The molal elevation constant for water is 0.56 K kg $m o l^{- 1}$ . Calculate the boiling point of solution made by dissolving 6.0 g of urea $(N H_{2} C O N H_{2})$ in 200 g of water.

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Solution

$△ = \frac{1000 K_{b} \times w}{m \times W}$
We know this elevation in boiling point is directly proportional to the molality of the solution.
ΔTbα molality
To remove the proportionality sign, we should bring a constant. That constant is the molal elevation constant.
$Δ T_{b} = K_{b} m$
$m = \frac{m a s s o f s o l u t e}{m a s s o f s o l v e n t (g)} \times 1000$

Given, $K_{b} = 0.56 K k g m o l^{- 1}, w = 6.0 g, W = 200 g$

$△ T_{b} = \frac{1000 \times 0.56 \times 6.0}{200 \times 60} = {0.28}^{o} C$

Thus, the boiling point of solution $= b o i l i n g p o i n t .$ of water $+ △ T_{b}$
$= (100^{o} C + {0.28}^{o} C) = {100.28}^{o} C$

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