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For liquid enthalpy of fusion is 1.435 k...

For liquid enthalpy of fusion is `1.435 kcal mol^(-1)` and molar entropy change is `5.26 cal mol^(-1) K^(-1)`. The melting point of the liquid is

A

`0^(@)C`

B

`-273^(@)C`

C

`173K`

D

`100^(@)C`

Text Solution

Verified by Experts

The correct Answer is:
A
`DeltaS_(fusion)=(DeltaH_(fusion))/(T_(mp))`
`5.26=(1.435xx1000)/(T_(mp))`
`T_(mp)=(1435)/(5.26)=273K` i.e., `0^(@)C`
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