Question

An ideal solution was prepared by dissolving some amount of cabe sugar (non-volatile) in $0.9moles$ 0f water.
The solution was then cooled just below its freezing temperature $\left(271K\right)$, where some ice get separated out.
The remaining aqueous solution registered a vapour pressure of $700torr$ at $373K$.
Calculate the mass of ice separated out, if the molar heat of fusion of water is $96kJ$.(integer value only)

Answer 1

Initial moles of $H_{2}O=0.9$
$\Delta T_f=6kJ$
$K_f=\frac{R{T}_f^{2}M}{1000\Delta H_f}=\frac{8.314\times (273{)}^2\times 18}{1000\times 6000}=1.86$
$\Delta T_f=k_f\times m$
$m=\frac{2}{1.86}=1.075$
So in $1000g{H}_{2}O\to 1.075mole$ solute
in $1g{H}_{2}O\to \frac{1.075}{1000}mole$ solute
in $0.9\times 18g{H}_{2}O\to \frac{1.075}{1000}\times 0.9\times 18mole$ solute
mole of solute $\left(n\right)=\mathrm{0.0174.15}$
$\frac{P^{\circ }-P_s}{P_s}=\frac{n}{N}=\frac{760-700}{0.0851}=0.0857$
moles of $H_{2}O\left(N\right)=\frac{0.017415}{0.0857}=0.2032$
moles of ice separated out $=0.9-0.2030=0.6968$
mass of ice separated out$=0.6968\times 18=12.54g$