Question

An ideal solution was prepared by dissolving some amount of cane sugar ( non-volatile) in 0.9 moles of water. The solution was then cooled just below its freezing temperature (271 K) where some ice get seperated out. The remaining aqueous solution registered a vapour pressure of 700 torr at 373 K. Calculate the mass of ice seperated out, if the molar heat of fusion of water is $6kJ$. $K_f=1.86Kkgmo{l}^{-1}$

Answer 1

$\Delta T_f=K_f\times m$
$m=1.07molk{g}^{-1}$

$m=\frac{\text{moles of solute}}{\text{weight of solvent (kg)}}$

$n=1.075\times \frac{0.9\times 18}{1000}$
$n=1.74\times {10}^{-2}$
$X_{solute}=\frac{760-700}{760}=0.079$
where X is mole fraction

Total moles = $\frac{1.74\times {10}^{-2}}{0.079}=0.22$
Moles of solvent ($H_{2}O$) = 0.2026
Mass of ice separated out
= (0.9 - 0.2026) $\times$ 18 = 12 g