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Molar heat of vapourisation of a liquid ...

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

A

323 K

B

`27^@C`

C

`164K`

D

0.3 K

Text Solution

Verified by Experts

The correct Answer is:
B
`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`
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