Class 9 Science Chapter 3 Question Answers - Atoms and Molecules

Q1: Glucose has the molecular formula C₆H₁₂O₆. Calculate :
(a) Its molecular mass. 
(b) The number of atoms in one molecule of glucose. 
(c) The number of gram molecules in 18 g of glucose.
Ans: The molecular mass of a compound is the sum of the atomic masses of all the atoms present in its molecular formula.
(a) Molecular mass of C₆H₁₂O₆  
= (6 × 12u) + (12 × 1u) + (6 × 16u)  
= 72u + 12u + 96u = 180u
(b) The number of atoms in one molecule of C₆H₁₂O₆  
= 6 atoms of C + 12 atoms of H   + 6 atoms of O  
= 6 + 12 + 6 = 24 atoms
(c) Number of gram molecules
= Mass of glucose (g) / Molecular mass of glucose (g)
= 18 g / 180 g/mol
= 0.1 mol

Q2: What is the use of the mole concept in chemistry?

Ans: The mole concept is extremely useful in chemistry for quantitative analysis:

1. It allows chemists to count atoms, molecules, or ions using mass.
2. One mole of any substance contains 6.022 × 10²³ particles (Avogadro's number).
3. One mole of any gas occupies 22.4 litres at standard temperature and pressure (STP).
4. It helps in calculating reactants and products in chemical reactions through balanced equations.
5. It is used to determine molar mass and establish relationships between mass, volume, and number of particles.

Q3: Give symbol and valency of: Potassium, Barium, Aluminium, Calcium, Cobalt, Fluorine, Lead, Zinc, Iodine, Sulphide.
Ans:

Q4: In a reaction, 6.83 g of lead chloride precipitated. The initial solution had 5 g each of lead nitrate and sodium chloride. What is the amount of sodium nitrate formed?

Ans: Initially, we have:

• 50 g of 10% lead nitrate solution = 5 g Pb(NO₃)₂ + 45 g water

• 50 g of 10% sodium chloride solution = 5 g NaCl + 45 g water Total mass before reaction = 5 g + 5 g + 90 g water = 100 g

After reaction, 6.83 g of lead chloride (PbCl₂) precipitates out. Assuming no mass is lost, Total mass after = 100 g Remaining components (water and NaNO₃) = 100 – 6.83 = 93.17 g

Water is 90 g, so: Mass of sodium nitrate formed = 93.17 – 90 = 3.17 g

Q5: Write a formula for the following : 
(a) Zinc sulphate, 
(b) Methane, 
(c) Ammonium carbonate.
Ans:

(a) Zinc sulphate 

Thus, Zn₂(SO₄)₂ and finally = ZnSO₄

(b) Methane

Thus, finally = CH₄

(c) Ammonium carbonate

Thus, finally = (NH₄)₂CO₃

Q6: Explain the law of multiple proportions.
Ans: According to the law of multiple proportions, when two elements combine to make one or more compounds, then the ratio of weights of these elements remains in a fixed ratio to one another.

For example: Hydrogen and oxygen combine to form water (H₂O) and hydrogen peroxide (H₂O₂) under different conditions. 2 grams of hydrogen combine with 16 grams of oxygen in the case of water, while 2 grams of hydrogen combine with 32 grams of oxygen to form hydrogen peroxide. Now, the weights of oxygen combine with a fixed weight of hydrogen in water and hydrogen peroxide, respectively ,are 16 and 32, which are in a simple ratio of 16: 32 or 1 : 2.

Q7: What are molecules? Give a brief explanation of the arrangement of the constituent atoms in the molecules.
Ans: A molecule is the smallest particle of an element or compound which is stable under normal conditions. And it can freely show all the properties of that element or compound. It may be made up of one, two or more atoms. A molecule with one atom is called a monoatomic molecule.
E.g. helium, neon, etc.

A molecule with two atoms is called a diatomic molecule. E.g. Cl₂, O₂.
Similarly, there are molecules containing three atoms (CO₂), four atoms (P₄) and so on.

Q8: The mass of one molecule of a substance is 4.65 × 10^-23 grams. What is its molecular mass?
Ans: Mass of 1 molecule of a substance = 4.65 × 10^-23 grams

Mass of 6.023 × 10²³ molecules of a substance  

= 4.65 × 10^-23 × 6.023 × 10²³

= 28 g/mol

Molecular mass of the substance = 28 g/mol

Q9: An element ₁₂X²⁴ loses two electrons to form a cation, which combines with the anion of element ₁₇Y³⁵ formed by gaining an electron. 
(a) Write the electronic configuration of element X. 
(b) Write the electronic configuration of the anion of element Y. 
(c) Write the formula for the compound formed by the combination of X and Y.
Ans: 

