If molar heat of vaporization is 9698 cals `mol^(-1)` then entropy of vaporization of water at `100^(@)C` will be

Options

A. 20.0 cals `mol^(-1) k^(-1)`
B. 24.0 cals `mol^(-1) k^(-1)`
C. 26.0 cals `mol^(-1) k^(-1)`
D. 28.0 cals `mol^(-1) k^(-1)`

Answer

To find the entropy of vaporization of water at \(100^\circ C\), we can use the formula:

\[ \Delta S_{vap} = \frac{\Delta H_{vap}}{T} \]

Where:
- \(\Delta S_{vap}\) = entropy of vaporization
- \(\Delta H_{vap}\) = molar heat of vaporization
- \(T\) = temperature in Kelvin

### Step 1: Convert the temperature from Celsius to Kelvin
The temperature given is \(100^\circ C\). To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273 \]

So,

\[ T = 100 + 273 = 373 \, K \]

### Step 2: Use the given molar heat of vaporization
The molar heat of vaporization is given as \(9698 \, \text{cal/mol}\).

### Step 3: Calculate the entropy of vaporization
Now, we can substitute the values into the formula:

\[ \Delta S_{vap} = \frac{9698 \, \text{cal/mol}}{373 \, K} \]

### Step 4: Perform the calculation
Now, we perform the division:

\[ \Delta S_{vap} = \frac{9698}{373} \approx 26 \, \text{cal/mol/K} \]

### Final Result
Thus, the entropy of vaporization of water at \(100^\circ C\) is approximately:

\[ \Delta S_{vap} \approx 26 \, \text{cal/mol/K} \]