1.

Molar heat of vaporization of a liquid is 6kJ mol^(-1). If the entropy change is 16 J mol^(-1) K^(-1), the boiling point of the liquid is

Answer»

`375^(@)C`
375 K
273 K
`102^(@)C`

Solution :`DeltaS=16 J mol^(-1) K^(-1)`
`T_(b.p.)=(DeltaH_("vapour"))/(DeltaS_("vapour"))=(6xx1000)/(16)=375 K`.


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