An aqueous solution of a non-volatile solute boils at $100.17^{\circ} \mathrm{C}$. The temperature at which…
An aqueous solution of a non-volatile solute boils at $100.17^{\circ} \mathrm{C}$. The temperature at which this solution will freeze $\left(ight.$ in $\left.{ }^{\circ} \mathrm{C}ight)$ is
$\begin{aligned} & \left(\mathrm{K}_{\mathrm{b}}\left(\mathrm{H}_2 \mathrm{O}ight)=0.512^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1},ight. \\ & \left.\qquad \mathrm{K}_{\mathrm{f}}\left(\mathrm{H}_2 \mathrm{O}ight)=1.86^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}ight)\end{aligned}$
$-0.62$
$-0.512$
$-1.24$
$-1.86$
Solution
Using formula,
$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{k}_{\mathrm{b}}$ molality
$(100.17-100)^{\circ} \mathrm{C}=0.512^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1} \times$ molality molality $=\frac{0.17}{0.512} \mathrm{~m}$
Now, depression in freezing point is given by, $\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{k}_{\mathrm{f}}$ molality
$\Delta \mathrm{T}_{\mathrm{f}}=1.86^{\circ} \mathrm{C} \mathrm{kg} \times \frac{0.17}{0.512} \mathrm{~m}=0.62^{\circ} \mathrm{C}$
Thus, the freezing point of the solution $=0^{\circ} \mathrm{C}-0.62^{\circ} \mathrm{C}=-0.62^{\circ} \mathrm{C}$.
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