CG Board Class 9 Science Chapter 9 Solutions: Chemical Formula and Mole Concept (रासायनिक सूत्र और मोल संकल्पना) – 2026-27

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CG SCERT (Chhattisgarh Board) Class 9 Science Solutions | 2026-27 Official Edition

Complete 100% textbook-grounded CG Board Class 9 Science Chapter 9 Solutions: Chemical Formula and Mole Concept (रासायनिक सूत्र और मोल संकल्पना) in Bilingual Format (English & Hindi Medium).

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Part 1

In-Text Questions, Tables & Numericals

Section 9.2

Molecular Weight & Formula Unit Mass Questions

Question 1

If the atomic weight of carbon is 12…


📝 Question:

**: If the atomic weight of carbon is 12 u, then what will be the molecular weight of carbon dioxide (CO2)?

💡 Solution:

If the atomic weight of carbon is 12 u, then what will be the molecular weight of carbon dioxide (CO2)?

  • Atomic weight of Carbon (C) = 12 u
  • Atomic weight of Oxygen (O) = 16 u
  • Molecular weight of CO2 = (Atomic weight of C) + 2 × (Atomic weight of O)
  • Molecular weight of CO2 = 12 + (2 × 16) = 12 + 32 = 44 u

Question 2

If the atomic weight of nitrogen is 14…


📝 Question:

**: If the atomic weight of nitrogen is 14 u, then what is the molecular weight of ammonia (NH3)?

💡 Solution:

If the atomic weight of nitrogen is 14 u, then what is the molecular weight of ammonia (NH3)?

  • Atomic weight of Nitrogen (N) = 14 u
  • Atomic weight of Hydrogen (H) = 1 u
  • Molecular weight of NH3 = 14 + (3 × 1) = 14 + 3 = 17 u

Question 3

Calculate the molecular weight of hydrogen Hsub2sub


📝 Question:

**: Calculate the molecular weight of hydrogen (H2).

💡 Solution:

Calculate the molecular weight of hydrogen (H2).

  • Atomic weight of Hydrogen (H) = 1 u
  • Molecular weight of H2 = 2 × 1 = 2 u

Question 4

Calculate the formula unit mass of the following…


📝 Question:

**: Calculate the formula unit mass of the following ionic compounds: Na_2O, MgCl2, CaCl2, CaO.

💡 Solution:

Calculate the formula unit mass of the following ionic compounds: Na_2O, MgCl2, CaCl2, CaO.

  1. Sodium Oxide (Na_2O)
  • Atomic weight of Na = 23 u, O = 16 u
  • Formula unit mass = (2 × 23) + 16 = 46 + 16 = 62 u
  1. Magnesium Chloride (MgCl2)
  • Atomic weight of Mg = 24 u, Cl = 35.5 u
  • Formula unit mass = 24 + (2 × 35.5) = 24 + 71 = 95 u
  1. Calcium Chloride (CaCl2)
  • Atomic weight of Ca = 40 u, Cl = 35.5 u
  • Formula unit mass = 40 + (2 × 35.5) = 40 + 71 = 111 u
  1. Calcium Oxide (CaO)
  • Atomic weight of Ca = 40 u, O = 16 u
  • Formula unit mass = 40 + 16 = 56 u

Section 9.3.1

Radical Table & Criss-Cross Method Questions

Question 1

Complete Table 1


📝 Question:

Complete Table 1

💡 Solution:

Name of CompoundRadicals Present: PositiveRadicals Present: NegativeNo. of Charges: PositiveNo. of Charges: NegativeFormula of Compound
Sodium carbonateNa^+CO32-12Na_2CO3
Ammonium hydroxideNH4^+OH^-11NH_4OH
Magnesium carbonateMg2+CO32-22MgCO3
Iron phosphateFe2+PO43-23Fe3(PO4)2

Question 2

Use the crisscross method to write down the…


📝 Question:

**: Use the criss-cross method to write down the formulae of the following ionic compounds

💡 Solution:

Use the criss-cross method to write down the formulae of the following ionic compounds

ammonium chloride, calcium hydroxide, magnesium sulphate, ammonium sulphate, calcium phosphate.

