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For a liquid, enthalpy of fusion is 1.43...

For a liquid, enthalpy of fusion is 1.435 kCal `mol^(-1)` and molar entropy change is 5.26 cal `mol^(-1) K^(-1)`. The freezing point of liquid in celcius will be ________

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The correct Answer is:
0
Tmp `= (Delta H "fusion")/(Delta S "fusion") = (1.435 xx 10^(3))/(5.26) = 272.81 k`
`= 0^(@)C`
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