If an aqueous solution contains 9 % and 1 % ( | vedclass

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(B) Mass of solute $X = 9 \ g$,Mass of solute $Y = 1 \ g$. Mass of solvent $= 100 - (9 + 1) = 90 \ g$. Moles of $X = \frac{9}{180} = 0.05 \ mol$. Mole

Vedclass · Accepted Answer

$100.4$

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(B) Mass of solute $X = 9 \ g$,Mass of solute $Y = 1 \ g$. Mass of solvent $= 100 - (9 + 1) = 90 \ g$. Moles of $X = \frac{9}{180} = 0.05 \ mol$. Moles of $Y = \frac{1}{50} = 0.02 \ mol$. Total moles of solute $= 0.05 + 0.02 = 0.07 \ mol$. Molality $(m) = \frac{	ext{Total moles of solute}}{	ext{Mass of solvent in kg}} = \frac{0.07}{0.090} \approx 0.777 \ mol \ kg^{-1}$. Elevation in boiling point $\Delta T_b = K_b 	imes m = 0.52 	imes 0.777 \approx 0.404 \ { }^{\circ} C$. Boiling point of solution $= 100 + 0.404 = 100.404 \ { }^{\circ} C \approx 100.4 \ { }^{\circ} C$. Thus,option $(B)$ is the correct answer.

If an aqueous solution contains $9 \%$ and $1 \%$ $(w/w)$ of two non-volatile non-electrolytes $X$ (molecular weight $180$) and $Y$ (molecular weight $50$) respectively,the boiling point of the solution in ${ }^{\circ} C$ is approximately $(K_b = 0.52 \ K \ kg \ mol^{-1})$

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