The entropy change for vaporisation of a liquid is `109.3 JK^(-1)mol^(-1)`. The molar heat of vaporisation of that liquid is `40.77kJ mol^(-1)`. Calculate the boiling point of that liquid. 

Explanation

Entropy change for vapourisation, `DeltaS=109.3JK^(-1)mol^(-1)`
Mohal heat of vapourisation of water is `40.77KJmol^(-1)`
Form the entropy change, `DeltaS=q_(rev)/t`
Boiling ponit, `T=q_(rev)/(DeltaS) (or ) T=(40.77xx1000)/(109.3)=373K`.
Boiling of liquid is `100^(@)C`.