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Elevation in Boiling Point

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Question

Calculate the molal elevation constant, $K_{b}$ for water and the boiling of $0.1$ molal urea solution. Latent heat of vaporization of water is $9.72 k c a l m o l^{- 1}$ at $373.15 K$

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Solution

Given, $m = 0.1, △ H_{v_{p}} = 9.72 K c a l; T = 373.15 K; R = 2 c a l / m o l$
$K_{b} = \frac{R T_{b}^{2}}{1000 △ H_{v_{p}}}; △ H_{v_{p}} \sim c a l / m o l$
$= \frac{2 \times (373.15)^{2}}{1000 \times 9.72} g / m o l$
$= 28.65 g / m o l$

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