# TEXTBOOK ANSWERS AND SOLUTIONS OF CBSE CLASS IX SCIENCE Chapter 3 Atoms And Molecules

1) 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.
ANSWER:-Total mass of Compound = 0.24 g
Mass of boron = 0.096 g
Mass of oxygen = 0.144 g
Thus, percentage of boron by weight
in the compound =0.096 / 0.24 x 100%= 40%
And, percentage of oxygen by weight
in the compound =0.144 / 0.24 x 100%= 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
ANSWER:-If 3 g of carbon is burnt in 50bg 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.
governed
by the law of constant
proportions.
3) What are polyatomic ions? Give examples?
ANSWER:-A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example,
Nitrate (NO 3- ),
hydroxide ion (OH -).
4) Write the chemical formulae of the following:
(a) Magnesium chloride MgCl 2
(b) Calcium oxide CaO
(c) Copper nitrate Cu (NO 3) 2
(d) Aluminium chloride AlCl 3
(e) Calcium carbonate CaCO 3
5) Give the names of the
elements present in the following compounds:
(a) Quick lime Calcium and oxygen
(b) Hydrogen bromide Hydrogen and bromine
(c) Baking powder Sodium, hydrogen,
carbon,
and oxygen
(d) Potassium sulphate Potassium, sulphur, and oxygen
6) Calculate the molar mass of
the following substances:
(a) Ethyne, C 2H2 Molar mass of ethyne, C 2H2 = 2
x 12 + 2 x 1 = 26 g
(b) Sulphur molecule, S 8 Molar mass of sulphur molecule, S 8 = 8 x 32 = 256 g
(c) Phosphorus molecule, P 4 (atomic
mass of phosphorus =
31) Molar mass of phosphorus molecule, P 4 = 4 x 31 = 124 g
(d) Hydrochloric acid, HCl Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5 g
(e) Nitric acid, HNO 3 Molar mass of nitric acid, HNO 3 = 1 + 14 + 3 x 16 = 63
g
7) 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 (Na 2SO 3)?
ANSWER 🙁 a) The mass of 1 mole of nitrogen atoms is 14 g.
(b) The mass of 4 moles of aluminium
atoms is (4 x 27) g
= 108 g
(c) The mass of 10 moles of Sodium
sulphite (Na 2SO 3) is
10 x [2 x 23 + 32 + 3 x 16]
g =
10 x 126 g = 1260 g
8) 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 = 1mole
Then, 12 g of oxygen gas = 12
/ 32 mole = 0.375 mole
(b) 18 g of water = 1 mole
Then, 20 g of water = 20 / 18 mole
= 1.111 mole
(c) 44 g of carbon dioxide = 1mole
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?
ANSWER 🙁 a) Mass of one mole of oxygen atoms = 16 g
Then, mass of 0.2 mole of oxygen
atoms = 0.2 x 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 x 18 g = 9 g
10) Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.
ANSWER;-1 mole of solid sulphur (S8) = 8 x 32 g = 256 g i.e., 256 g of solid sulphur contains
= 6.022 x 1023 molecules
Then, 16 g of solid sulphur contains
= 6.022 x 1023 / 256
= 16 molecules = 3.76375
x 10 22 molecules
11) 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)
ANSWER:-mole of aluminium oxide (Al 2O 3) = 2 x 27 + 3 x 16 = 102 g i.e., 102 g of Al 2O 3= 6.022 x 1023 molecules of Al 2O3
Then, 0.051 g of Al 2O 3contains = 6.022 x 1023 / 102 x 0.051
molecules = 3.011 x 1020 molecules of Al 2O3
The number of aluminium ions (Al3+)
present in one molecule
of aluminium oxide is 2.
Therefore, the number of aluminium
ions (Al 3+) present
in 3.011 x 1020molecules (0.051 g) of aluminium oxide (Al 2O 3) = 2 x 3.011 x 10 20= 6.022
x 1020