Question

Phenol associates in benzene to a certain extent to form a dimer. A solution containing $20 \times 10^{-3}\,kg$ of phenol in $1.0\,kg$ of benzene has its freezing point depressed by $0.69 \,K$ Calculate the fraction of phenol that has dimerised. $(K_{f}$ for benzene $= 5.12 \,K\, kg\, mol^{-1})$

• A. $28.5\%$
• B. $32.6\%$
• C. $73.3\%$
• D. $82\%$

Answer 1

Answer: (C)

Observed mol. mass $=\frac{1000\times K_{f}\times w}{W\times\Delta T_{f}}$
$=\frac{1000\times5.12\times20\times10^{-3}}{1\times0.69}=148.4$
Normal mol. mass of phenol = 94
$i=\frac{\text{Normal mol. mass}}{\text{Observed mol. mass}}=\frac{94}{148.4}=0.633$
Since, phenol is dimerized
$\therefore n=2$
Also, $i=1+\left[\frac{1}{n}-1\right]\alpha=1+\left[\frac{1}{2}-1\right]\alpha=1-\frac{\alpha}{2}$
$\Rightarrow 0.633=1-\frac{\alpha}{2}$
$\Rightarrow \frac{\alpha}{2}=0.367$
$\Rightarrow \alpha=0.733$
$\alpha=0.733\times100=73.3\%$