Back to Directory
NEET CHEMISTRYEasy

The enthalpy of fusion of water is 1.435 kcal/mol1.435 \text{ kcal/mol}. The molar entropy change for the melting of ice at 0C0^\circ\text{C} is:

A

10.52 cal/(mol K)10.52 \text{ cal/(mol K)}

B

21.04 cal/(mol K)21.04 \text{ cal/(mol K)}

C

5.260 cal/(mol K)5.260 \text{ cal/(mol K)}

D

0.526 cal/(mol K)0.526 \text{ cal/(mol K)}

Step-by-Step Solution

For a phase transition like melting of ice, the entropy change (ΔfusS\Delta_{fus} S) is given by the formula: ΔfusS=ΔfusHTf\Delta_{fus} S = \frac{\Delta_{fus} H}{T_f} where ΔfusH\Delta_{fus} H is the enthalpy of fusion and TfT_f is the freezing/melting point in Kelvin. Given: ΔfusH=1.435 kcal/mol=1435 cal/mol\Delta_{fus} H = 1.435 \text{ kcal/mol} = 1435 \text{ cal/mol} Tf=0C=273 KT_f = 0^\circ\text{C} = 273 \text{ K} Substituting the values: ΔfusS=1435 cal/mol273 K5.256 cal/(mol K)\Delta_{fus} S = \frac{1435 \text{ cal/mol}}{273 \text{ K}} \approx 5.256 \text{ cal/(mol K)} This value is closest to 5.260 cal/(mol K)5.260 \text{ cal/(mol K)}.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started
Solved: CHEMISTRY Question for NEET | Sushrut