Question

A solution containing 0.1 mole of naphthalene and 0.9 mole of benzene is cooled out until some benzene freezes out. The solution is then decanted off from the solid and warmed up to 353 K where its vapour pressure was found to be 670 torr. The freezing point and boiling point of benzene are 278.5 K and 353 K respectively and its enthalpy of fusion is 10.67 kJ $mo{l}^{-1}$. Calculate the depression in freezing point for original solution to the nearest integer. Assume ideal behaviour

Answer 1

$P^0=760mm$ at boiling point 353 K and $\frac{P^0-P_S}{P_S}=\frac{w\times M}{m\times W}$

$\frac{760-670}{670}=\frac{w_N\times M_B}{m_N\times W_B}=\frac{0.1\times 78}{W_B}$

$\therefore W_B=58.06g$

Also in original solution $W_B=0.9\times 78=70.2g$

$\therefore C_6{H}_6$ frozen out $=70.2-58.06=12.14g$

Now, $\Delta T$ (for original solution)$=K_f\times molality$

$=\frac{R{T}_b^2}{1000l}\times molality(molality=\frac{0.1}{58.06}\times 1000)$

$=\frac{8.314\times (278.5{)}^2}{1000\times 136.8}\times \frac{1000\times 0.1}{58.06}(1=\frac{10.67\times {10}^3}{78}=136.8J/g)$

$=8.12K$