If 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 solution in ${ }^{\circ} \mathrm{C}$ approximately is $\left(K_b=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$

Question

If 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 solution in ${ }^{\circ} \mathrm{C}$ approximately is $\left(K_b=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$, 101.4, 100.4, 102.4, 100.8

MARKS App · Accepted Answer

Given, Mass of non-volatile solute $X=9 \mathrm{~g}$ Mass of non-volatile solute $Y=\mathbf{l} \mathbf{g}$ Molar mass of $X=180 \mathrm{~g} \mathrm{~mol}^{-1}$ Molar mass of $Y=50 \mathrm{~g} \mathrm{~mol}^{-1}$ Mass of solvent (approx.) $=100$ $\begin{aligned} K_b & =0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \ \Delta T_b & =K_b \times M=K_b\left[M_{(X)}+M_{(Y)}\right] \ & =K_b\left[\frac{w_{(X)}}{M_{(X)}}+\frac{w_{(Y)}}{M_{(Y)}}\right] \times \frac{1000}{100} \ & =0.52\left[\frac{9}{180}+\frac{1}{50}\right] 10 \ & =0.52[0.05+0.02] 10 \ & =0.52[0.07] 10=0.36 \end{aligned}$ Thus, boiling point $=100+0.36$ $=100.36 \text { or } 100.4$ Hence, option (b) is the correct answer.

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