Entropy of vaporisation of water at `100^(@)C`, if molar heat of vaporisation is `9710 cal mol^(-1)` will be

Options

A. `20 cal mol^(-1)K^(-1)`
B. `26 cal mol^(-1)K^(-1)`
C. `24 cal mol^(-1)K^(-1)`
D. `28 cal mol^(-1)K^(-1)`

Explanation

To calculate the entropy of vaporization of water at \(100^\circ C\), given that the molar heat of vaporization is \(9710 \, \text{cal mol}^{-1}\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Formula for Entropy Change**:
The entropy change (\(\Delta S\)) during a phase change can be calculated using the formula:
\[ \Delta S = \frac{Q_{\text{rev}}}{T} \]
where \(Q_{\text{rev}}\) is the heat absorbed or released reversibly and \(T\) is the absolute temperature in Kelvin.

2. **Convert Temperature to Kelvin**:
The given temperature is \(100^\circ C\). To convert this to Kelvin, use the formula:
\[ T(K) = T(°C) + 273 \]
Therefore,
\[ T = 100 + 273 = 373 \, K \]

3. **Substitute the Values**:
The heat of vaporization (\(Q_{\text{rev}}\)) is given as \(9710 \, \text{cal mol}^{-1}\). Now, substitute the values into the entropy formula:
\[ \Delta S = \frac{9710 \, \text{cal mol}^{-1}}{373 \, K} \]

4. **Calculate the Entropy Change**:
Now perform the division:
\[ \Delta S = \frac{9710}{373} \approx 26.032 \, \text{cal K}^{-1} \text{mol}^{-1} \]

5. **Final Result**:
The calculated entropy of vaporization of water at \(100^\circ C\) is approximately:
\[ \Delta S \approx 26.032 \, \text{cal K}^{-1} \text{mol}^{-1} \]

### Conclusion:
Since the options provided are 20, 26, 24, and 28, we can round \(26.032\) to \(26\). Thus, the correct answer is:
\[ \Delta S \approx 26 \, \text{cal K}^{-1} \text{mol}^{-1} \]