NCERT Solutions Class 9th Science Chapter – 3 Atoms and Molecules Question & Answer

May 26, 2022

NCERT Solutions Class 9th Science Chapter 3 Atoms and Molecules

Textbook	NCERT
Class	9^th
Subject	Science
Chapter	3^rd
Chapter Name	Atoms and Molecules
Category	Class 9^th Science
Medium	English
Source	Last Doubt

NCERT Solutions Class 9th Science Chapter – 3 Atoms and Molecules Question & Answer What is the difference atom and molecule,What is atom and molecule with example,How many atoms in a molecule, What are 3 differences between atoms and molecules, What are 5 examples of molecules,What is called atom,What are 3 examples of atoms,What is a molecule example,Is water an atom,What color is an atom,What is atom made of molecule.

NCERT Solutions Class 9th Science Chapter 3 Atoms and Molecules

Chapter – 3

Atoms and Molecules

Question & Answer

Page: 27

Question 1. In a reaction 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass carbonate.

Answer- In the reactant side we have sodium carbonate and ethanoic acid
Mass of sodium carbonate = 5.3g (given)
Mass of ethanoic acid = 6g (given)

Now total mass before reaction that is mass of reactants = (5.3g + 6g) = 11.3gNow let us check product side Mass of carbon dioxide = 2.2g (given)
Mass of water = 0.9g (given)
Mass of sodium ethanoate = 8.2g (given)

Now total mass after reaction that is mass of products = (8.2g + 2.2g + 0.9g) = 11.3gTotal mass before reaction = total mass after reaction Therefore the given observation is in agreement with the law of conservation of mass .Total mass before reaction = total mass after reaction
Therefore the given observation is in agreement with the law of conservation of mass.

Question 2. Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?

Answer- Ratio of H : O by mass in water is:
Hydrogen : Oxygen —> H₂O
∴ 1 : 8 = 3 : x
x = 8 x 3
x = 24 g

∴ 24 g of oxygen gas would be required to react completely with 3 g of hydrogen gas.

Question 3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Answer- The postulate of Dalton’s atomic theory that is the result of the law of conservation of mass is

• the relative number and kinds of atoms are constant in a given compound.
• Atoms cannot be created nor destroyed in a chemical reaction.

Question 4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

Answer– The relative number and kinds of atoms are constant in a given compound.

Page: 30

Question 1. Define the atomic mass unit.

Answer- One atomic mass unit is equal to exactly one-twelfth (1/12th) the mass of one atom of carbon-12. The relative atomic masses of all elements have been found with respect to an atom of carbon – 12.

Question 2. Why is it not possible to see an atom with naked eyes?
Answer– Atom is too small to be seen with naked eyes. It is measured in nanometres.
1 m = 10⁹ nm

Page: 34

Question 1. Write down the formulae of
(i) Sodium oxide
(ii) Aluminium chloride
(iii) Sodium sulphide
(iv) Magnesium hydroxide

Answer- (i) Sodium oxide : Na₂O
(ii) Aluminium Chloride : AlCl₃
(iii) Sodium sulfide : Na₂S
(iv) Magnesium Hydroxide : Mg(OH)₂

Question 2. What is meant by the term chemical formula?

Answer– The chemical formula of the compound is a symbolic representation of its composition, e.g., chemical formula of sodium chloride is NaCl.

Question 3. How many atoms are present in a

(i) H₂S molecule and
(ii) P0₄^3- ion?

Answer- (i) H₂S —> 3 atoms are present
(ii) P0₄^3- —> 5 atoms are present

Page: 35

Question 1.Calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.

