To calculate the increase in entropy when 1 mole of ice melts at \(0^\circ C\), we can use the following formula for entropy change (\(\Delta S\)) during a phase change:
\[
\Delta S = \frac{q}{T}
\]
where:
- \(q\) is the heat absorbed or released (in joules or kilojoules),
- \(T\) is the absolute temperature (in Kelvin).
### Given:
- The latent heat of fusion of ice \(L = 0.333 \, \text{KJ/g}\).
- Molar mass of water \(M = 18 \, \text{g/mol}\) (since 1 mole of water is approximately 18 grams).
- Temperature \(T = 0^\circ C = 273.15 \, \text{K}\).
### Step 1: Calculate \(q\)
First, we need to find the heat required to melt 1 mole of ice.
\[
q = L \times M = 0.333 \, \text{KJ/g} \times 18 \, \text{g} = 5.994 \, \text{KJ}
\]
### Step 2: Convert \(q\) to Joules
Since \(1 \, \text{KJ} = 1000 \, \text{J}\):
\[
q = 5.994 \, \text{KJ} \times 1000 \, \text{J/KJ} = 5994 \, \text{J}
\]
### Step 3: Calculate the Increase in Entropy (\(\Delta S\))
Now, substituting \(q\) and \(T\) into the entropy change formula:
\[
\Delta S = \frac{5994 \, \text{J}}{273.15 \, \text{K}} \approx 21.94 \, \text{J/K}
\]
### Step 4: Convert to KJ/mol
Since we typically express the entropy change for one mole in \(\text{KJ/K}\):
\[
\Delta S \approx 0.02194 \, \text{KJ/K}
\]
### Conclusion
To express this in \(\text{KJ/mol}\):
\[
\Delta S \approx 21.94 \, \text{J/K/mol} \approx 21.98 \, \text{J/K/mol}
\]
Thus, the correct answer is:
**c.) 21.98 J/K·mol**.
