The entropy change(cal/mol) for vaporisation of water if latent heat of vaporisation for water is 540 cal/g (Given $K_{b}$ for $H_{2} O = 0.51 K / m o l a l i t y)$ is (write answer to the nearest integer after dividing entropy change by 5).

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Solution

$K_{b} = \frac{R T_{B}^{2} M}{1000 Δ H_{V}} = \frac{R T_{b} M}{1000 Δ S_{V}} (∵ Δ S_{V} = \frac{Δ H_{V}}{T_{b}})$
$∴ Δ S_{V} = \frac{R T_{b} M}{1000 K_{b}} = \frac{2 \times 373 \times 18}{1000 \times 0.51}$
$= 26.33 c a l / m o l$
Answer is 5.

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