The entropy if vaporisation of benzene is `85JK^(-1) mol^(-1)`. When 117g benzene vaporizes at its's normal boiling point, the entropy change in surrounding is:

Options

A. `-85 JK^(-1)`
B. `-85xx1.5 JK^(-1)`
C. `85 xx1.5 JK^(-1)`
D. None of these

Explanation

To solve the problem, we need to calculate the entropy change in the surroundings when 117 g of benzene vaporizes at its normal boiling point.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Entropy of vaporization of benzene, \( \Delta S_{vap} = 85 \, \text{J K}^{-1} \text{mol}^{-1} \)
- Mass of benzene, \( m = 117 \, \text{g} \)
- Molar mass of benzene, \( M = 78 \, \text{g mol}^{-1} \)

2. **Calculate the Number of Moles of Benzene**:
\[ n = \frac{m}{M} = \frac{117 \, \text{g}}{78 \, \text{g mol}^{-1}} = 1.5 \, \text{mol} \]

3. **Calculate the Entropy Change of the System**:
The entropy change of the system when 1.5 moles of benzene vaporizes is:
\[ \Delta S_{system} = n \times \Delta S_{vap} = 1.5 \, \text{mol} \times 85 \, \text{J K}^{-1} \text{mol}^{-1} = 127.5 \, \text{J K}^{-1} \]

4. **Calculate the Entropy Change of the Surroundings**:
For a reversible process, the entropy change of the surroundings is equal to the negative of the entropy change of the system:
\[ \Delta S_{surroundings} = -\Delta S_{system} = -127.5 \, \text{J K}^{-1} \]

### Final Answer:
The entropy change in the surroundings when 117 g of benzene vaporizes at its normal boiling point is:
\[ \Delta S_{surroundings} = -127.5 \, \text{J K}^{-1} \]