Question

The entropy change for the fusion of 1 mol of a solid which melts at $27^{\circ}C$ is (latent heat of fusion $= 600\mathrm{cal}\mathrm{mol}^{-1}$)

• A. 2 cal $\mathrm{K}^{-1}$
• B. 22.2 cal $\mathrm{K}^{-1}$
• C. 180 k cal K
• D. 0.5 cal $\mathrm{K}^{-1}$

Hint:

Entropy of fusion $\Delta S_{fus} = \frac{\Delta H_{fus}}{T_{fus}}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
$\Delta S_{fus} = \frac{\Delta H_{fus}}{T_{fus}}$.
Step 2: Detailed Explanation:
$\Delta S_{fus} = \frac{600}{27 + 273} = \frac{600}{300} = 2$ cal K$^{-1}$.
Step 3: Final Answer:
Entropy change is 2 cal K$^{-1}$.