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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

Verified by Experts

The correct Answer is:
C
`Delta_(f)H^(@) =1.435 kcal//mol=1435 cal//mol`
`DeltaS=Delta_(f)^(@)H=(1435)/(273)=5.256 cal K^(-1)mol^(-1)`
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