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Elevation in Boiling Point

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Question

A solution containing $2.5 \times 10^{- 3} k g$ of a solute dissolved in $75 \times 10^{- 3} k g$ of water boils at $373.535 K .$ The molar mass of the solute is _____ $g m o l^{- 1}$ . [nearest integer] (Given : $K_{b} (H_{2} O) = 0.52 K k g m o l^{- 1}$ and boiling point of water $= 373.15 K)$

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Solution

$W_{s o l u t e} = 2.5 \times 10^{- 3} k g$ = 2.5 g
$W_{s o l v e n t} = 75 \times 10^{- 3} k g$
Boiling point of aqueous solution $T_{2}$ = 373.535
Boiling point of water $T_{1}$ = 373.15K
So, elevation in B.P.;
$Δ T_{b} = 373.535 - 373.15$
$= 0.385 K$
$K_{b} (H_{2} O) = 0.52 K k g m o l^{- 1}$
$Δ T_{b} = \frac{K_{b} \times W_{s o l u t e}}{M_{s o l u t e} \times W_{s o l v e n t}}$

$M_{s o l u t e} = \frac{0.52 \times 2.5}{75 \times 10^{- 3} \times 0.385}$
$= 45.02$
$\approx 45$

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