If aqueous solution contains \(9 \%\) and \(1 \%\) \((w / w)\) of two non-volatile non-electrolytes \(X\)…
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
Solution
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.