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 solid ( in cal `K^(-1))` at the same tempertre would be `:`

Options

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

Explanation

To find the entropy change for the fusion of one mole of the solid at its melting point, we can use the formula for entropy change during a phase transition:

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

where:
- \(\Delta S\) is the change in entropy,
- \(L_f\) is the latent heat of fusion,
- \(T\) is the temperature in Kelvin.

### Step-by-step Solution:

1. **Identify the given values:**
- Latent heat of fusion, \(L_f = 600 \, \text{cal mol}^{-1}\)
- Melting point (temperature), \(T = 300 \, \text{K}\)

2. **Substitute the values into the formula:**
\[ \Delta S = \frac{600 \, \text{cal mol}^{-1}}{300 \, \text{K}} \]

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 one mole of the solid at the melting point is \(2 \, \text{cal K}^{-1} \text{ mol}^{-1}\).

### Final Answer:
\[ \Delta S = 2 \, \text{cal K}^{-1} \text{ mol}^{-1} \]