Question 1: Give examples.
a. Positive radicals
Examples: Ammonium (NH₄⁺), Hydronium (H₃O⁺)
b. Basic radicals
Examples: Sodium (Na⁺), Potassium (K⁺), Calcium (Ca²⁺), Aluminium (Al³⁺)
c. Composite radicals
Examples: Nitrate (NO₃⁻), Sulphate (SO₄²⁻), Carbonate (CO₃²⁻), Phosphate (PO₄³⁻)
d. Metals with variable valency
Examples: Iron (Fe²⁺, Fe³⁺), Copper (Cu⁺, Cu²⁺), Mercury (Hg⁺, Hg²⁺), Tin (Sn²⁺, Sn⁴⁺)
e. Bivalent acidic radicals
Examples: Sulphate (SO₄²⁻), Carbonate (CO₃²⁻), Sulphite (SO₃²⁻)
f. Trivalent basic radicals
Examples: Aluminium (Al³⁺), Iron (Fe³⁺), Chromium (Cr³⁺)
Question 2: Write symbols of the following elements and the radicals obtained from them, and indicate the charge on the radicals.
| Element | Symbol | Radical | Charge |
| Mercury | Hg | Hg⁺, Hg²⁺ | +1, +2 |
| Potassium | K | K⁺ | +1 |
| Nitrogen | N | N³⁻ (Nitride) | -3 |
| Copper | Cu | Cu⁺, Cu²⁺ | +1, +2 |
| Sulphur | S | S²⁻ (Sulphide) | -2 |
| Carbon | C | C⁴⁺ (Carbide) | +4 |
| Chlorine | Cl | Cl⁻ (Chloride) | -1 |
| Oxygen | O | O²⁻ (Oxide) | -2 |
Question 3: Write the steps in deducing the chemical formulae of the following compounds.
General Steps:
- Write the symbols of the elements/radicals.
- Write the valency below each symbol.
- Cross-multiply the valencies to get the ratio of atoms.
- Write the formula using the ratio.
a. Sodium sulphate
- Sodium (Na⁺, valency=1), Sulphate (SO₄²⁻, valency=2)
- Cross-multiply: Na₂(SO₄)₁ → Na₂SO₄
b. Potassium nitrate
- Potassium (K⁺, valency=1), Nitrate (NO₃⁻, valency=1)
- Cross-multiply: K₁(NO₃)₁ → KNO₃
c. Ferric phosphate
- Ferric (Fe³⁺, valency=3), Phosphate (PO₄³⁻, valency=3)
- Cross-multiply: Fe₁(PO₄)₁ → FePO₄
d. Calcium oxide
- Calcium (Ca²⁺, valency=2), Oxide (O²⁻, valency=2)
- Cross-multiply: Ca₁O₁ → CaO
e. Aluminium hydroxide
- Aluminium (Al³⁺, valency=3), Hydroxide (OH⁻, valency=1)
- Cross-multiply: Al₁(OH)₃ → Al(OH)₃
Question 4: Write answers to the following questions and explain your answers.
a. Explain the monovalency of the element sodium.
- Sodium has atomic number 11. Its electron configuration is 2,8,1.
- It has 1 electron in its outermost shell and tends to lose this electron to achieve a stable octet.
- Thus, it forms a +1 ion (Na⁺) and shows a valency of 1 (monovalency).
b. M is a bivalent metal. Write down the steps to find the chemical formulae of its compounds formed with the radicals: sulphate and phosphate.
- Bivalent metal M has valency 2.
- With sulphate (SO₄²⁻, valency=2):
Cross-multiply: M₂(SO₄)₂ → Simplify to MSO₄. - With phosphate (PO₄³⁻, valency=3):
Cross-multiply: M₃(PO₄)₂ → M₃(PO₄)₂.
c. Explain the need for a reference atom for atomic mass. Give some information about two reference atoms.
- Atomic masses are relative and need a standard for comparison.
- Initially, hydrogen (H=1) was used as it is the lightest element.
- Later, oxygen (O=16) was adopted since it forms compounds with many elements.
- Currently, carbon-12 (C-12=12) is used as the standard because its mass can be measured accurately using mass spectrometry.
d. What is meant by Unified Atomic Mass?
