NCERT Solutions for Class 9 Chemistry Chapter 3 - Atoms and Molecules
Study the law of conservation of mass from your textbook with supporting NCERT Solutions for CBSE Class 9 Chemistry Chapter 3 Atoms and Molecules. With TopperLearning’s expert solutions, you can learn to calculate molecular mass of oxygen, sulphur and other molecules.
Revise CBSE Class 9 Chemistry topics in the syllabus such as mole concept, polyatomic ions, atomic mass etc. using our textbook solutions. To ensure top performance in your exam, you need to practise these topics thoroughly. You can also use our revision notes, self-assessment tests and concept videos to achieve academic excellence.
Chapter 3 - Atoms and Molecules Exercise 32
The total mass of the reactants = 5.3 + 6 = 11.3 g
The total mass of the products = 2.2 + 0.9 + 8.2 = 11.3 g
Thus, the total mass of the reactants is same as that of the products. Thus, the mass was neither created nor destroyed during the chemical reaction. Therefore, the mass was conserved during the reaction of sodium carbonate and ethanoic acid.
Chapter 3 - Atoms and Molecules Exercise 33
Concept insight: Recall the law of constant proportions and then apply that in a chemical substance, the elements are always present in definite proportions by mass.
Concept insight: For answering this question you should know both the law of conservation of mass and the postulates of Dalton's atomic theory.
Concept insight: For answering this question, you should know both law of definite proportions and the postulates of Dalton's atomic theory.
Chapter 3 - Atoms and Molecules Exercise 35
Concept insight: In this answer, write the proper definition of atomic mass unit.
Concept insight: For answering this question, you should recall the size of an atom.
Chapter 3 - Atoms and Molecules Exercise 39
Concept insight: For answering this question, write the cation and anion with the proper charges, cross multiply and write the formula.
(ii) Calcium chloride
(iii) Potassium sulphate
(iv) Potassium nitrate
(v) Calcium carbonate
(ii) Five (One phosphorus and four oxygen)
Concept insight: Count the total number of atoms of each type and add them.
Chapter 3 - Atoms and Molecules Exercise 40
Molecular mass of O2 = Mass of 2O atoms = 16u + 16u = 32u
Molecular mass of Cl2 = Mass of 2Cl atoms = 35.5u + 35.5u = 71u
Molecular mass of CO2 = Mass of C atom + 2 x Mass of O atom = 12u + (2 16u) = 44u
Molecular mass of CH4 = Mass of C atom + 4 x Mass of H atom = 12u + (4 1 u) = 16u
Molecular mass of NH3 = Mass of N atom + 3 x Mass of H atom = 14u + (3 1 u) = 17u
Molecular mass of CH3OH = Mass of C atom + 3 x Mass of H atom + Mass of O atom + Mass of H atom = 12u + (3 1 u) + 16 u + 1 u = 32u
Concept insight: For calculating the molecular mass, add the atomic masses of all the atoms present in the formula.
Formula unit mass of Na2O = (2 x 23 u) + 16 u = 62 u
Formula unit mass of K2CO3 = (2 x 39 u) + 12 u + (3 x 16 u) = 138 u
Chapter 3 - Atoms and Molecules Exercise 42
i.e. 6.022 x 1023 atoms of carbon weigh 12 gram.
Therefore,
1 atom of carbon weighs 12 / (6.022 x 1023) = 1.993 x 10-23 gram.
Concept insight: For answering questions related to mole concept apply the relation
Chapter 3 - Atoms and Molecules Exercise 43
Thus, by the law of definite proportions, when 3.00 g of carbon is burnt in 50.00 g of oxygen, only 8.00 g of oxygen will be used to produce 11.00 gram of carbon dioxide. The remaining 42.00 g of oxygen will remain unreacted. Law of constant proportion is governed.
Concept insight: Relate the given data in the question to the laws and then identify which law is governing the answer.
Chapter 3 - Atoms and Molecules Exercise 44
Concept insight: Recall the definition of polyatomic ions and give examples of it. It is an important question from exam point of view.
(b) CaO
(c) Cu(NO3)2
(d) AlCl3
(e) CaCO3
Concept insight: For answering this question, write the cation and anion with the proper charges, cross multiply and write the formula.
(b) Hydrogen and bromine
(c) Sodium, hydrogen, carbon, and oxygen
(d) Potassium, sulphur, and oxygen
Concept insight: For answering this question, write the formula of the compound first, see which all atoms are present in it and then write their names.


(b) Molecular mass of sulphur molecule, S8 = 8 x Mass of S = 8

(c) Molecular mass of phosphorus molecule, P4 = 4 x Mass of P = 4

(d) Molecular mass of HCl = Mass of H + Mass of Cl = 1 u + 35.5u = 36.5u
(e) Molecular mass of HNO3 = Mass of H + Mass of N + 3 x Mass of O = 1 u + 14u + 3

Concept insight: For calculating the molecular mass, add the atomic masses of all the atoms present in the formula.
(b) Mass of 4 moles of aluminium atoms = 4 x Mass of 1 mole of Al atoms = 4 x molecular mass of aluminum atoms in grams = 4 x 27 = 108 g
(c) Mass of 10 moles of sodium sulphite = 10 x Mass of 1 mole of Na2SO3 = 10 x molecular mass of Na2SO3 in grams
Molecular mass of sodium sulphite, Na2SO3 = 2 x 23 + 32 + 3 x 16 = 126 g
Thus, the mass of 10 moles of sodium sulphite = 10 x 126 = 1260 g
Concept insight: For answering questions related to mole concept, apply the relation
For an atom ,
1 mole = Gram atomic mass = 6.022

For a molecule,
1 mole = Gram atomic mass = 6.022

Thus, the mass of 0.2 mole of oxygen atoms = 16

(b) The mass of 1 mole of water molecules = 18 g
Thus, the mass of 0.5 mole of water molecules = 18

Concept insight: For answering questions related to mole concept, apply the relation:
1 mole = Gram atomic mass = 6.022

For a molecule,
1 mole = Gram molecule mass = 6.022

1 mole = Gram atomic mass = 6.022

For a molecule,
1 mole = Gram molecule mass = 6.022

1 mole = Gram atomic mass = 6.022

For a molecule,
1 mole = Gram molecule mass = 6.022

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