A compound X of sodium forms a white powder .It is a consituent of baking powder and used in some antacids .when heated it gives a compound Y which is anhydrous and absorbs water to become hydrated salt .When this salt kept open in air looses water molecules in a process called efflorescence .When dissolved in water it forms a strong base and a weak acid Z.
What is Y ?

Options

A. `NaHCO_(3)`
B. `Na_(2)CO_(3)`
C. `Na_(2)CO_(3).10H_(2)O`
D. `NaCl`

Explanation

To solve the question step by step, we need to analyze the information provided about the compound X and its transformations:

### Step 1: Identify Compound X
- The question states that compound X is a sodium compound that forms a white powder, is a constituent of baking powder, and is used in some antacids.
- The chemical formula for baking powder is sodium bicarbonate, which is NaHCO₃.

**Hint:** Look for common sodium compounds used in baking and antacids.

### Step 2: Determine the Transformation of X
- When heated, compound X (NaHCO₃) decomposes to form compound Y.
- The decomposition reaction is:
\[ 2 \text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O} \]
- From this reaction, we see that compound Y is sodium carbonate, which is Na₂CO₃.

**Hint:** Consider what happens to baking soda when it is heated.

### Step 3: Identify the Properties of Compound Y
- The question mentions that compound Y is anhydrous and can absorb water to become a hydrated salt.
- Anhydrous sodium carbonate (Na₂CO₃) can absorb water to form hydrated sodium carbonate (Na₂CO₃·10H₂O).

**Hint:** Think about how certain salts can absorb moisture from the air.

### Step 4: Understand Efflorescence
- The question states that when the hydrated salt is kept open in air, it loses water molecules in a process called efflorescence.
- This is a characteristic behavior of hydrated salts, where they lose water to the environment.

**Hint:** Recall the term for the loss of water from hydrated salts.

### Step 5: Analyze the Dissolution of Y
- When dissolved in water, compound Y (Na₂CO₃) forms a strong base and a weak acid.
- The dissolution reaction can be represented as:
\[ \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} \rightarrow 2 \text{NaOH} + \text{H}_2\text{CO}_3 \]
- Here, NaOH is the strong base, and H₂CO₃ (carbonic acid) is the weak acid.

**Hint:** Identify the products formed when sodium carbonate is dissolved in water.

### Conclusion
- Based on the analysis, we conclude that the compound Y is anhydrous sodium carbonate, which has the formula Na₂CO₃.

**Final Answer:** Y is Na₂CO₃ (anhydrous sodium carbonate).