NCERT Solutions Class 9 Science Chapter 3 Atoms and Molecules

NCERT Solutions Class 9 Science 

The NCERT Solutions in English Language for Class 9 Science Chapter – 3 (Atoms and Molecules) has been provided here to help the students in solving the questions from this exercise.

Chapter – 3 (Atoms and Molecules)

Questions

1. In a reaction, 5.3g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g of water and 8.2 g of sodium acetate. Show that these observations are in agreement with the law of conservation of mass.
Sodium carbonate + acetic acid → Sodium acetate + carbon dioxide +  water
Answer –
Sodium carbonate (5.3g) + acetic acid (6g) → Sodium acetate (8.2g) + carbon dioxide (2.2g) +  water (0.9g)
Na₂​CO₃ ​ +  CH₃​COOH → CH₃​COONa + CO₂ ​ +  H₂​O
Mass of reactants =5.3 + 6 = 11.3 gm
Mass of products = 2.2 + 0.9 + 8.2 = 11.3 gm
This shows, that during a chemical reaction, the mass of reactant = mass of the product.

2. Hydrogen and oxygen combine in a 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 –   We know hydrogen and water mix in a ratio 1 : 8.
This means that 1g of Hydrogen reacts with 8g of Oxygen to form. H₂O
1g Hydrogen → 8g Oxygen
3g Hydrogen → 3 × 8g Oxygen
= 24g Oxygen
Thus, 24g of oxygen gas would be required to react completely with 3 g of hydrogen gas.

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 which is a result of the law of conservation of mass is, “Atoms are indivisible particles, which can neither be created nor be destroyed”.

4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer – The postulate of Dalton’s atomic theory that can explain the law of definite proportions is – the relative number and kinds of atoms are equal in given compounds.

Questions

1. Define the atomic mass unit.
Answer – An atomic mass unit is a unit of mass used to express the weight subatomic particles, where one unit is equal to exactly one-twelfth the mass of a carbon-12 atom.

Questions

1. Write down the formulae of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide
Answer –  The following are the formulae:
(i) sodium oxide – Na₂O
(ii) aluminium chloride – AlCl₃
(iii) sodium sulphide – Na₂S
(iv) magnesium hydroxide – Mg (OH)₂

2. Write down the names of compounds represented by the following formulae:
(i) Al₂(SO₄)₃
(ii) CaCl₂
(iii) K₂SO₄
(iv) KNO₃
(v) CaCO₃
Answer –  The names of the compounds for each of the following formulae:
(i) Al₂(SO₄)₃ – Aluminium sulphate
(ii) CaCl₂ – Calcium chloride
(iii) K₂SO₄ – Potassium sulphate
(iv) KNO₃ – Potassium nitrate
(v) CaCO₃ – Calcium carbonate

3. What is meant by the term chemical formula?
Answer –  The chemical formula is the symbolic representation of a chemical compound.
Example: Chemical formula of sodium chloride is $NaCl$

4. How many atoms are present in a
(i) H₂S molecule and
(ii) PO₄^3- ion?
Answer –  The number of atoms present is as follows:
(i) H₂S molecule has 2 atoms of hydrogen and 1 atom of sulphur hence 3 atoms in total.
(ii) PO₄^3- ion has 1 atom of phosphorus and 4 atoms of oxygen hence 5 atoms in total.

Questions

1. Calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.
Answer –  The following are the molecular masses:

• The molecular mass of H₂ – 2 × atoms atomic mass of H = 2 × 1u = 2u
• The molecular mass of O₂ – 2 × atoms atomic mass of O = 2 × 16u = 32u
• The molecular mass of Cl₂ – 2 × atoms atomic mass of Cl = 2 × 35.5u = 71u
• The molecular mass of CO₂ – atomic mass of C + 2 × atomic mass of O = 12 + ( 2 × 16)u = 44u
• The molecular mass of CH₄ – atomic mass of C + 4 × atomic mass of H = 12 + ( 4 × 1)u = 16u
• The molecular mass of C₂H₆ – 2 × atomic mass of C + 6 × atomic mass of H = (2 × 12) + (6 × 1)u
  = 24 + 6
  = 30u
• The molecular mass of C₂H₄– 2 × atomic mass of C + 4 × atomic mass of H = (2 × 12) + (4 × 1)u
  = 24 + 4
  = 28u
• The molecular mass of NH₃ – atomic mass of N + 3 × atomic mass of H = (14 + 3 × 1)u
  = 17u
• The molecular mass of CH₃OH – atomic mass of C + 3 × atomic mass of H + atomic mass of O + atomic mass of H = (12 + 3 × 1 + 16 + 1)u
  =(12 + 3 + 17)u
  = 32u

2. Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃,
given atomic masses of Zn = 65u, Na = 23 u,  K=39u, C = 12u, and O=16u.
Answer –  Given:
The formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O
= 65u + 16u
= 81u

The formula unit mass of Na₂O = 2 × Atomic mass of Na + Atomic mass of O
= (2 × 23)u + 16u
= 46u + 16u
= 62u

The formula unit mass of K₂CO₃ = 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of O
= (2 × 39)u + 12u + (3 × 16)u
= 78u + 12u + 48u
= 138u

Questions

1. If one mole of carbon atoms weighs 12grams, what is the mass (in grams) of 1 atom of carbon?
Answer – 1 mole of carbon atoms means 6.022 × 10²³ carbon atoms. In this case 1 mole of carbon atoms weighs 12 grams. This means that the mass of 6.022 × 10²³ atoms of carbon is 12 grams.
Now, 6.022 × 10²³ atoms of carbon have mass = 12 g
1 atom of carbon has mass =
= 1.99 × 10^-23 g
Thus, the absolute mass of 1 atom of carbon is 1.99 × 10^-23 grams.

2. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given the atomic mass of Na = 23u, Fe = 56 u)?
Answer –
(a) Moles of sodium = Mass of sodium/Molar mass of sodium
= 100/23
= 4.34
N₀ = 6.022 × 10²³ atoms
= 4.34 × 6.022 × 10²³ atoms
= 26.14 × 10²³ atoms

(b) Moles of iron = Mass of iron/Molar mass of iron
= 100/56
= 1.78
N₀ = 6.022 × 10²³ atoms
= 1.78 × 6.022 × 10²³
= 10.72 ×10²³ atoms
Therefore, the number of atoms is more in 100 g of Na than in 100 g of Fe.

Exercise

1. A 0.24g sample of a compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144g of oxygen. Calculate the percentage composition of the compound by weight.
Answer –
Mass of the sample compound = 0.24g,
mass of boron = 0.096g,
mass of oxygen = 0.144g
Percentage of boron = Mass of boron in compound/Mass of compound × 100
=
= 40%
Percentage of oxygen = 100 – percentage of boron
= 100 – 40 = 60%

2. When 3.0g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00g of carbon is burnt in 50.00 g of oxygen?  Which law of chemical combination will govern your answer?
Answer –
C (3g) + O₂ (8g)  ​→  CO₂ (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 $\text{law of constant proportions}$. 3 g of carbon must also combine with 8 g of oxygen only. This means that $(50-8)=42g$ of oxygen will remain unreacted.

3. What are polyatomic ions? Give examples.
Answer – Polyatomic ions are ions that contain more than one atom, but they behave as a single unit.
Example – CO₃^2-, H₂PO₄^–

4. Write the chemical formula of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
Answer –  The following are the chemical formula of the above-mentioned list:
(a) Magnesium chloride – MgCl₂
(b) Calcium oxide – CaO
(c) Copper nitrate – Cu(NO₃)₂
(d) Aluminium chloride – AlCl₃
(e) Calcium carbonate – CaCO₃

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 –  The following are the names of the elements present in the following compounds:
(a) Quick lime – Calcium and oxygen (CaO)
(b) Hydrogen bromide – Hydrogen and bromine (HBr)
(c) Baking powder – Sodium, Carbon, Hydrogen, Oxygen (NaHCO₃)
(d) Potassium sulphate – Sulphur, Oxygen, Potassium (K₂SO₄)

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
Answer –  Listed below is the molar mass of the following substances:
(a) Molar mass of Ethyne C₂H₂ = 2 × Mass of C + 2 × Mass of H
= (2 × 12) + (2 × 1)
= 24 + 2
= 26g

(b) Molar mass of Sulphur molecule S₈ = 8 × Mass of S
= 8  × 32
= 256g

(c) Molar mass of  Phosphorus molecule, P₄ = 4 × Mass of P
= 4 × 31
= 124g

(d) Molar mass of Hydrochloric acid, HCl = Mass of H + Mass of Cl
= 1 + 35.5
= 36.5g

(e) Molar mass of Nitric acid, HNO₃ = Mass of H + Mass of Nitrogen + 3 × Mass of O
= 1 + 14 + 3 × 16
= 63g

