Chapter 3, Atoms and Molecules
Class 9, NCERT (CBSE) Science Text Book
Exercise Questions Solved
Q.1: A 0.24 gm sample of compound of oxygen and boron was found by analysis to contain 0.096 gm of boron and 0.144 gm of oxygen. Calculate the percentage composition of the compound by weight.
Ans: % of boron in sample = (0.096 ÷ 0.24) × 100 = 40%
% of oxygen in sample = (0.144 ÷ 0.24) × 100 = 60%
The sample of compound contains 40% boron and 60% oxygen by weight.
Q.2: When 3.0 gm of carbon s burnt in 8.0 gm oxygen, 11.0 gm of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.0 gm of carbon is burnt in 50.00 gm of oxygen ? Which law of chemical combination will govern your answer ?
Ans: When 3.0 gm of carbon s burnt in 8.0 gm oxygen, 11.0 gm of carbon dioxide is produced. It means all of carbon and oxygen are combined in the ratio of 3 : 8 to form carbon dioxide. Thus when there is 3 gm carbon and 50 gm oxygen, then also only 8 gm of oxygen will be used and 11 gm of carbon dioxide will be formed. The remaining oxygen is not used.
This indicates law of definite proportions which says that in compounds, the combining elements are present in definite proportions by mass.
Q.3: What are polyatomic ions ? Give examples.
Ans: When two or more atoms combine together and behave like one entity with a net charge, then it is called polyatomic ion. For example, oxygen atom and hydrogen atom combine to form hydroxide ion (OH–). One carbon atom and three hydrogen atom combine to form carbonate ion (CO3–2).
Q.4: Write chemical formulae of the following:
(a) Magnesium chloride (b) Calcium oxide (c) Calcium nitrate (d) Aluminium chloride (e) Calcium carbonate
Ans: (a) MgCl2 (b) CaO (c) Cu(NO3)2 (d) AlCl3 (e) CaCO3
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
Ans: (a) Calcium and oxygen (b) Hydrogen and bromine (c) Sodium, hydrogen, carbon and oxygen (d) Potassium, sulphur and oxygen.
Q.7: What is the mass of:
(a) 1 mole of nitrogen atoms.
(b) 4 moles of aluminium atoms.
(c) 10 moles of sodium sulphite (Na2SO3).
Ans:
(a) 1 mole of nitrogen atoms = 14u = 14gm
(b) 4 moles of aluminium atoms = 4 × 27 = 108u = 108gm
(c) 1mole of sodium sulphite, Na2SO3 = 2 × 23 + 1 × 32 + 3 × 16 = 126u = 126gm
10 moles of sodium sulphite = 126 x 10 = 1260u = 1260gm.
Q.8: Convert into mole
(a) 12 gm of oxygen gas
(b) 20 gm of water
(c) 22 gm of carbon dioxide
Ans:
(a) 32 gm of oxygen gas = 1 mole
12 gm of oxygen gas = 12 ÷ 32 = 0.375 mole.
(b) 18 gm of water = 1 mole
20 gm of water = 20 ÷ 18 = 1.1 mole.
(c) 44 gm of carbon dioxide = 1 mole
22 gm carbon dioxide = 22 ÷ 44 = 0.5 mole.
Q.9: What is the mass of :
(a) 0.2 mole of oxygen atoms
(b) 0.5 mole of water molecules
Ans:
(a) 1 mole of oxygen atoms = 16 gm
0.2 moles of oxygen atoms = 16 x 0.2 = 3.2 gm
(b) 1 mole of water molecules = 18 gm
0.5 mole of water molecules = 18 x 0.5 = 9 gm.
Q.10: Calculate the number of molecules of sulphur (S8) present in 16 gm of solid sulphur.
Ans: 1 mole of S8 = 32 x 8 = 256 gm
1 mole of S8 = 6.023 x 1023 molecules
So, 256 gm of S8 = 6.023 x 1023 S8 molecules
Or, 16 gm S8 = (6.023 x 1023 ÷ 256) x 16 = 3.76 x 1022 molecules.
Q.11: Calculate the number of aluminium ions present in 0.51 gm of aluminium oxide.
Ans: 1 mole of aluminium oxide, Al2O3 = 2 x 27 + 3 x 16 = 102u = 102 gm
That is 102 gm Al2O3 has 6.023 x 1023 Al2O3 molecules
Or, 0.51 gm Al2O3 has = (6.023 x 1023 x 0.51) ÷ 102 = 3.01 x 1021 Al2O3 molecules.
We know 1 molecule of Al2O3 has 2Al+++ ions.
Hence, 0.51 gm Al2O3 has = 2 x 3.01 x 1021 Al+++ ions
= 6.023 x 1021 aluminium ions.
Further study on this chapter
=> Class 9, NCERT (CBSE) Science (Chemistry): Chapter 3, Atoms and Molecules | In Text Questions [Read]
=> Class IX, NCERT (CBSE) | Chapter 3, Atoms and Molecules | Extrascore Multiple Choice Questions (MCQ) [Read]
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