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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

`173 K`

D

`100^(@)C`

Text Solution

Verified by Experts

`Delta_(fus)S^(Theta) = (Delta_(fus)H^(Theta))/(T_(mp))`
`5.26 = (1.435 xx 1000)/(T_(mp))`
`T_(mp) = (1435)/(5.26) = 273 K, i.e., 0^(@)C`
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