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

Answer

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, given that the entropy of vaporization of benzene is 85 J K⁻¹ mol⁻¹.

### Step-by-Step Solution:

1. **Calculate the number of moles of benzene (C₆H₆)**:
- The molar mass of benzene (C₆H₆) is approximately 78 g/mol.
- Use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{117 \text{ g}}{78 \text{ g/mol}} = 1.5 \text{ moles} \]

2. **Calculate the total entropy change for the system**:
- The entropy of vaporization is given as 85 J K⁻¹ mol⁻¹.
- For 1.5 moles, the total entropy change (ΔS_system) is:
\[ \Delta S_{\text{system}} = \text{Number of moles} \times \text{Entropy of vaporization} = 1.5 \text{ moles} \times 85 \text{ J K}^{-1} \text{ mol}^{-1} = 127.5 \text{ J K}^{-1} \]

3. **Determine the entropy change in the surroundings**:
- At the boiling point, the system and surroundings are in equilibrium, so:
\[ \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} = 0 \]
- Therefore, we can express the entropy change in the surroundings as:
\[ \Delta S_{\text{surroundings}} = -\Delta S_{\text{system}} = -127.5 \text{ J K}^{-1} \]

### Final Answer:
The entropy change in the surroundings when 117 g of benzene vaporizes at its normal boiling point is **-127.5 J K⁻¹**.

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