Question

Pure benzene freezes at $5.3^{\circ} C$. A solution of $0.223 \,g$ of phenylacetic acid $(C_6H_5CH_2COOH)$ in $4.4\, g$ of benzene $(k_f = 5.12\,kg\,mol^{-1})$ freezes at $4.47^{\circ} C$. From this observation, one can conclude that

• A. phenylacetic acid exists as such in benzene
• B. phenylacetic acid undergoes partial ionisation in benzene
• C. phenylacetic acid undergoes complete ionisation in benzene
• D. phenylacetic acid dimerises in benzene

Answer 1

Answer: (D)

Depression in freezing point, $\Delta T_{f}=k_{f} m$
(where $m=$ molality)
$=5.12 \times \frac{0.223 \times 1000}{4.4 \times 136}$
$=5.12 \times 0.372$
$=1.91^{\circ} C$
$\therefore $ Theoretical $\Delta T_{f}=1.91$
Actual $\Delta T_{f}=5.3^{\circ} C -4.47^{\circ} C$
$=0.83^{\circ} C$
$\therefore i=\frac{0.83}{1.91}=0.43$
$\because i < 1 $
$ \therefore $ Association will take place.
Thus, it can be said that phenylacetic acid dimerises in benzene.