  1. Ammonium chloride: Ions are NH4^+ (valency 1) and Cl^- (valency 1). Criss-crossing valencies gives ratio 1:1 → Formula: NH_4Cl.
  2. Calcium hydroxide: Ions are Ca2+ (valency 2) and OH^- (valency 1). Criss-crossing valencies gives ratio 1:2 → Formula: Ca(OH)2.
  3. Magnesium sulphate: Ions are Mg2+ (valency 2) and SO42- (valency 2). Simplifying ratio 2:2 to 1:1 → Formula: MgSO4.
  4. Ammonium sulphate: Ions are NH4^+ (valency 1) and SO42- (valency 2). Criss-crossing valencies gives ratio 2:1 → Formula: (NH4)_2SO4.
  5. Calcium phosphate: Ions are Ca2+ (valency 2) and PO43- (valency 3). Criss-crossing valencies gives ratio 3:2 → Formula: Ca3(PO4)2.

Section 9.5.1

Mole Concept Reaction Questions

Question

Explain the following reactions using the mole concept


📝 Question:

**: Explain the following reactions using the mole concept

💡 Solution:

Explain the following reactions using the mole concept

(a) C + 1/2 O2 → CO

(b) 1/3 Al_2O3 + 2 HCl → 2/3 AlCl3 + H_2O

(a) C + 1/2 O2 → CO

  • 1 mole (12 g) of carbon atoms reacts with 1/2 mole (16 g) of oxygen gas molecules (O2) to form 1 mole (28 g) of carbon monoxide (CO) gas molecules.

(b) 1/3 Al_2O3 + 2 HCl → 2/3 AlCl3 + H_2O

  • 1/3 mole of solid aluminium oxide (Al_2O3) reacts completely with 2 moles of hydrochloric acid (HCl) molecules to form 2/3 mole of aluminium chloride (AlCl3) and 1 mole (18 g) of water (H_2O) molecules.

Part 2

End-of-Chapter Exercises

Exercise Question 1

Choose the correct option

(i) The molecular weight of CH_3OH is:

(a) 32 u

(b) 29 u

(c) 25 u

(d) 20 u

  • Correct Option / : (a) 32 u.
  • Step-by-step Calculation:

12 + (3 × 1) + 16 + 1 = 12 + 3 + 16 + 1 = 32 u.

(ii) Basic radical is:

(a) Positively charged ion /

(b) Negatively charged ion /

  • Correct Option / : (a) Positively charged ion / .

(iii) Ions that have a negative charge are:

(a) Positively charged ion /

(c) Acidic radicals /

  • Correct Option / : (c) Acidic radicals / .

(iv) The value of Avogadro’s number is:

(a) 6.022 × 1023

(b) 6.022 × 1022

(c) 6.022 × 1024

(d) 60.22 × 1023

  • Correct Option / : (a) 6.022 × 1023.

(v) The volume of any gas at 0°C and 1 atm pressure (STP) is:

(a) 11.2 litre

(b) 22.4 litre

(c) 100 litre

(d) 33.8 litre

  • Correct Option / : (b) 22.4 litre.
Exercise Question 2

Fill in the blanks

(i) Unit formula mass is used for ……………………………………….. compounds.

–

(ii) The molecular weight of C12H22O11 is ………………………………………….

–

  • Step-by-step Calculation: (12 × 12) + (22 × 1) + (11 × 16) = 144 + 22 + 176 = 342 u.

(iii) The number of elements in PO43- is ……………………………………………………

–

  • Explanation: Phosphorus (P) and Oxygen (O) are the two distinct elements present.

(iv) The number of atoms in one mole of carbon is ……………………………………….

–

Answer / : 6.022 × 1023.

