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The enthalpy of fusion of water is 1.435...

The enthalpy of fusion of water is `1.435 kcal//"mole"`. The molar entropy change for melting of ice at `0^(@)C` is

A

10.52 cal/(mol K)

B

21.04 cal/(mol K)

C

5.260 cal/(mol K)

D

0.526 cal/(mol K)

Text Solution

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The correct Answer is:
D
`DeltaH_f = (1.435 ` kcal / mol
`Delta s = (DeltaH_f)/(T_f)= (1.435 xx 10^3)/(273) = 5.26` cal / mol K.
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