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NEET CHEMISTRYThermodynamicsEasy

Question

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)}.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThermodynamicsenthalpyfusionkcalmolentropychange

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