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Molar heat of vaporization of a liquid i...

Molar heat of vaporization of a liquid is 6 `KJ mol^(-1)`. If its entropy change is `16 JK^(-1)mol^(-1)`, then boiling point of the liquid is

A

`375^(@)C`

B

375 K

C

273 K

D

`102^(@)C`

Text Solution

Verified by Experts

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