At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If = 0.52, the boiling point of this solution will be [2016]

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Solution

Q.
At 100°C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If $K_{b}$ = 0.52, the boiling point of this solution will be [2016]

1 102 °C

2 103 °C

3 101 °C

4 100 °C

Ans.
(3)

Given: $W_{B} = 6.5$ $g, W_{A} = 100$ $g$

$p_{s} = 732$ $mm, K_{b} = 0.52, T_{b}^{0} = 100 ° C, p^{0} = 760$ $mm$

$\frac{p^{o} - p_{s}}{p^{o}} = \frac{n_{2}}{n_{1}} \Rightarrow \frac{760 - 732}{760} = \frac{n_{2}}{100 / 18}$

$\Rightarrow n_{2} = \frac{28 \times 100}{760 \times 18} = 0.2046$ $mol$

$Δ T_{b} = K_{b} \times m$

$T_{b} - T_{b}^{o} = K_{b} \times \frac{n_{2} \times 1000}{W_{A} (g)}$

$T_{b} - 100 ° C = \frac{0.52 \times 0.2046 \times 1000}{100} = 1.06$

$T_{b} = 100 + 1.06 = 101.06 ° C$

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