Answer- The total sum of the masses of the atoms or elements present in the molecule is termed as molecular mass.

i) Molecular mass of H₂= 2 x Atomic mass of H
= 2 x 1Molecular mass of H₂ = 2 u

ii) Molecular mass of O₂= 2 x Atomic mass of O
= 2 x 16Molecular mass of O₂= 32 u

iii) Molecular mass of Cl₂= 2 x Atomic mass of Cl
= 2 x 35.5
Molecular mass of Cl₂ = 71 u

iv) Molecular mass of CO₂= Atomic mass of C + 2 x Atomic mass of O
= 12 + 2 x 16
Molecular mass of CO₂ = 44 u

v) Molecular mass of CH₄= Atomic mass of C + 4 x Atomic mass of H
= 12 + 4 x 1
Molecular mass of CH₄ = 16 u

vi) Molecular mass of C₂H₆= 2 x Atomic mass of C + 6 x Atomic mass of H
= 2 x 12 + 6 x 1
Molecular mass of C₂H₆ = 30 u

vii) Molecular mass of C₂H₄= 2 x Atomic mass of C + 4 x Atomic mass of H
= 2 x 12 + 4 x 1
Molecular mass of C₂H₄ = 28 u

viii) Molecular mass of NH₃= Atomic mass of N + 3 x Atomic mass of H
= 14 + 3 x 1
Molecular mass of NH₃ = 17 u

ix) Molecular mass of CH₃OH = Atomic mass of C + 4 x Atomic mass of H + 1 x Atomic mass of
= 12 + 4 x 1 + 16
Molecular mass of CH₃OH = 32 u

Question 2.Calculate the formula unit masses of ZnO, Na₂O, K₂C0₃, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.

Answer- The formula unit mass of
(i) ZnO = 65 u + 16 u = 81 u
(ii) Na₂O = (23 u x 2) + 16 u = 46 u + 16 u = 62 u
(iii) K₂C0₃ = (39 u x 2) + 12 u + 16 u x 3
= 78 u + 12 u + 48 u = 138 u

Page: 35

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.

Answer- Given: Mass of the sample compound
= 0.24g, mass of boron
= 0.096g, mass of oxygen
= 0.144gTo calculate percentage composition of the compound:Percentage of boron = mass of boron / mass of the compound x 100
= 0.096g / 0.24g x 100 = 40%
Percentage of oxygen = 100 – percentage of boron
= 100 – 40 = 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?

Answer – C + O2 → CO2
3g 8g 11g
Total mass of reactants = mass of carbon + mass of oxygen = 3+8 = 11g
Total mass of reactants = Total mass of products
Hence, the law of conservation of mass is proved.
Further, it also shows carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8.
Thus it also proves the law of constant proportions. 3 g of carbon must also combine with 8 g of oxygen only.

This means that (50 − 8) = 42g of oxygen will remain unreacted.

Question 3. What are poly atomic ions? Give examples.

Answer– The ions which contain more than one atoms (same kind or may be of different kind) and behave as a single unit are called polyatomic ions e.g., OH^–, SO₄^2-, CO₃^2-.

Question 4. Write the chemical formulae of the following:

(a) Magnesium chloride

Answer- (a) Magnesium chloride
Symbol —> Mg Cl
Change —> +2 -1
Formula —> MgCl₂

(b) Calcium oxide

Answer- Calcium oxide
Symbol —> Ca O
Charge —> +2 -2
Formula —> CaO

(c) Copper nitrate

Answer- Copper nitrate
Symbol —> Cu NO
Change + 2 – 1
Formula – 4 CU(N0₃)₂

(d) Aluminium chloride

Answer- Aluminium chloride
Symbol —> Al Cl
Change —> +3 -1
Formula —> AlCl₃

(e) Calcium carbonate.

Answer- Calcium carbonate
Symbol —> Ca CO₃
Change —> +2 -2
Formula —> CaC0₃

Question 5. Give the names of the elements present in the following compounds:

(a) Quick lime

Answer– Quick lime —> Calcium oxide
Elements —> Calcium and oxygen

(b) Hydrogen bromide

Answer– Hydrogen bromide
Elements —> Hydrogen and bromine

(c) Baking powder

Answer- Baking powder —> Sodium hydrogen carbonate
Elements —> Sodium, hydrogen, carbon and oxygen

(d) Potassium sulphate.

