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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

5.260 cal/(mol K)

B

0.526 cal / (mol K)

C

10.52 cal / (mol K)

D

21.04 cal/(mol K)

Text Solution

Verified by Experts

The correct Answer is:
A
`Delta S= (DeltaH)/(T)=(1.435xx1000)/(273)=5.26 ("Cal")/("mol"xx K)`
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