(v) The mass of one mole of water is …………………………………………

–

SECTION / EXERCISE

Exercise Question 3

  • English:
  • Basic Radicals: Positively charged cations obtained upon the ionization of bases (e.g., NaOH → Na^+ + OH^-, where Na^+ is a basic radical).
  • Acidic Radicals: Negatively charged anions obtained upon the ionization of acids (e.g., HCl → H^+ + Cl^-, where Cl^- is an acidic radical).
  • Hindi:
SECTION / EXERCISE

Exercise Question 4

  • English:

Clusters or groups of atoms of different elements that carry a net fixed positive or negative charge and act as a single unit in chemical reactions are called polyatomic ions.

Examples: Ammonium ion (NH4^+), Carbonate ion (CO32-), Sulphate ion (SO42-), Hydroxide ion (OH^-), Nitrate ion (NO3^-).

  • Hindi:
Exercise Question 5

Connect the appropriate pairs

Column AColumn BAnswer /
(i) Number of atoms in one mole(i) 14 gram(v) 6.022 × 1023
(ii) Acidic radical(ii) Mg2+(iii) SO42-
(iii) Basic radical(iii) SO42-(ii) Mg2+
(iv) The weight of one mole of nitrogen atoms(iv) 2 mol(i) 14 gram
(v) The number of moles in 88 g of CO2(v) 6.022 × 1023(iv) 2 mol
SECTION / EXERCISE

Exercise Question 6

  • English:
  • Definition: The formula unit mass of an ionic compound is the sum of the atomic masses of all atoms/ions present in its formula unit. It is used for ionic solids that do not exist as discrete molecules.
  • Calculation for MgSO4:
  • Atomic mass of Magnesium (Mg) = 24 u
  • Atomic mass of Sulphur (S) = 32 u
  • Atomic mass of Oxygen (O) = 16 u
  • Formula unit mass of MgSO4 = 24 + 32 + (4 × 16) = 24 + 32 + 64 = 120 u.
  • Hindi:
  • MgSO4
SECTION / EXERCISE

Exercise Question 7

If the formula of the carbonate of a metal is M_2CO3, then what will be the formula of its nitrate? / M_2CO3 , ?

Step-by-step Solution /

  • English:
  1. Carbonate ion is CO32- with a valency of 2.
  2. In formula M_2CO3, two atoms of metal M combine with one CO32- ion.
  3. Therefore, valency of metal M is +1 (M^+ ion).
  4. Nitrate ion is NO3^- with a valency of 1.
  5. Criss-crossing valency of M^+ (1) and NO3^- (1) yields the formula MNO3.
  • Hindi:
SECTION / EXERCISE

Exercise Question 8

If the formula of aluminium sulphate is Al2(SO4)3, then what is the charge on the Al ion and what is the formula of zinc sulphate? / Al2(SO4)3 , Al ?

Step-by-step Solution /

  • English:
  1. Charge on Al ion: In Al2(SO4)3, SO4 has a charge of -2. Three SO42- ions carry a total charge of -6. For charge balance, two Al ions must carry a total charge of +6. Hence, each Al ion carries a charge of +3 (Al3+).
  2. Formula of zinc sulphate: Zinc ion is Zn2+ (valency 2) and sulphate ion is SO42- (valency 2). Criss-crossing gives ratio 2:2 which simplifies to 1:1 → Formula: ZnSO4.
  • Hindi:
SECTION / EXERCISE

Exercise Question 9

The ion X has two positive charges on it (X2+). Write the formulae of the nitrate, sulphate and phosphate of X. / X (X2+) X ,

Step-by-step Solution /

  • English:
  • Given ion: X2+ (Valency = 2)
  1. Nitrate of X: Nitrate ion is NO3^- (Valency = 1). Criss-crossing valencies yields X(NO3)2.
  2. Sulphate of X: Sulphate ion is SO42- (Valency = 2). Simplifying 2:2 ratio yields XSO4.
  3. Phosphate of X: Phosphate ion is PO43- (Valency = 3). Criss-crossing valencies yields X3(PO4)2.
  • Hindi:
SECTION / EXERCISE

