Latent heat of fusion of ice is `0.333 kJ g^(-1)` . The increase in entropy when 1 mole water melts at `0^(@)C` will be

Options

A. `21.96 kJ ^(-1) mol^(-1)`
B. `21.98 kcal K ^(-1) mol^(-1)`
C. `21.96 J K ^(-1) mol^(-1)`
D. `21.98 cal K ^(-1) mol^(-1)`

Answer

To find the increase in entropy when 1 mole of water melts at \(0^\circ C\), we can follow these steps:

### Step 1: Understand the given data
- Latent heat of fusion of ice = \(0.333 \, \text{kJ/g}\)
- We need to convert this to joules:
\[ 0.333 \, \text{kJ/g} = 333 \, \text{J/g} \]

### Step 2: Calculate the mass of 1 mole of ice
- The molecular weight of water (H₂O) is \(18 \, \text{g/mol}\).
- Therefore, the mass of 1 mole of ice is:
\[ \text{Mass} = 18 \, \text{g} \]

### Step 3: Convert the temperature to Kelvin
- The temperature given is \(0^\circ C\).
- To convert this to Kelvin:
\[ T = 0 + 273 = 273 \, \text{K} \]

### Step 4: Use the formula for change in entropy
- The change in entropy (\(\Delta S\)) can be calculated using the formula:
\[ \Delta S = \frac{n \cdot \Delta H}{T} \]
where:
- \(n\) = number of moles
- \(\Delta H\) = latent heat of fusion (in joules)
- \(T\) = temperature in Kelvin

### Step 5: Substitute the values into the formula
- Here, \(n = 1 \, \text{mol}\), \(\Delta H = 333 \, \text{J/g} \times 18 \, \text{g} = 5994 \, \text{J}\), and \(T = 273 \, \text{K}\).
- Now substituting these values:
\[ \Delta S = \frac{1 \cdot 5994}{273} \]

### Step 6: Calculate the change in entropy
- Performing the calculation:
\[ \Delta S = \frac{5994}{273} \approx 21.96 \, \text{J/K} \]

### Conclusion
- The increase in entropy when 1 mole of water melts at \(0^\circ C\) is approximately \(21.96 \, \text{J/K}\).