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A solution of nonvolatile solute is obtained by dissolving $1.5 \mathrm{~g}$ in $30 \mathrm{~g}$ solvent has boiling point elevation $0.65 \mathrm{~K}$. Calculate the molal elevation constant if molar mass of solute is $150 \mathrm{~g} \mathrm{~mol}^{-1}$.
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The correct answer is:
$1.95 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
$\begin{aligned} \mathrm{M}_2=\frac{1000 . \mathrm{K}_{\mathrm{b}} \mathrm{W}_2}{\Delta \mathrm{T}_{\mathrm{b}} \mathrm{W}_1} & \\ \mathrm{~K}_{\mathrm{b}}=\frac{\mathrm{M}_2^2 \times \Delta \mathrm{T}_{\mathrm{b}} \times \mathrm{W}_1}{1000 \times \mathrm{W}_2} & =\frac{150 \times 0.65 \times 30}{1000 \times 1.5} \\ & =1.95 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\end{aligned}$
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