EXERCISES (Text Book)
1. A 0.24 g sample of compound of Oxygen and Boron was found by analysis to contain 0.096 g if Boron and 0.144 g of Oxygen. Calculate the percentage composition of the compound by weight.
Ans: 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 = 0.096 × 1000.24% = 40%
Thus, percentage of Oxygen by weight in the compound = 0.144 × 1000.24% = 60 %.
2. 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.
Ans: Carbon + Oxygen → Carbon 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 unreactive.
Here only 11 g of carbon dioxide will be formed.
The above answer is governed by the law of constant proportions.
3. What are polyatomic ions. Give examples.
Ans: A polyatomic ion is a group of atoms carrying a charge (positive or negative).
eg: ammonium ion (4+), hydroxide ion (OH−), carbonate ion (3−−), sulphate ion (4−−).
4. Write the chemical formula of the following:
(a) Magnesium chloride (b) Calcium oxide (c) Copper nitrate
(d) Aluminium chloride (e) Calcium carbonate
Ans: (a) Magnesium chloride → MgCl2
(b) Calcium oxide → CaO
(c) Copper nitrate → Cu (NO3)2
(d) Aluminium chloride → AlCl3
(e) Calcium carbonate → CaCO3.
5. Give the names of the elements present in the following compounds:
(a) Quick lime (b) Hydrogen bromide (c) Baking powder
(d) Potassium sulphate compound
Ans: The elements present in:
a) Quick lime (CaO) are Calcium, oxygen;
b) Hydrogen bromide (HBr) are Hydrogen, bromine;
c) Baking powder (NaHCO3) are Sodium, hydrogen, carbon, oxygen;
d) Potassium sulphate (K2SO4 ) are Potassium, sulphur, oxygen
6. 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
Ans: (a) Molar mass of ethyne, C2H2 = 2 × 12 + 2 × 1 = 28 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
7. What is the mass of
(a) 1 mole of nitrogen atoms.
(b) 4 mole of aluminium atoms (Atomic mass of aluminium = 27).
(c) 10 moles of sodium sulphite (Na2SO3).
Ans: (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
8. Convert into mole.
(a) 12 g of oxygen gas (b) 12 g of water (c) 22 g of carbon dioxide
Ans: (a) 32 g of oxygen gas = 1 mole .
Then, 12g of oxygen gas = 12/32 mole
= 0.375 mole
(b) 18 g of water = 1 mole.
Then, 20 g of water = 20/18 mole
= 1.11 moles (approx.)
(c) 44 g of carbon dioxide = 1 mole.
Then, 22 g of carbon dioxide = 22/44 mole
= 0.5 mole.
9. What is the mass of: (a) 0.2 mole of oxygen atoms. (b) 0.5 mole of water molecules.
Ans: (a) Mass of one mole of oxygen atoms = 16 g.
Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16 g = 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
10. Calculate the number of molecules of sulphur (S8) present in 16g of solid sulphur.
Ans: 1 mole of solid sulphur (S8) = 8 × 32 g = 256 g .
And 256 g of solid sulphur contains = 6.022 × 1023 molecules.
Then, 16 g of solid sulphur contains (6.022 × 1023 × 16 ) /256 molecules
= 3.76 × 1022 molecules (approx.)
11. Calculate the number of aluminium ions present in 0.051g 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 = 27u)
Ans: 1 mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16 = 102g .
And 102g of (Al2O3) = 6.022 × 1023 molecules of Al2O3.
Then, 0.051 g of (Al2O3) contains = 6.022×1023102×0.051 molecules
= 3.011 × 1020 molecules of Al2O3
The number of aluminium ions (Al3 +) present in one molecules of aluminium oxide is 2.
The number of aluminium ions (Al3 +) present in 3.11 × 1020 molecules (0.051g) of aluminium oxide Al2 O3
= 2 × 3.011 × 1020
= 6.022 × 1020