The entropy change involved in the conversion of 1 mole of liquid water at $373 K$ to vapour will be:
Given: $△ H_{v a p} = 2.257 k J / g$

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

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

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

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

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Solution

The correct option is B $108.9 J K^{- 1} m o l^{- 1}$
The enthalpy change of vaporisation for 1 mole of water can be calculated as follows:
$△ H_{v a p}$ $=$ $2.257 k J / g \times 18 g / m o l = 40.6 k J m o l^{- 1}$
so,
$△ S_{v a p} = \frac{△ H_{v a p}}{T} = \frac{40.6 \times 10^{3} J m o l^{- 1}}{373 K} = 108.9 J K^{- 1} m o l^{- 1}$

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