Question

The molar volume of liquid benzene (density$=0.877gm{L}^{-1}$) increases by a factor of $2750$ as it vaporises at ${20}^{\circ }C$ while in equilibrium with liquid benzene. At ${27}^{\circ }C$ when a non-volatile solute (that does not dissociate) is dissolved in $54.6c{m}^3$ of the benzene vapour pressure of this solution, is found to be $98.88mmHg$.

Calculate the freezing point of the solution.
Given: Enthalpy of vaporisation of benzene $\left(l\right)=394.57J/g$
Molal depression constant for benzene$=5.12Kkg.mo{l}^{-1}$
Freezing point of benzene$=278.5K$.

Answer 1

At ${20}^{\circ }C$:
For$C_6{H}_6\to V=\frac{78}{0.877}\times 2750mL$
$PV=1\times 0.0821\times 293$
$P=74.74mmHg$
If vapour pressure of benzene at ${27}^{\circ }C$ is $P_1$ then
$ln\frac{P_1}{P}=\frac{\Delta H_V}{R}\left[\frac{1}{T}-\frac{1}{T_1}\right]$
$ln\frac{P_1}{74.74}=\frac{394.57\times 78}{8.314}\left[\frac{1}{293}-\frac{1}{300}\right]$
$P_1=100.364mmHg$
$m=\frac{P^{\circ }-P_s}{P_s}\times \frac{1000}{M_{solvent}}$
$m=\frac{100.364-98.88}{98.88}\times \frac{1000}{78}=0.1924$
$\Delta T_f=k_f\times m=5.12\times 0.1924={0.985}^{\circ }C$
$T_f=278.5-0.985={277.51}^{\circ }C$