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A solution of $6 \mathrm{~g}$ of solute in $100 \mathrm{~g}$ of water boils at $100.52^{\circ} \mathrm{C}$.…
A solution of $6 \mathrm{~g}$ of solute in $100 \mathrm{~g}$ of water boils at $100.52^{\circ} \mathrm{C}$. The molal elevation constant of water is $0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. What is molar mass of solute?
$60 \mathrm{~g} \mathrm{~mol}^{-1}$ $120 \mathrm{~g} \mathrm{~mol}^{-1}$ $90 \mathrm{~g} \mathrm{~mol}^{-1}$ $180 \mathrm{~g} \mathrm{~mol}^{-1}$ Solution
$\begin{aligned}
& \mathrm{W}_2=6 \mathrm{~g}, \mathrm{~W}_1=100 \mathrm{~g}, \mathrm{~K}_{\mathrm{b}}=0.52 \mathrm{Kg} \mathrm{mol}^{-1}, \mathrm{~T}_{\mathrm{b}} 100.52^{\circ} \mathrm{C} \\
& \therefore \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{T}_{\mathrm{b}}-\mathrm{T}_{\mathrm{b}}^{\circ}=(100.52+273)-(100+273) \\
& =0.52 \mathrm{k}
\end{aligned}$
Now, $\mathrm{M}_2=\frac{1000 \cdot \mathrm{K}_{\mathrm{b}} \cdot \mathrm{W}_2}{\Delta \mathrm{T}_{\mathrm{b}} \cdot \mathrm{W}_1}=\frac{1000 \mathrm{~g} \mathrm{~kg}^{-1} \times 0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \times 6 \mathrm{~g}}{0.52 \mathrm{~K} \times 100 \mathrm{~g}}$
$\therefore \mathrm{M}_2=60 \mathrm{~g} \mathrm{~mol}^{-1}$
Asked in: MHT CET 2021 (20 Sep Shift 2)
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