nitrogen atoms?

(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?

(c) 10 moles of sodium sulphite (Na2SO3)?

(a) The mass of 1 mole of nitrogen atoms is 14 g.

(b) The mass of 4 moles of aluminium atoms is (4 × 27) g = 108 g

(c) The mass of 10 moles of sodium sulphite (Na2SO3) is

10 × [2 × 23 + 32 + 3 × 16] g = 10 × 126 g = 1260 g

Comments

Convert into mole.

(a) 12 g of oxygen gas

(b) 20 g of water

(c) 22 g of carbon dioxide

(a) 32 g of oxygen gas = 1 mole

Then, 12 g of oxygen gas == 0.375 mole

(b) 18 g of water = 1 mole

Then, 20 g of water == 1.11 moles (approx)

(c) 44 g of carbon dioxide = 1 mole

Then, 22 g of carbon dioxide == 0.5 mole

Comments

What is the mass of:

(a) 0.2 mole of oxygen atoms?

(b) 0.5 mole of water molecules?

(a) Mass of one mole of oxygen atoms = 16 g

Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 3.2 g

(b) Mass of one mole of water molecule = 18 g

Then, mass of 0.5 mole of water molecules = 0.5 × 18 g = 9 g

Comments

Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.

1 mole of solid sulphur (S8) = 8 × 32 g = 256 g

i.e., 256 g of solid sulphur contains = 6.022 × 1023 molecules

Then, 16 g of solid sulphur contains =

= 3.76 × 1022 molecules (approx)

Comments

Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.

(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)

 

1 mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16

= 102 g

i.e., 102 g of Al2O3 = 6.022 × 1023 molecules of Al2O3

Then, 0.051 g of Al2O3 contains =

= 3.011 × 1020 molecules of Al2O3

The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.

Therefore, the number of aluminium ions (Al3+) present in 3.011 × 1020 molecules (0.051 g ) of aluminium oxide (Al2O3) = 2 × 3.011 × 1020

= 6.022 × 1020

Comments

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