Answer– Potassium sulphate
Elements —> Potassium, sulphur and oxygen

Question 6. Calculate the molar mass of the following substances.

(a) Ethyne, C₂H₂
Answer– The molar mass of the following: [Unit is ‘g’]
Ethyne, C₂H₂ = 2 x 12 + 2 x 1 = 24 + 2 = 26 g

(b) Sulphur molecule, S₈
Answer– Sulphur molecule, S₈ = 8 x 32 = 256 g

(c) Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31)
Answer- Phosphorus molecule, P₄= 4 x 31 = i24g

(d) Hydrochloric acid, HCl
Answer– Hydrochloric acid, HCl = 1 x 1 + 1 x 35.5 = 1 + 35.5 = 36.5 g

(e) Nitric acid, HNO₃
Answer– Nitric acid, HN0₃ = 1 x 1 + 1 x 14 + 3 x 16 = 1 + 14 + 48 = 63 g

Question 7. What is the mass of

(a) 1 mole of nitrogen atoms?
Answer– Mass of 1 mole of nitrogen atoms = 14 g

(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
Answer– 4 moles of aluminium atoms
Mass of 1 mole of aluminium atoms = 27 g
∴ Mass of 4 moles of aluminium atoms = 27 x 4 = 108 g

(c) 10 moles of sodium sulphite (Na₂S0₃)?
Answer– 10 moles of sodium sulphite (Na₂SO₃)
Mass of 1 mole of Na₂SO₃ = 2 x 23 + 32 + 3 x 16 = 46 + 32 + 48 = 126 g
∴ Mass of 10 moles of Na₂SO₃ = 126 x 10 = 1260 g

Question 8. Convert into mole.

(a) 12 g of oxygen gas
Answer- (a) Given mass of oxygen gas = 12 g
Molar mass of oxygen gas (O₂) = 32 g
Mole of oxygen gas 12/32 = 0.375 mole

(b) 20 g of water
Answer– Given mass of water = 20 g
Molar mass of water (H₂O) = (2 x 1) + 16 = 18 g
Mole of water = 20/18 = 1.12 mole

(c) 22 g of Carbon dioxide.
Answer- Given mass of Carbon dioxide = 22 g
Molar mass of carbon dioxide (CO₂) = (1 x 12) + (2 x 16)
= 12 + 32 = 44 g
∴ Mole of carbon dioxide = 22/44 = 0.5 mole

Question 9. What is the mass of:

(a) 0.2 mole of oxygen atoms?
Answer– (a) Mole of Oxygen atoms = 0.2 mole
Molar mass of oxygen atoms = 16 g
Mass of oxygen atoms = 16 x 0.2 = 3.2 g

(b) 0.5 mole of water molecules?
Answer- Mole of water molecule = 0.5 mole
Molar mass of water molecules = 2 x 1 + 16= 18 g .
Mass of H₂O = 18 x 0.5 = 9 g

Question 10. Calculate the number of molecules of sulphur (S₈) present in 16 g of solid sulphur.

Answer– 1 mole of solid sulphur (S₈) = 8 × 32_g = 256_gi.e., 256g of solid sulphur contains = 6.022 × 10²³ molecules Then, 16g of solid sulpur contains 6.022×10²³ /256 × 16 molecules
= 3.76 × 10²² molecules (approx)

Question 11. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)

Answer- Molar mass of Al₂O₃ = 2 × mass of O = 2×27+3×16 =102g1 mole of Al₂O₃ contains 2 moles of Aluminium ionsSo, 102 g of Al₂O₃ contains 2×6.022×10²³ Aluminium ions
Hence,0.051 g of Al₂O₃ will contain = 2 × 6.022 ×10²³ × 0.051/102 Aluminium ions
= 6.022×10²⁰ Aluminium ions

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