The heat of solution of one mole of `Na_2SO_4.10H_2O ` is -78.7 kJ ,and that of dehydration , `-81.6 kJ. ` The heat of solution (kJ) of anhydrous sodium sulphate is

Options

A. `+2.9 kJ`
B. `+160.3 kJ`
C. `-2.9 kJ`
D. `-160.3 kJ`

Answer

To find the heat of solution of anhydrous sodium sulfate (Na₂SO₄), we can use the given data about the heat of solution of hydrated sodium sulfate (Na₂SO₄·10H₂O) and the heat of dehydration.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Heat of solution of Na₂SO₄·10H₂O = -78.7 kJ
- Heat of dehydration (Na₂SO₄·10H₂O → Na₂SO₄ + 10H₂O) = -81.6 kJ

2. **Write the Reaction:**
- The process can be represented as:
\[ \text{Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O} \rightarrow \text{Na}_2\text{SO}_4 + 10\text{H}_2\text{O} \]
- The heat of dehydration is the energy change when the hydrated salt loses water.

3. **Apply Hess's Law:**
- According to Hess's law, the total heat change for a reaction is the sum of the heat changes for the individual steps.
- The heat of solution of anhydrous sodium sulfate can be calculated as:
\[ \text{Heat of solution of Na}_2\text{SO}_4 = \text{Heat of solution of Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O} + \text{Heat of dehydration} \]

4. **Substitute the Values:**
- Plugging in the values:
\[ \text{Heat of solution of Na}_2\text{SO}_4 = -78.7 \, \text{kJ} + (-81.6 \, \text{kJ}) \]
- This simplifies to:
\[ \text{Heat of solution of Na}_2\text{SO}_4 = -78.7 \, \text{kJ} + 81.6 \, \text{kJ} \]
- Note that the heat of dehydration is negative, so it adds to the overall heat of solution.

5. **Calculate the Result:**
\[ \text{Heat of solution of Na}_2\text{SO}_4 = -78.7 + 81.6 = 2.9 \, \text{kJ} \]

### Final Answer:
The heat of solution of anhydrous sodium sulfate is **2.9 kJ**.