If `Delta H_("vaporisation")` of substance `X (l)` (molar mass `: 30 g//mol`) is 300 J/g at it's boling point 300 K , then molar entropy change for reversible condensation process is

Options

A. `30 J//mol.K`
B. `-300 J//mol.K`
C. `-30 J//mol.K`
D. none of these

Explanation

To solve the problem, we need to calculate the molar entropy change for the reversible condensation process of substance X given the enthalpy of vaporization and its boiling point.

### Step-by-Step Solution:

1. **Identify the given data:**
- Molar mass of substance X, \( M = 30 \, \text{g/mol} \)
- Enthalpy of vaporization, \( \Delta H_{\text{vaporization}} = 300 \, \text{J/g} \)
- Boiling point, \( T = 300 \, \text{K} \)

2. **Convert the enthalpy of vaporization from J/g to J/mol:**
\[ \Delta H_{\text{vaporization}} \text{ (in J/mol)} = \Delta H_{\text{vaporization}} \text{ (in J/g)} \times M \text{ (in g/mol)} \]
\[ \Delta H_{\text{vaporization}} = 300 \, \text{J/g} \times 30 \, \text{g/mol} = 9000 \, \text{J/mol} \]

3. **Use the formula for molar entropy change:**
The molar entropy change (\( \Delta S \)) during a phase change at constant temperature can be calculated using the formula:
\[ \Delta S = \frac{\Delta H}{T} \]
Since we are considering condensation, we take the negative of the enthalpy of vaporization:
\[ \Delta S = -\frac{\Delta H_{\text{vaporization}}}{T} \]

4. **Substitute the values into the formula:**
\[ \Delta S = -\frac{9000 \, \text{J/mol}}{300 \, \text{K}} = -30 \, \text{J/mol/K} \]

5. **Conclusion:**
The molar entropy change for the reversible condensation process of substance X is:
\[ \Delta S = -30 \, \text{J/mol/K} \]

### Final Answer:
The molar entropy change for the reversible condensation process is \(-30 \, \text{J/mol/K}\).

---