NCERT Solutions for Class 9 Chemistry CBSE Chapter 3: Atoms and Molecules

Atoms and Molecules Exercise 27

Solution 1

The reactants in this reaction are sodium carbonate and ethanoic acid and the products are carbon dioxide, water and sodium ethanoate.
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.

Concept insight: Firstly, recall the law of conservation of mass. And then check whether it is being followed by the data which is given or not.

Atoms and Molecules Exercise 28

Solution 2

The law of constant proportions states that in a chemical substance, the elements are always present in definite proportions by mass. Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water.

1g of hydrogen reacts with oxygen = 8g

Thus, in order to form water, 3 x 8 = 24 g of oxygen will react completely with 3 g of hydrogen gas.

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.

Solution 3

The second postulate of Dalton's atomic theory which states 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.

Concept insight: For answering this question you should know both the law of conservation of mass and the postulates of Dalton's atomic theory.

Solution 4

The postulate of Dalton's atomic theory which states that "The relative number and kinds of atoms are constant in a given compound" can explain the law of definite proportions.

Concept insight: For answering this question, you should know both law of definite proportions and the postulates of Dalton's atomic theory.

Atoms and Molecules Exercise 30

Solution 1

Atomic mass unit is a mass unit equal to exactly one-twelfth the mass of a carbon-12 atom. It is used to measure the masses of subatomic particles.

Concept insight: In this answer, write the proper definition of atomic mass unit.

Solution 2

The dimension of atoms is very, very small and is of the order of 1 nm. Thus, we cannot see an atom with naked eyes.

Concept insight: For answering this question, you should recall the size of an atom.

Atoms and Molecules Exercise 34

Solution 1

Concept insight: For answering this question, write the cation and anion with the proper charges, cross multiply and write the formula.

Solution 2

(i) Aluminium sulphate
(ii) Calcium chloride
(iii) Potassium sulphate
(iv) Potassium nitrate
(v) Calcium carbonate

Concept insight: For writing the name of the compound, write the name of the cation first and then the anion.

Solution 3

The chemical formula of a compound is a symbolic representation of its composition.

Concept insight: For this question, write the proper definition of chemical formula.

Solution 4

(i) Three (Two hydrogen atoms and one sulphur atom)
(ii) Five (One phosphorus and four oxygen)

Concept insight: Count the total number of atoms of each type and add them.

Atoms and Molecules Exercise 35

Solution 1

Molecular mass of H₂ = Mass of 2H atoms = 1u + 1u = 2u
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.

Solution 2

Formula unit mass of ZnO = 65 u + 16 u = 81 u
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

Concept insight: For calculating the formula unit mass, add the atomic masses of all the atoms present in the formula.

Atoms and Molecules Exercise 36

Solution 1

Concept insight:

Always remember,

Solution 2

3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.
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.

Solution 3

Polyatomic ions are those ions that are formed from groups of joined atoms. These atoms can be of the same type or of different type. Some examples of polyatomic ions are NO₃^-, NH₄⁺, OH^-, SO₄^2-, and SO₃^2-.

Concept insight: Recall the definition of polyatomic ions and give examples of it. It is an important question from exam point of view.

Solution 4

(a) MgCl₂
(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.

Solution 5

(a) Calcium and oxygen
(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.

Solution 6

(a) Molecular mass of ethyne, C₂H₂ = 2 x Mass of C + 2 x Mass of H = 2

12u + 2

u = 26u
(b) Molecular mass of sulphur molecule, S₈ = 8 x Mass of S = 8

32u = 256u
(c) Molecular mass of phosphorus molecule, P₄ = 4 x Mass of P = 4

31u = 124u
(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

16u = 63u

Concept insight: For calculating the molecular mass, add the atomic masses of all the atoms present in the formula.

Class 9 NCERT Solutions Chemistry Chapter 3 - Atoms and Molecules

Atoms and Molecules Exercise 27

Solution 1

Atoms and Molecules Exercise 28

Solution 2

Solution 3

Solution 4

Atoms and Molecules Exercise 30

Solution 1

Solution 2

Atoms and Molecules Exercise 34

Solution 1

Solution 2

Solution 3

Solution 4

Atoms and Molecules Exercise 35

Solution 1

Solution 2

Atoms and Molecules Exercise 36

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

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Class 9 NCERT Solutions Chemistry Chapter 3 - Atoms and Molecules

Atoms and Molecules Exercise 27

Solution 1

Atoms and Molecules Exercise 28

Solution 2

Solution 3

Solution 4

Atoms and Molecules Exercise 30

Solution 1

Solution 2

Atoms and Molecules Exercise 34

Solution 1

Solution 2

Solution 3

Solution 4

Atoms and Molecules Exercise 35

Solution 1

Solution 2

Atoms and Molecules Exercise 36

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

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