Molar heat of vapourisation of a liquid is $6 k J . m o l^{- 1}$ . If the entropy change is $16 J m o l^{- 1} K^{- 1}$ , the boiling point of the liquid is:

375^{o} C

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375 K

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273 K

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102^{o} C

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Solution

The correct option is B $375 K$
Since boiling occurs at equilibrium conditions so $Δ G = 0$

Now,

$T = \frac{Δ H}{Δ S}$
$T = \frac{6000 J / m o l}{16 J / m o l / K}$
$T = 375 K$
Hence, the boiling point of the liquid is $375 K$ .

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