1.

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

Answer»

A) `375^(@)C`
B) `375 K`
C) `273 K`
D) `102^(@)C`

Solution :`/_\S = 16 J mol E^(-1) K^(-1)`
`T_(b.p.)=(/_\H_("vapour"))/(/_\S_("vapour")=((6xx1000)/(16))=375k`


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