The enthalpy of vaporization of chloroform is `29.4 kJmol^(-1)` at its normal boiling point of `61.7^(@)` C. What is the entropy of condensation of chlorofom at this temperature?

Options

A. `-57.3 `R
B. `-10.6` R
C. `-1.18` R
D. 10.6 R

Explanation

To find the entropy of condensation of chloroform at its normal boiling point, we can use the relationship between enthalpy of vaporization and entropy change. The entropy of condensation is equal in magnitude but opposite in sign to the entropy of vaporization.

### Step-by-Step Solution:

1. **Identify the given values:**
- Enthalpy of vaporization (ΔH_vap) = 29.4 kJ/mol
- Boiling point of chloroform (T_b) = 61.7 °C

2. **Convert the enthalpy of vaporization to Joules:**
\[ \Delta H_{vap} = 29.4 \, \text{kJ/mol} = 29.4 \times 10^3 \, \text{J/mol} = 29400 \, \text{J/mol} \]

3. **Convert the boiling point from Celsius to Kelvin:**
\[ T_b = 61.7 \, °C + 273.15 = 334.85 \, K \]

4. **Calculate the entropy of vaporization (ΔS_vap):**
The entropy change for vaporization can be calculated using the formula:
\[ \Delta S_{vap} = \frac{\Delta H_{vap}}{T_b} \]
Substituting the values:
\[ \Delta S_{vap} = \frac{29400 \, \text{J/mol}}{334.85 \, K} \approx 87.8 \, \text{J/(mol·K)} \]

5. **Calculate the entropy of condensation (ΔS_cond):**
The entropy of condensation is the negative of the entropy of vaporization:
\[ \Delta S_{cond} = -\Delta S_{vap} = -87.8 \, \text{J/(mol·K)} \]

### Final Answer:
The entropy of condensation of chloroform at its boiling point of 61.7 °C is approximately:
\[ \Delta S_{cond} \approx -87.8 \, \text{J/(mol·K)} \]