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Ethanol boils at $78.4^{\circ}C$, the enthalpy of vaporization of ethanol is $42.4\: kJ/mol$. Calculate entropy of vaporization of ethanol.

$\begin {array} {1 1} (a)\;540.816\: J/K-mol & \quad (b)\;0.1206\: J/K-mol \\ (c)\; 120.66 \: J/K-mol & \quad (d)\;-540.816 \: J/K-mol \end {array}$


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$ \Delta S_{vap} =\large\frac{ \Delta H_{vap}}{T_b}$
$=\large\frac{ 42.4 \times 1000\: J/mol}{ 351.4K}$$ = 120.66\: J/K-mol$
Ans : (c)
answered Mar 13, 2014 by thanvigandhi_1

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