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Question
The latent heat of vaporization of a liquid of molar mass, 80 g/mol and boiling point, 127°C is 8 kcal/mol. The ebullioscopic constant of the liquid is
Options
A.3.2 K-kg/mol
B.0.04 K-kg/mol
C.0.32 K-kg/mol
D.0.52 K-kg/mol
Solution
$K_{b} = \frac{RT^{2}.M}{1000.\Delta H_{\text{vap}}} = \frac{2 \times (400)^{2} \times 80}{1000 \times 8 \times 10^{3}} = 3.2\text{ K.Kg/mol}$
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