Statement-1 : Sodium meta aluminate on boiling with ammonium chloride produces white gelatinous precipitate.
Statement-2 :Aluminium hydroxide is formed which is not soluble in water

Options

A. Both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation of Statement-1.
B. Both Statement-1 and Statement-2 are true and Statement-2 is not correct explanation of Statement-1.
C. Statement-1 is true but Statement-2 is false.
D. Statement-1 is false but Statement-2 is true.

Explanation

To solve the question, let's analyze the two statements provided:

**Statement 1**: Sodium meta aluminate on boiling with ammonium chloride produces a white gelatinous precipitate.

**Statement 2**: Aluminium hydroxide is formed which is not soluble in water.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- Sodium meta aluminate (NaAlO2) and ammonium chloride (NH4Cl) are the reactants involved in the reaction.

2. **Understand the Reaction**:
- When sodium meta aluminate is boiled with ammonium chloride, a chemical reaction occurs that leads to the formation of aluminum hydroxide and sodium hydroxide.
- The reaction can be represented as:
\[ \text{NaAlO}_2 + \text{NH}_4\text{Cl} \rightarrow \text{Al(OH)}_3 \downarrow + \text{NaOH} + \text{NH}_3 + \text{HCl} \]
- Here, Al(OH)3 is aluminum hydroxide, which precipitates out of the solution.

3. **Formation of Precipitate**:
- Aluminum hydroxide (Al(OH)3) is known to be a white gelatinous precipitate. This confirms Statement 1.

4. **Solubility of Aluminum Hydroxide**:
- It is important to note that aluminum hydroxide is not soluble in water. This confirms Statement 2.

5. **Conclusion**:
- Both statements are true. Statement 2 provides the correct explanation for Statement 1, as it describes the nature of the precipitate formed in the reaction.

### Final Answer:
Both Statement 1 and Statement 2 are true, and Statement 2 is the correct explanation of Statement 1.