Question

The freezing point of benzene decreases by ${0.45}^{\circ }C$ when 0.2g of acetic acid is added to 20g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be
($K_f$ for benzene = 5.12K kg $mo{l}^{-1}$)

• A. 64.6%
• B. 80.4%
• C. 74.6%
• D. 94.6%

Answer 1

The correct option is D 94.6%

Let the degree of association of acetic acid $\left(C{H}_{3}COOH\right)$ in benzene is $\alpha$, then
$2C{H}_{3}COOH\rightleftharpoons (C{H}_{3}COOH{)}_2$
$\begin{array}{ccc}& & \\ Initialmoles& 1& 0\\ Molesatequilibrium& 1-\alpha & \frac{\alpha }{2}\end{array}$
$\therefore Totalmoles=1-\alpha +\frac{\alpha }{2}=1-\frac{\alpha }{2}ori=1-\frac{\alpha }{2}$
Now, depression in freezing point $\left(\Delta T_f\right)$ is given as
$\Delta T_f=i{K}_{f}m$ . . . . (i)
Where, $K_f$ - molal depression constant or cryoscopic constant.
m = Molality
$Molality=\frac{numberofmolesofsolute}{weightofsolvent\left(inkg\right)}=\frac{0.2}{60}\times \frac{100}{20}$
Putting the values in Eq.(i)
$\therefore 0.45=[1-\frac{\alpha }{2}]\left(5.12\right)\left[\frac{1000}{20}\right]$
$1-\frac{\alpha }{2}=\frac{0.45\times 60\times 20}{5.12\times 0.2\times 1000}$
$\Rightarrow 1-\frac{\alpha }{2}=0.527$
$\Rightarrow \frac{\alpha }{2}=1-0.527$
$\therefore \alpha =0.946$
Thus, percentage of association = 94.6%