On dissolving 10.8 g glucose $(m . w t . = 180)$ in 240 g of water, its boiling point increases by ${0.13}^{o} C$ . Calculate the molecular elevation constant of water.

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Solution

$△ T = \frac{100 K^{'} \times w}{W \times m}$
or $K^{'} = \frac{△ T \times W \times m}{100 \times w}$
Given, $△ T = {0.13}^{o} C, W = 240 g, m = 180$ and $w = 10.8 g$
$K^{'} = \frac{0.13 \times 240 \times 180}{100 \times 10.8} = 5.2$

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