A colourless solid A dissolves in water. The aqueous solution gives a white precipitate B when NaOH or `NH_4` OH is added. B dissolves in excess of NaOH but not in excess of NH,OH. BaCl2 solution added to a solution of A gives a white precipitate C which is insoluble in dilute HCI. A may be 

Options

A. `AI_(2)(SO_(4))_(3)`
B. `ZnSO_4`
C. `SnSO_4`
D. All of the above

Answer

To solve the given problem step by step, we will analyze the information provided and identify the possible identity of the colorless solid A.

### Step 1: Identify the properties of A
- A is a colorless solid that dissolves in water.
- When NaOH or NH₄OH is added to the solution of A, it forms a white precipitate B.

**Hint:** Think about common colorless solids that are soluble in water and can react with bases to form precipitates.

### Step 2: Analyze the precipitate B
- The white precipitate B dissolves in excess NaOH but not in excess NH₄OH.

**Hint:** Consider compounds that can form precipitates with hydroxides and have different solubility properties in NaOH and NH₄OH.

### Step 3: Identify the possible identity of A
- The formation of a white precipitate B suggests that A could be a salt containing a metal that forms a hydroxide.
- Common candidates for A could be aluminum sulfate (Al₂(SO₄)₃), zinc sulfate (ZnSO₄), or tin(II) sulfate (SnSO₄).

### Step 4: Examine the behavior of B with NaOH and NH₄OH
- If A is Al₂(SO₄)₃, then:
- Al₂(SO₄)₃ + 6 NaOH → 2 Al(OH)₃ (white precipitate B) + 3 Na₂SO₄
- Al(OH)₃ is insoluble in NH₄OH but soluble in excess NaOH, forming NaAlO₂.

- If A is ZnSO₄, then:
- ZnSO₄ + 2 NaOH → Zn(OH)₂ (white precipitate B) + Na₂SO₄
- Zn(OH)₂ is soluble in excess NaOH, forming Na₂ZnO₂, and is also insoluble in excess NH₄OH.

- If A is SnSO₄, then:
- SnSO₄ + 2 NaOH → Sn(OH)₂ (white precipitate B) + Na₂SO₄
- Sn(OH)₂ is also soluble in excess NaOH and insoluble in excess NH₄OH.

**Hint:** Check the solubility of the hydroxides formed with NaOH and NH₄OH for each candidate.

### Step 5: Analyze the reaction with BaCl₂
- When BaCl₂ is added to the solution of A, it gives a white precipitate C that is insoluble in dilute HCl.

- For Al₂(SO₄)₃:
- Al₂(SO₄)₃ + 3 BaCl₂ → 3 BaSO₄ (white precipitate C) + 2 AlCl₃
- BaSO₄ is insoluble in dilute HCl.

- For ZnSO₄:
- ZnSO₄ + BaCl₂ → BaSO₄ (white precipitate C) + ZnCl₂
- BaSO₄ is also insoluble in dilute HCl.

- For SnSO₄:
- SnSO₄ + BaCl₂ → BaSO₄ (white precipitate C) + SnCl₂
- BaSO₄ remains insoluble in dilute HCl.

**Hint:** Consider the formation of BaSO₄ as a common precipitate for all candidates.

### Conclusion
Based on the analysis:
- A could be Al₂(SO₄)₃, ZnSO₄, or SnSO₄, as all three meet the criteria for forming precipitate B and C as described in the question.

**Final Answer:** A may be Al₂(SO₄)₃, ZnSO₄, or SnSO₄.