(a) X = 2, 8, 2

(b) Y^– = 2, 8, 8

(c) XY₂

Q10: Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃ given atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u, and 0 = 16u.
Ans: Formula unit mass of ZnO = 1 × 65u + 1 × 16u = 81u  

Formula unit mass of Na₂O = 2 × 23u + 1 × 16u = 62u  

Formula unit mass of K₂CO₃ = 2 × 39u + 1 × 12u + 3 × 16u = 138u

Q11: 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)?
Ans: 

(a) 1 mole of nitrogen atoms  

= 1 × gram atomic mass of nitrogen atom  

= 1 × 14 g = 14 g

(b) 4 moles of aluminium atoms   

= 4 × gram atomic mass of aluminium atoms  

= 4 × 27 g = 108 g

(c) 10 moles of sodium sulphite (Na₂SO₃)   

= 10 (2 × gram atomic mass of Na   + 1 × gram atomic mass of sulphur   + 3 × gram atomic mass of oxygen)  

= 10 (2 × 23 g + 1 × 32 g + 3 × 16g)  

= 10 (46 g + 32 g + 48 g)  

= 10 × 126 g = 1260 g

Q12: Give the postulates of Dalton’s atomic theory.
Ans: John Dalton proposed his atomic theory in 1808. It explains the nature of matter in terms of atoms. The main postulates of Dalton’s atomic theory are:

1. All matter is made up of indivisible particles called atoms.
2. Atoms of a given element are identical in mass and properties.
3. Atoms of different elements have different masses and chemical properties.
4. Atoms combine in simple whole-number ratios to form compounds.
5. Atoms can neither be created nor destroyed in a chemical reaction.
6. The relative number and types of atoms in a given compound remain constant.

These postulates laid the foundation for modern chemistry, although some parts were later modified with the discovery of subatomic particles.

Q13: (a) Give one point of difference between an atom and an ion. 
(b) Give one example each of a polyatomic cation and an anion. 
(c) Identify the correct chemical name of FeSO₃: Ferrous sulphate, Ferrous sulphide, Ferrous sulphite.
(d) Write the chemical formula for the chloride of magnesium.

Ans: 

(a) An atom is a neutral particle containing equal numbers of protons and electrons. An ion is a charged particle formed when an atom loses or gains electrons. Positive ions are called cations, and negative ions are called anions.

(b) (i) Polyatomic cation: Ammonium ion, (NH₄)⁺

 (ii) Polyatomic anion: Sulphate ion, (SO₄)^2–

(c) FeSO₃ contains Fe²⁺ and SO₃²⁻ (sulphite ion), so the correct name is Ferrous sulphite.

(d) Magnesium has a valency of 2, and chlorine has a valency of 1. Therefore, the formula is MgCl₂ (Magnesium chloride).

Q14: When 3.0 g of magnesium is burnt in 2.00 g of oxygen, 5.00 g of magnesium oxide is produced. What mass of magnesium oxide will be formed when 3.00 g of magnesium is burnt in 5.00 g of oxygen? Which law of chemical combination will govern your answer? State the law.
Ans: According to the given reaction: 3.00 g magnesium + 2.00 g oxygen → 5.00 g magnesium oxide. This shows that 3 g of magnesium reacts completely with 2 g of oxygen.

Now, when 3.00 g magnesium is burnt in 5.00 g oxygen, only 2 g of oxygen will be used (as per the fixed ratio), and 3 g of oxygen will remain unused.

So, mass of magnesium oxide formed = 3 + 2 = 5.00 g

Law involved: This illustrates the Law of Definite Proportions, which states that a chemical compound always contains the same elements in the same fixed ratio by mass.

Q15: (a) Calculate the number of molecules of SO₂ present in 44 g of it.
(b) If one mole of oxygen atoms weighs 16 grams, find the mass of one atom of oxygen in grams.
Ans:
(a) Molecular mass of SO2 = Atomic mass of S    + 2 × Atomic mass of O
= 32 + 2 × 16
= 64u
Molar mass = 64 g
Number of molecules, N =
= 44/64 x 6.022 x 10²³
= 4.14 × 10²³ molecules
(b) One mole of oxygen contains 6.022 × 10²³ atoms of oxygen
Mass of one atom of oxygen =
= 2.66 × 10^–23 g

Q16: Sodium is represented as ₂₃Na¹¹.
(a) What is its atomic mass?
(b) Write its gram atomic mass.
(c) How many atoms of Na will be there in 11.5 g of the sample?
Ans:
(a) Atomic mass = 23u
(b) Gram atomic mass = 23 g
(c) Given mass = 11.5 g  
Molar mass = 23 g Number of atoms (N) =

= 3.011 × 10²³ atoms