Question

Phenol associates in benzene to a certain extent to form dimer. A solution containing $2.0 × 10^{-2}\, kg$ of phenol in 1.0 kg of benzene has its freezing point decreased by 0.69 K. The degree of association of phenol is ($K_f$ for benzene $= 5.12\, K\, kg\, mol^{-1}$)

• A. 73.4%
• B. 50.1%
• C. 42.3%
• D. 25.1%

Answer 1

Answer: (A)

$M_{2}\left(obs\right) = \frac{K_{f}\cdot w\cdot1000}{W\cdot\Delta T_{f}}$
$= \frac{5.12\times2.0\times 10^{-1}\times 1000}{1.0\times 0.69} = 148.4$
Calculated molecular mass of phenol $= 94$
$i = \frac{M_{2}\left(cal\right)}{M_{2}\left(obs\right)} = \frac{94}{148.4} = 0.633$
$\begin{matrix}2C_{6}H_{5}OH&{\rightleftharpoons}&\left(C_{6}H_{5}OH\right)_{2}\\ 1-\alpha&&\alpha/2\end{matrix}$
Total species $=\left(1-\alpha\right) + \frac{\alpha}{2} = 1- \frac{\alpha}{2}$
$i = \frac{1-\alpha /2}{1} \quad$ or $\quad \frac{\alpha}{2} = 1-i$
or $\quad\alpha = 2\left(1- i\right) = 2\left(1- 0.633\right) = 0.734 = 73.4\%$