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If $K_b$ denote molal elevation constant of water, then boiling point of an aqueous solution containing $36…
If $K_b$ denote molal elevation constant of water, then boiling point of an aqueous solution containing $36 \mathrm{~g}$ glucose $($ molar mass $=180)$ per $\mathrm{dm}^3$ is:
$\left(100+\mathrm{K}_{\mathrm{b}}\right)^{\circ} \mathrm{C}$ $\left(100+2 \mathrm{~K}_{\mathrm{b}}\right)^{\circ} \mathrm{C}$ $\left(100+\frac{\mathrm{K}_{\mathrm{b}}}{10}\right){ }^{\circ} \mathrm{C}$ $\left(100+\frac{2 \mathrm{~K}_{\mathrm{b}}}{10}\right){ }^{\circ} \mathrm{C}$ Solution
The aqueous solution contains $36 \mathrm{~g}$ glucose per $\mathrm{dm}^3$, so mass of solute $\mathrm{W}_2$ is $36 \mathrm{~g}$.
Assuming that the density of solution is $1 \mathrm{~g} / \mathrm{dm}^3$, the mass of solvent (water) is $1000 \mathrm{~g}$.
$\begin{aligned}
& \Delta \mathrm{T}_{\mathrm{b}}=\frac{1000 \mathrm{~K}_{\mathrm{b}} \mathrm{W}_2}{\mathrm{M}_2 \mathrm{~W}_1} \\
\Delta \mathrm{T}_{\mathrm{b}} & =\frac{1000 \mathrm{~g} \mathrm{~kg}^{-1} \times \mathrm{K}_{\mathrm{b}} \times 36 \mathrm{~g}}{180 \mathrm{~g} \times 1000 \mathrm{~g}} \\
\Delta \mathrm{T}_{\mathrm{b}} & =\frac{2 \mathrm{~K}_{\mathrm{b}}}{10} \\
\Delta \mathrm{T}_{\mathrm{b}} & =\mathrm{T}_{\mathrm{b}}-\mathrm{T}_{\mathrm{b}}^0 \\
& \mathrm{~T}_{\mathrm{b}}=\mathrm{T}_{\mathrm{b}}^0+\Delta \mathrm{T}_{\mathrm{b}} \\
\therefore \quad & \mathrm{T}_{\mathrm{b}}=\left(100+\frac{2 \mathrm{~K}_{\mathrm{b}}}{10}\right)^{\circ} \mathrm{C}
\end{aligned}$
Asked in: MHT CET 2023 (11 May Shift 1)
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