Exercise Question 10

Calculate the molecular weights of the following: PCl5, H_2O2, S8, HCl, NH3. / : PCl5, H_2O2, S8, HCl, NH3

Step-by-step Numerical Solution /

  • English:
  1. Phosphorus pentachloride (PCl5):
  • Atomic mass: P = 31 u, Cl = 35.5 u
  • Molecular weight = 31 + (5 × 35.5) = 31 + 177.5 = 208.5 u
  1. Hydrogen peroxide (H_2O2):
  • Atomic mass: H = 1 u, O = 16 u
  • Molecular weight = (2 × 1) + (2 × 16) = 2 + 32 = 34 u
  1. Sulphur molecule (S8):
  • Atomic mass: S = 32 u
  • Molecular weight = 8 × 32 = 256 u
  1. Hydrogen chloride (HCl):
  • Atomic mass: H = 1 u, Cl = 35.5 u
  • Molecular weight = 1 + 35.5 = 36.5 u
  1. Ammonia (NH3):
  • Atomic mass: N = 14 u, H = 1 u
  • Molecular weight = 14 + (3 × 1) = 17 u
  • Hindi:
SECTION / EXERCISE

Exercise Question 11

Step-by-step Numerical Solution /

  • English:
  • Formula: Mass (m) = Number of moles (n) × Molar Mass (M)
  1. 5 moles of Ammonia (NH3):
  • M = 17 g/mol ⇒ m = 5 × 17 = 85 g
  1. 0.5 moles of Water (H_2O):
  • M = 18 g/mol ⇒ m = 0.5 × 18 = 9 g
  1. 1.50 moles of Na^+ ions:
  • M = 23 g/mol ⇒ m = 1.50 × 23 = 34.5 g
  1. 0.2 moles of Oxygen atoms (O):
  • M = 16 g/mol ⇒ m = 0.2 × 16 = 3.2 g
  • Hindi:
SECTION / EXERCISE

Exercise Question 12

Calculate the number of moles in the following: 12 g of O2, 22 g of CO2. / : 12 g O2, 22 g CO2

Step-by-step Numerical Solution /

  • English:
  • Formula: Number of moles (n) = (Given mass (m))/(Molar mass (M))
  1. 12 g of Oxygen gas (O2):
  • Given mass m = 12 g
  • Molar mass M = 2 × 16 = 32 g/mol
  • n = 12/32 = 3/8 = 0.375 moles
  1. 22 g of Carbon dioxide (CO2):
  • Given mass m = 22 g
  • Molar mass M = 12 + (2 × 16) = 44 g/mol
  • n = 22/44 = 0.5 moles
  • Hindi:
SECTION / EXERCISE

Exercise Question 13

**Determine the chemical formulae of the following using the crisscross method:

Step-by-step Solution /

  • English:
  1. Iron sulphate (Iron(II) sulphate):
  • Positive ion = Fe2+ (valency 2), Negative ion = SO42- (valency 2)
  • Ratio 2:2 simplifies to 1:1 → Formula: FeSO4
  • (Note: If Iron(III) sulphate is considered, Fe3+ and SO42- → Fe2(SO4)3)
  1. Copper nitrate (Copper(II) nitrate):
  • Positive ion = Cu2+ (valency 2), Negative ion = NO3^- (valency 1)
  • Criss-crossing valencies yields → Formula: Cu(NO3)2
  1. Sodium sulphide:
  • Positive ion = Na^+ (valency 1), Negative ion = S2- (valency 2)
  • Criss-crossing valencies yields → Formula: Na_2S
  1. Magnesium hydrogencarbonate:
  • Positive ion = Mg2+ (valency 2), Negative ion = HCO3^- (valency 1)
  • Criss-crossing valencies yields → Formula: Mg(HCO3)2
  • Hindi:
  • , Fe3+ SO42- → Fe2(SO4)3)*

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