NCERT Solutions for Class 9th Science
Chapter 3 – ATOMS AND MOLECULES
(Complete Downloadable Chapter Solution PDF file is at the bottom of the page)
Q.1 In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water
Ans The reaction is given by:
Na2CO3 + CH3COOH → CH3COONa + CO2 + H2O
(5.3 g) (6 g) (8.2 g) (2.2 g) (0.9 g)
Total Mass of reactants = 5.3 + 6 = 11.3 g
Total Mass of products = 8.2 + 2.2 + 0.9 = 11.3 g
Since total mass of reactants is equal to the total mass of products, the observations of this reaction are in agreement with the law of conservation of mass.
Q.2 Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of
oxygen gas would be required to react completely with 3 g of hydrogen gas?
Ans Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water i.e. 8g of oxygen gas is required to react completely with 1 g of hydrogen gas. Hence, mass of oxygen gas required to react completely with 3g of hydrogen gas = 3 x 8 = 24 g. The reaction of the same is:
H2 + O2 → H2O
(2 g) (16 g) (18 g)
Q.3 Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Ans Dalton’s postulate that “ Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction” is the result of the law of conservation of mass.
Q.4 Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Q.5 Define the atomic mass unit.
Q.6 Why is it not possible to see an atom with naked eyes?
Q.7 Write down the formulae of
(i) sodium oxide (ii) aluminium chloride (iii) sodium suphide (iv) magnesium hydroxide
Q.8 Write down the names of compounds represented by the following formulae:
(i) Al2(SO4)3 (ii) CaCl2 (iii) K2SO4 (iv) KNO3 (v) CaCO3
Q.9 What is meant by the term chemical formula?
Q.10 How many atoms are present in a
(i) H2S molecule and (ii) PO43– ion?
Q.11 Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.
Ans Molecular mass of H2 = 1 x 2 = 2 u
Molecular mass of O2 = 16 x 2 = 32 u
Molecular mass of Cl2 = 35.5 x 2 = 71 u
Molecular mass of CO2 = 12 + 16 x 2 = 44 u
Molecular mass of CH4 = 12 + 1 x 4 = 16 u
Molecular mass of C2H6 = 12 x 2 + 1 x 6 = 30 u
Molecular mass of C2H4 = 12 x 2 + 1 x 4 = 28 u
Molecular mass of NH3 = 14 + 1 x 3 = 17 u
Molecular mass of CH3OH = 12 + 1 x 3 + 16 + 1 = 32 u
Q.12 Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of
Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.
Q.13 If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1 atom of carbon?
Q.14 Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given,
atomic mass of Na = 23 u, Fe = 56 u)?
EXERCISES
Q.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.
Ans Weight of the compound of oxygen and boron = 0.24 g
Weight of boron in compound = 0.096 g
Therefore, % of boron in compound = (0.096/0.24) x 100 = 40%
Weight of oxygen in compound = 0.144 g
Therefore, % of oxygen in compound = (0.144/0.24) x 100 = 60%
Q.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 combination will govern your answer?
Ans The chemical reaction will be:
C + O2 → CO2
(Carbon) (Oxygen) (Carbon dioxide)
3 g of carbon is burnt in 8 g of oxygen (i.e. 3 g of carbon reacts with 8 g of oxygen) to produce 11 g of carbon dioxide. When 3 g of carbon is burnt in 50 g of oxygen, 11 g of carbon dioxide will be formed. This is because 3 g of carbon will react with 8 g of oxygen to produce 11 g of carbon dioxide and the remaining 42 g of oxygen will be left unreacted. This is governed by Law of Constant Proportions.
Q.3 What are polyatomic ions? Give examples.
Ans Clusters of atoms that act as an ion are called polyatomic ions. They carry a fixed charge on them. Examples of polyatomic ions are hydroxide (OH -), nitrate (NO3 -) etc.
Q.4 Write the chemical formulae of the following.
(a) Magnesium chloride (b) Calcium oxide (c) Copper nitrate (d) Aluminium chloride
(e) Calcium carbonate
Q.5 Give the names of the elements present in the following compounds.
(a) Quick lime (b) Hydrogen bromide (c) Baking powder (d) Potassium sulphate
Q.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
Q.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 (Na2SO3)?
Q.8 Convert into mole.
(a) 12 g of oxygen gas (b) 20 g of water (c) 22 g of carbon dioxide
Q.9 What is the mass of:
(a) 0.2 mole of oxygen atoms? (b) 0.5 mole of water molecules?
Q.10 Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.
Ans Molar mass of S8 = 32 x 8 = 256 g
So, no. of moles in 16 g of solid sulphur = 16/256 = 0.0625 mole
Now, No. of molecules of sulphur (S8) in 1 mole = 6.022 x 1023
» no. of molecules of sulphur (S8) in 16 g of solid sulphur i.e. 0.0625 mole
= 0.0625 x 6.022 x 1023 = 3.76 x 1022 molecules
Q.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)