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
govern your answer?

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. The above answer is
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:

ANSWER:-

(a) Magnesium chloride► MgCl 2

(b) Calcium oxide ► CaO

(d) Aluminium chloride ► AlCl 3

(e) Calcium carbonate ► CaCO 3

5) Give the names of the
elements present in the following compounds:

ANSWER:-

(a) Quick lime ► Calcium and oxygen

(b) Hydrogen bromide ► Hydrogen and bromine

(d) Potassium sulphate ► Potassium, sulphur, and oxygen

6) Calculate the molar mass of
the following substances:

ANSWER:-

(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

(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)?

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

8) Convert into mole.

(a) 12 g of oxygen gas

(b) 20 g of water

ANSWER:-

(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

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