A substance 'X' (1.5 g) dissolved in 150 g | vedclass

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Question

(A) The elevation in boiling point is given by $\Delta T_{b} = K_{b} 	imes m$,where $m$ is the molality.$0.5 = 5.0 	imes m \implies m = 0.1 \ mol \

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) The elevation in boiling point is given by $\Delta T_{b} = K_{b} 	imes m$,where $m$ is the molality.$0.5 = 5.0 	imes m \implies m = 0.1 \ mol \ kg^{-1}$.Molality $m = \frac{n_{X}}{W_{Y} (	ext{in } kg)} = \frac{n_{X}}{0.150 \ kg} = 0.1 \implies n_{X} = 0.015 \ mol$.The relative lowering in vapour pressure is given by $\frac{P^{o} - P_{s}}{P^{o}} = X_{X} = \frac{n_{X}}{n_{X} + n_{Y}}$.$n_{Y} = \frac{150 \ g}{300 \ g \ mol^{-1}} = 0.5 \ mol$.Since the solution is dilute,$n_{X} + n_{Y} \approx n_{Y} = 0.5 \ mol$.Relative lowering $= \frac{0.015}{0.5} = 0.03 = 3 	imes 10^{-2}$.Thus,the value is $3$.

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