Entropy change involved in conversion of one mole of liquid water at $373 K$ to vapour at the same temperature (latent heat of vaporisation of water $= 2.257 k J g^{- 1})$ .

30.7 J K^{- 1} m o l^{- 1}

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60.3 J K^{- 1} m o l^{- 1}

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9.8 J K^{- 1} m o l^{- 1}

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108.9 J K^{- 1} m o l^{- 1}

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Solution

The correct option is D $108.9 J K^{- 1} m o l^{- 1}$
$△ H_{v a p} = 2.257 k J g^{- 1} = (2.257 \times 18) k J m o l^{- 1} = 40.626 k J m o l^{- 1} △ S_{v a p} = \frac{40.626}{373} = 0.1089 k J K^{- 1} m o l^{- 1}$

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Similar questions

373 K

△ H_{v a p} = 2.257 k J / g

△_{v a p} H = 40.98 k J m o l^{- 1} .

H_{2} O_{(l)} \to H_{2} O_{(g)}

100^{o} C

40.63 k J m o l^{- 1}

373 K

= 2.257 k J g^{- 1})

100^{\circ} C

40.63 {kJ mol}^{- 1}

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