The aqueous solution of salt gives white ppt with lead acetate solution which is insoluble in hot water and nitric acid .The salt contains

Options

A. `CI^(Θ)`
B. `Br^(2+)`
C. `CO_(3)^(2+)`
D. `SO_(4)^(2-)`

Explanation

To solve the problem, we need to analyze the information given about the salt in the aqueous solution and its reaction with lead acetate.

### Step-by-Step Solution:

1. **Identify the Reaction**: The salt in the aqueous solution gives a white precipitate when treated with lead acetate. This indicates that the anion in the salt is likely to form an insoluble compound with lead ions (Pb²⁺).

2. **Consider Possible Anions**: The options provided are:
- Cl⁻ (Chloride)
- Br⁻ (Bromide)
- CO₃²⁻ (Carbonate)
- SO₄²⁻ (Sulfate)

3. **Analyze the Precipitate**: Lead acetate (Pb(CH₃COO)₂) reacts with certain anions to form precipitates. The white precipitate formed is likely to be lead sulfate (PbSO₄), which is known to be white and insoluble in water and nitric acid.

4. **Check Solubility**:
- Lead chloride (PbCl₂) and lead bromide (PbBr₂) are soluble in hot water.
- Lead carbonate (PbCO₃) is also soluble in hot water.
- Lead sulfate (PbSO₄) is insoluble in both hot water and nitric acid.

5. **Conclusion**: Since the white precipitate formed is insoluble in hot water and nitric acid, the only anion that fits this criterion is sulfate (SO₄²⁻).

Thus, the salt contains **SO₄²⁻ ions**.

### Final Answer:
The salt contains SO₄²⁻ ions.

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