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The enthalpy of fusion of water is 1.435 kcal / mol. The molar entropy change for the melting of ice at 0°C is

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21.04 cal/mol/K

5.260 cal/mol/K

0.526 cal/mol/K

10.52 cal/mol/K

answer is C.

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Detailed Solution

${ΔH}_{fus} = 1 .435 kcal / mol$
${ΔS}_{fus} = \frac{{ΔH}_{fus}}{T_{fus}} = \frac{1 .435 \times 10^{3}}{273} = 5 .26 cal / mole / K$

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