Molar heat of vapourisation of a liquid is 4.8 kJ "mol"^(-1) . If the entropy change is 16 J mol^(-1) K^(-1) , the boiling point of the liquid is

Molar heat of vapourisation of a liquid is 4.8 kJ "mol"^(-1) . If the

1.	Molar heat of vapourisation of a liquid is 4.8 kJ "mol"^(-1) . If the entropy change is 16 J mol^(-1) K^(-1) , the boiling point of the liquid is
Answer» 323 K `27^@C` `164K` 0.3 K Solution :`DeltaS_V=(DeltaH_V)/T_b` `T_b=(DeltaH_V)/(DeltaS_V)="4800 J MOL"^(-1) /("16 J mol"^(-1) K^(-1))=300 K = 27^@C`