The enthalpy of fusion of water is`1.435kcal//mol`.The molar entropy change for the melting of ice at`0^(@)C`is

Options

A. 10.52 cal/mol K
B. 21.04 cal/mol K
C. 5.260 cal/mol K
D. 0.526 cal/mol K

Explanation

To find the molar entropy change for the melting of ice at \(0^\circ C\), we can use the formula for entropy change (\( \Delta S \)) during a phase transition, which is given by:

\[ \Delta S = \frac{\Delta H}{T} \]

where:
- \( \Delta H \) is the enthalpy change (in calories),
- \( T \) is the temperature (in Kelvin).

### Step-by-Step Solution:

1. **Identify the given data**:
- The enthalpy of fusion of water (\( \Delta H_f \)) is given as \( 1.435 \, \text{kcal/mol} \).
- The temperature is given as \( 0^\circ C \).

2. **Convert the enthalpy from kilocalories to calories**:
- Since \( 1 \, \text{kcal} = 1000 \, \text{cal} \), we convert:
\[ \Delta H_f = 1.435 \, \text{kcal/mol} \times 1000 \, \text{cal/kcal} = 1435 \, \text{cal/mol} \]

3. **Convert the temperature from Celsius to Kelvin**:
- The conversion from Celsius to Kelvin is done by adding 273.15:
\[ T = 0^\circ C + 273.15 = 273.15 \, \text{K} \]

4. **Substitute the values into the entropy change formula**:
- Now, we substitute \( \Delta H_f \) and \( T \) into the formula:
\[ \Delta S = \frac{\Delta H_f}{T} = \frac{1435 \, \text{cal/mol}}{273.15 \, \text{K}} \]

5. **Calculate the molar entropy change**:
- Performing the calculation:
\[ \Delta S \approx \frac{1435}{273.15} \approx 5.26 \, \text{cal/mol} \cdot \text{K} \]

### Final Answer:
The molar entropy change for the melting of ice at \(0^\circ C\) is approximately \(5.26 \, \text{cal/mol} \cdot \text{K}\).

---