The melting point of a solid is 300K and its latent heat of fusion is 600 cal `mol^(-1)`. The entropy change for the fusion of 1 mole of the soli (in cal `K^(-1)`) at the same temperature would be:

Options

A. 200
B. 2
C. 0.2
D. 20

Answer

To find the entropy change (ΔS) for the fusion of 1 mole of the solid at its melting point, we can use the formula:

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

where:
- ΔS is the change in entropy,
- ΔH is the latent heat of fusion, and
- T is the temperature in Kelvin.

### Step-by-Step Solution:

1. **Identify the given values**:
- Melting point (T) = 300 K
- Latent heat of fusion (ΔH) = 600 cal/mol

2. **Substitute the values into the formula**:
\[ \Delta S = \frac{600 \, \text{cal/mol}}{300 \, \text{K}} \]

3. **Perform the calculation**:
\[ \Delta S = \frac{600}{300} = 2 \, \text{cal K}^{-1} \text{mol}^{-1} \]

4. **Conclusion**:
The entropy change for the fusion of 1 mole of the solid at 300 K is:
\[ \Delta S = 2 \, \text{cal K}^{-1} \text{mol}^{-1} \]

### Final Answer:
The entropy change for the fusion of 1 mole of the solid at 300 K is **2 cal K\(^{-1}\) mol\(^{-1}\)**.

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