Solution A is a sodium hydroxide solution. Solution B is a weak acid. Solution C is dilute sulphuric acid. Which solution will
(i) liberate sulphur dioxide from sodium sulphite?
(ii) give a white precipitate with zinc sulphate solution?
(iii) contain solute molecules and ions?

Answer

To solve the given question, we need to analyze the three solutions and their reactions with other compounds based on the information provided.

### Step-by-Step Solution:

1. **Identify the Solutions:**
- Solution A: Sodium hydroxide (NaOH) - a strong base.
- Solution B: A weak acid (e.g., acetic acid, CH₃COOH).
- Solution C: Dilute sulfuric acid (H₂SO₄) - a strong acid.

2. **Determine which solution liberates sulfur dioxide from sodium sulfite:**
- Sodium sulfite (Na₂SO₃) reacts with dilute sulfuric acid (H₂SO₄) to produce sodium sulfate (Na₂SO₄), sulfur dioxide (SO₂), and water (H₂O).
- **Reaction:** Na₂SO₃ + H₂SO₄ → Na₂SO₄ + SO₂ + H₂O
- Therefore, **Solution C (dilute sulfuric acid)** will liberate sulfur dioxide from sodium sulfite.

3. **Determine which solution gives a white precipitate with zinc sulfate:**
- Sodium hydroxide (NaOH) reacts with zinc sulfate (ZnSO₄) to form zinc hydroxide (Zn(OH)₂), which is a white precipitate.
- **Reaction:** ZnSO₄ + 2NaOH → Zn(OH)₂ (white precipitate) + Na₂SO₄
- Therefore, **Solution A (sodium hydroxide)** will give a white precipitate with zinc sulfate solution.

4. **Determine which solution contains solute molecules and ions:**
- A weak acid, such as acetic acid (CH₃COOH), partially dissociates in water to form ions (CH₃COO⁻ and H₃O⁺) and also contains undissociated molecules.
- Therefore, **Solution B (the weak acid)** will contain both solute molecules and ions.

### Summary of Answers:
- (i) **Solution C** (dilute sulfuric acid) liberates sulfur dioxide from sodium sulfite.
- (ii) **Solution A** (sodium hydroxide) gives a white precipitate with zinc sulfate solution.
- (iii) **Solution B** (the weak acid) contains solute molecules and ions.