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For water, $∆ H_{Vap} = 41 KJ {mol}^{- 1}$ ,then the molar entropy of vapourisation at 1 atm pressure is approximately

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$410 J K^{- 1} {mol}^{- 1}$

$110 {JK}^{- 1} {mol}^{- 1}$

$41 {JK}^{- 1} {mol}^{- 1}$

$11 {JK}^{- 1} {mol}^{- 1}$

answer is B.

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Detailed Solution

$∆ H_{vap} = 41 kJ / mol ∆ S_{vap} = ? H_{2} O_{(l)} ⇌ H_{2} O_{(g)} T_{b} = 100^{o} C = 373 K ∆ S_{vap} = \frac{∆ H_{V}}{T_{b}} = \frac{41000}{373} = 110 J / K / mol$

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