Atoms and Molecules NCERT Solutions
Hello readers ! Here, class 9, Chapter – 03 Atoms and Molecules NCERT Solutions are given.
Question 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: Total mass of compound of oxygen and boron = 0.24 g
Mass of boron in this compound = 0.096 g
So, % composition of boron in the compound by weight = (Mass of boron / Total mass of compound) × 100
= (0.096 / 0.24) × 100 = 40 %
Mass of oxygen in given compound = 0.144 g
So, % composition of oxygen in given compound by weight = (Mass of oxygen / Total mass of compound) × 100
= (0.144 / 0.24) × 100 = 60 %
Question 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: C (3 g) + O2 (8 g)→ CO2 (11 g)
3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 g of carbon dioxide. Here, carbon and oxygen are in a fixed ratio by mass, which is 3 : 8. According to the law of definite proportions, this ratio will always be constant.
When 3.0 g of carbon is burnt in the presence of 50.0 g of oxygen, only 8.0 g of oxygen will be utilised to produce 11.0 g of carbon dioxide because in carbon dioxide the ratio of carbon and oxygen, will always be maintained according to the law of definite proportions. The remaining 42.0 g (50.0 g – 8.0 g = 42.0 g) of oxygen will remain unreacted.
Key concept ☞ Law of constant proportions / Law of definite proportions: “In a chemical substance the elements are always present in definite proportions by mass”. For example, In water, the ratio of the mass of hydrogen to the mass of oxygen is always 1:8, whatever the source of water. Thus, if 9 g of water is decomposed, 1 g of hydrogen and 8 g of oxygen are always obtained.
Table of Contents
Question 3:
What are polyatomic ions? Give examples.
Ans: ‘Poly’ means many ‘Atomic’ is related to atom
Poly + atomic = Polyatomic = many atom or group of atoms together
So, a polyatomic ion is an ion that contains more than one atom in it. Atoms are grouped together by covalent bond between them.
For example: NH4+ ( Ammonium ion) it constitute one nitrogen atom and 4 hydrogen atom.
CO32− ( Carbonate ion ) it constitute one carbon atom and 3 oxygen atom.
NO2− :: Nitrite HCO3− :: Hydrogen carbonate
Question 4:
Write the chemical formulae 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
Question 5:
Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.
Answer: Quick lime (CaO) — elements present — Calcium, Oxygen
Hydrogen bromide (HBr) –elements present– Hydrogen, Bromine
Baking powder (NaHCO3) — elements present — Sodium, Hydrogen, Carbon, Oxygen
Potassium Sulphate (K2 SO4) –elements present– Potassium, Sulphur, Oxygen
Question 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: Molar Mass of,
(a) Ethyne (C2H2) = 2 × Atomic Mass of C atom + 2 × Atomic Mass of H atom = 2 × 12 + 2 × 1 = 2 4 + 2 = 26 g/mol
(b) Sulphur dioxide (S8) = 8 × Atomic Mass of Sulphur atom = 8 × 32 = 256 g/mol
(c) Phosphorus molecule (P4 ) = 4 × Atomic Mass of P atom = 4 × 31 = 124 g/mol
(d) Hydrochloric acid (HCl) = 1 × Atomic Mass of H atom + 1 × Atomic Mass of Cl atom = 1 + 35.5 = 36.5 g/mol
(e) Nitric acid (HNO3 ) = 1 × Atomic Mass of H atom + 1 × Atomic Mass of N atom + 3 × Atomic Mass of O atom = 1 × 1 + 1 × 14 + 3 × 16 = 63 g/mol
Question 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) ?
Answer: (a) Mass of 1 mole of nitrogen atoms = 1 × Atomic mass of N = 1 × 14 = 14 g
(b) Mass of 4 moles of aluminium atoms = 4 × Atomic mass of Al = 4 × 27 = 108 g
(c) Mass of 10 moles of sodium sulphite (Na2SO3) = 10 × Mass of Na2SO3
10 × [ 2 × Atomic Mass of Na atom + 1 × Atomic Mass of S atom + 3 × Atomic Mass of O atom] = 10 × [2 × 23 + 1 × 32 + 3 × 16 ] = 10 × [46 + 32 + 48] = 1260 g
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