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 concept of latent heat of fusion
The latent heat of fusion is the amount of energy required to change a substance from solid to liquid at its melting point without changing its temperature. For ice, this is given as \(0.333 \, \text{kJ/g}\).

### Step 2: Convert latent heat of fusion to Joules
Since we need to work in consistent units, we convert the latent heat from kilojoules to joules:
\[ 0.333 \, \text{kJ/g} = 0.333 \times 10^3 \, \text{J/g} = 333 \, \text{J/g} \]

### Step 3: Calculate the latent heat for 1 mole of water
The molar mass of water (H₂O) is \(18 \, \text{g/mol}\). Therefore, the total latent heat for 1 mole of water is:
\[ \text{Latent heat for 1 mole} = 333 \, \text{J/g} \times 18 \, \text{g} = 5994 \, \text{J} \]

### Step 4: Convert the temperature to Kelvin
The melting point of ice is \(0^\circ C\), which in Kelvin is:
\[ T = 0 + 273 = 273 \, \text{K} \]

### Step 5: Calculate the increase in entropy
The change in entropy (\(\Delta S\)) during the phase change can be calculated using the formula:
\[ \Delta S = \frac{\Delta H}{T} \]
Where \(\Delta H\) is the latent heat and \(T\) is the temperature in Kelvin. Substituting the values we have:
\[ \Delta S = \frac{5994 \, \text{J}}{273 \, \text{K}} \approx 21.96 \, \text{J/K} \]

### Final Result
The increase in entropy when 1 mole of water melts at \(0^\circ C\) is approximately:
\[ \Delta S \approx 21.96 \, \text{J/K} \]