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' $\mathrm{x}$ ' $\mathrm{g}$ of urea (molar mass $60 \mathrm{gmol}^{-1}$ ) is completely dissolved in ' $\mathrm{y}$ ' $\mathrm{g}$ of pure water and the solution boiled at $373.202 \mathrm{~K}$. If the boiling point of pure water at $1.013 \mathrm{bar}$ is $373.15 \mathrm{~K}$, then $\mathrm{x}: \mathrm{y}$ is $\left(\mathrm{K}_{\mathrm{b}}\left(\mathrm{H}_2 \mathrm{O}\right)=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$
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The correct answer is:
$6.0 \times 10^{-3}$
$\begin{aligned} & \text {} \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{T}_{\mathrm{b}}-\mathrm{T}_{\mathrm{b}}^{\circ}=\mathrm{K}_{\mathrm{b}} \times \text { molality of solute } \\ & =\mathrm{K}_{\mathrm{b}} \times\left[\frac{\mathrm{m}_{\text {solute }} \times 1000}{\mathrm{~m}_{\text {solvent }} \times \mathrm{M}_{\text {solute }}}\right] \\ & 373.202-373.150=0.52\left[\frac{\mathrm{x} \times 1000}{\mathrm{y} \times 60}\right] \\ & 0.052=\frac{\mathrm{x}}{\mathrm{y}} \times \frac{0.52 \times 1000}{60} \\ & \Rightarrow \frac{\mathrm{x}}{\mathrm{y}}=0.006=6.0 \times 10^{-3}\end{aligned}$
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