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 –  The mass of the above-mentioned list is as follows:
(a) Atomic mass of nitrogen atoms = 14u
Mass of 1 mole of nitrogen atoms = Atomic mass of nitrogen atom
Atomic mass of nitrogen atom = 14u
Therefore, Mass of 1 mole of nitrogen atoms = 14 g

(b) Atomic mass of aluminium = 27u
Mass of 1 mole of Aluminium atoms = Atomic mass of Aluminium
Atomic mass of Aluminium = 27 u
Mass of 1 mole of aluminium atom = 27 g
Therefore, mass of 4 moles of aluminium atoms = 4 × 27 = 108 g

(c) Mass of 1 mole of sodium sulphite Na₂SO₃ = Molecular mass of sodium sulphite
= 2 × Mass of Na + Mass of S + 3 × Mass of O
=  (2 × 23) + 32 + (3 × 16)
= 46 + 32 + 48
= 126g
Therefore, mass of 10 moles of Na₂SO₃  = 10 × 126 = 1260g

8. Convert into a mole.
(a) 12g of oxygen gas
(b) 20g of water
(c) 22g of carbon dioxide
Answer –
(a) Atomic mass of Oxygen = 16 u
Molecular mass of Oxygen (O₂) = 16 × 2 = 32 g
Now, 1 mole of oxygen gas = Molecular mass of oxygen in grams = 32 g
Also, 32g of oxygen gas = 1 mole
12 g of Oxygen gas = × 12 = 0.375 mole
Hence, 12 g of Oxygen gas = 0.375 mole

(b) Molecular mass of water (H₂0) = 1 ×2 + 16 = 18u.
Now, 1 mole of water = Molecular mass of water in grams = 18 g
If 18 g of water = 1 mole
Then 20 g of water =  × 20 mole = 1.12 moles
Hence, 20 g of water = 1.12 moles

(c) Molecular mass of carbon dioxide (CO₂) = 12 + 16 × 2 = 12 + 32 = 44 u.
Now, 1 mole of carbon dioxide = Molecular mass of carbon dioxide in grams = 44g
If 44g of carbon dioxide = 1 mole
Then, 22 g of carbon dioxide = = 0.5 mole
Hence, 22 g of carbon dioxide = 0.5 mole

9. What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Answer –
(a)  The atomic mass of oxygen (O) = 16 u.
The mass of 1 mole of oxygen atoms = 16g.
Now, 1 mole of oxygen atoms = 16g
So, 0.2 mole of oxygen atoms = 16 × 0.2g = 3.2g
Hence, 0.2 mole of oxygen atom = 3.2 g.

(b) The molecular mass of water (H₂0)= 18u,
The mass of 1 mole of water molecules = 18 g
Now, 1 mole of water molecules = 18g
0.5 mole of water molecules = 18 × 0.5g = 9g
Hence, 0.5 mole of water molecules = 9 g

10. Calculate the number of molecules of sulphur (S₈) present in 16g of solid sulphur.
Answer –
Molecular mass of Sulphur (S₈) = 8xMass of Sulphur = 8 × 32 = 256g
Mass given = 16g
Number of moles = mass given/ molar mass of sulphur
= 16/256
= 0.0625 moles
Also, 1 mole of sulphur molecules = 6.023 × 10²³ molecules
0.0625 mole of sulphur molecules = 6.023 × 10²³ × 0.0625 molecules
= 3.76 × 10²² molecules
Therefore, the number of molecules present in 16 g sulphur is 3.76 × 10²²

11. Calculate the number of aluminium ions present in 0.051g 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 = 27u)
Answer –
1 mole of Al₂O₃ = Formula mass of Al₂O₃ in grams
= Mass of Al × 2 + Mass of O × 3
= 27 × 2 + 16 × 3
= 54 + 48
= 102 grams
Now, 1 mole of Al₂O₃ contains 2 moles of Al.
So, Mass of Al in 1 mole of Al₂O₃ = Mass of Al × 2
= 27 × 2
= 54 grams
Now, 102 g aluminium oxide contains = 54 g Al
So, 0.051 g aluminium oxide contains = 54/102 × 0.051 g Al
= 0.027 g Al
The atomic mass of aluminium is given to be 27 u. This means that 1 mole of aluminium atoms (or aluminium ions) has a mass of 27 grams, and it contains 6.022 × 10²³ aluminium ions.
Now, 27 g of aluminium has ions = 6.022 × 10²³
So, 0.027 g of aluminium has ions = 6.022 × 10²³/27 × 0.027
= 6.022 × 10²⁰
Thus, the number of aluminium ions (Al³⁺) in 0.051 gram of aluminium oxide is 6.022 × 10²⁰.

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