Molar heat of vaporisation of a liquid is $6\ kJmol^{- 1}$. If the entropy change is $16\ J\ mol^{- 1}K^{- 1}$, the boiling point of the liquid is

Question

Dr. Abubacker · Accepted Answer

(c)
$\mathrm{\Delta}S = 16$ J $mol^{- 1}K^{- 1}$
$T_{b.p} = \frac{\mathrm{\Delta}H_{vapour}}{\mathrm{\Delta}S_{vapour}} = \frac{6 	imes 1000}{16}$
$= 375K$

Q. Molar heat of vaporisation of a liquid is $6\ kJmol^{- 1}$. If the entropy change is $16\ J\ mol^{- 1}K^{- 1}$, the boiling point of the liquid is

Explanation

Reviewed by Dr. Abubacker

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