The enthalpy of vaporisation of per mole of ethanol (b.p. = ${79.5}^{o} C$ and $△ S$ = $109.8 J K^{- 1} m o l^{- 1})$

27.35 K J / m o l

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32.19 K J / m o l

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38.70 K J / m o l

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42.37 K J / m o l

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Solution

The correct option is C $38.70 K J / m o l$
Assuming the process to be at equilibrium.
$Δ G = 0$
$Δ H = T . Δ S$
$Δ H = 352.5 K \times 109.8 J K^{- 1} m o l^{- 1}$
$Δ H = 38.7 k J / m o l$

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Δ S = 109.8 J K^{- 1} {m o l}^{- 1}

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