- Unified Atomic Mass (u) is a standard unit of mass for atoms and molecules.
- 1 u is defined as exactly 1/12th the mass of a carbon-12 atom.
- It is used to express atomic and molecular masses.
e. Explain with examples what is meant by a ‘mole’ of a substance.
- A mole is the amount of substance that contains as many entities (atoms, molecules, etc.) as there are atoms in exactly 12g of carbon-12.
- This number is Avogadro’s number (6.022 × 10²³).
- Example: 1 mole of oxygen atoms = 16g, 1 mole of water molecules = 18g.
Question 5: Write the names of the following compounds and deduce their molecular masses.
| Compound | Name | Molecular Mass |
| Na₂SO₄ | Sodium sulphate | 2(23) + 32 + 4(16) = 46 + 32 + 64 = 142 u |
| K₂CO₃ | Potassium carbonate | 2(39) + 12 + 3(16) = 78 + 12 + 48 = 138 u |
| CO₂ | Carbon dioxide | 12 + 2(16) = 12 + 32 = 44 u |
| MgCl₂ | Magnesium chloride | 24 + 2(35.5) = 24 + 71 = 95 u |
| NaOH | Sodium hydroxide | 23 + 16 + 1 = 40 u |
| AlPO₄ | Aluminium phosphate | 27 + 31 + 4(16) = 27 + 31 + 64 = 122 u |
| NaHCO₃ | Sodium bicarbonate | 23 + 1 + 12 + 3(16) = 23 + 1 + 12 + 48 = 84 u |
Question 6: Two samples ‘m’ and ‘n’ of slaked lime were obtained from two different reactions. The details about their composition are as follows:
‘sample m’ mass: 7g
Mass of constituent oxygen: 2g
Mass of constituent calcium: 5g
‘sample n’ mass: 1.4g
Mass of constituent oxygen: 0.4g
Mass of constituent calcium: 1.0g
Which law of chemical combination does this prove? Explain.
- This proves the Law of Constant Proportion (or Definite Proportion).
- The law states that a chemical compound always contains the same elements in the same proportion by mass, regardless of its source or method of preparation.
- In both samples, the ratio of calcium to oxygen is:
- Sample m: Ca:O = 5:2 = 2.5:1
- Sample n: Ca:O = 1.0:0.4 = 2.5:1
- Since the ratio is the same, the law is verified.
Question 7: Deduce the number of molecules of the following compounds in the given quantities.
Formula: Number of molecules = (Given mass / Molar mass) × Avogadro’s number (6.022 × 10²³)
a. 32g oxygen (O₂)
- Molar mass of O₂ = 32g/mol
- Number of molecules = (32/32) × 6.022 × 10²³ = 6.022 × 10²³ molecules
b. 90g water (H₂O)
- Molar mass of H₂O = 18g/mol
- Number of molecules = (90/18) × 6.022 × 10²³ = 5 × 6.022 × 10²³ = 3.011 × 10²⁴ molecules
c. 8.8g carbon dioxide (CO₂)
- Molar mass of CO₂ = 44g/mol
- Number of molecules = (8.8/44) × 6.022 × 10²³ = 0.2 × 6.022 × 10²³ = 1.2044 × 10²³ molecules
d. 7.1g chlorine (Cl₂)
- Molar mass of Cl₂ = 71g/mol
- Number of molecules = (7.1/71) × 6.022 × 10²³ = 0.1 × 6.022 × 10²³ = 6.022 × 10²² molecules
Question 8: If 0.2 mol of the following substances are required, how many grams of those substances should be taken?
Formula: Mass = Number of moles × Molar mass
a. Sodium chloride (NaCl)
- Molar mass = 23 + 35.5 = 58.5g/mol
- Mass = 0.2 × 58.5 = 11.7g
b. Magnesium oxide (MgO)
- Molar mass = 24 + 16 = 40g/mol
- Mass = 0.2 × 40 = 8.0g
c. Calcium carbonate (CaCO₃)
- Molar mass = 40 + 12 + 3(16) = 100g/mol
- Mass = 0.2 × 100 = 20.0g