$10 g$ of non-volatile solute is dissolved in $180 g$ of $H_{2} O$ resulting in lowering of vapour pressure by $0.5$ %. Determine the boiling point of a solution if $K_{b}$ of water is 0.52 K kg mol $^{- 1}$ .

{100.01}^{\circ} C

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{100.15}^{\circ} C

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{100.23}^{\circ} C

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{100.32}^{\circ} C

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Solution

The correct option is C ${100.15}^{\circ} C$
$Δ p = p^{\circ} x_{2}$
$0.50 = 100 \times \frac{n_{2}}{n_{1}} = 100 \times \frac{w \times M}{m \times W}$

Here,
$w =$ mass of solute = 10g
$W =$ mass of solvent = 180g
$M =$ molar mass of solvent =18g/mol
$m =$ molar mass of solute =?

Substituting the value we get, $m = 200 g / m o l$
$Δ T_{b} = k_{b} m = K_{b} \frac{w}{W} \times \frac{1000}{m}$
$= 0.52 \times \frac{10}{180} \times \frac{1000}{200} = 0.14 \approx 0.15$
$T_{b} = {100.15}^{\circ} C$ .

Hence, the correct option is $B$

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10 g of non-volatile solute is dissolved in 180 g of H2O resulting in lowering of vapour pressure by 0.5%. Determine the boiling point of a solution if Kb of water is 0.52 K kg mol−1.

$10 g$ of non-volatile solute is dissolved in $180 g$ of $H_{2} O$ resulting in lowering of vapour pressure by $0.5$ %. Determine the boiling point of a solution if $K_{b}$ of water is 0.52 K kg mol $^{- 1}$ .