Question

$2$g of benzoic acid $\left(C_6{H}_{5}COOH\right)$ dissolved in $25$g of benzene shows a depression in freezing point equal to $1.62$K. Molal depression constant for benzene is $4.9$ kg m${}^{-1}$. What is the percentage association of acid if it forms a dimer in solution?

• A. $99.2$ %
• B. $97.4$%
• C. $86.0$%
• D. $87.5$%

Answer 1

The correct option is A $99.2$ %

Given,

$W_B=2g,K_f=4.9KKgmo{l}^{-1}$

$W_A=25g,\Delta T_f=1.62K$

Now,

$\Delta T_f=K_f\times \frac{W_B}{M_B}\times \frac{1000}{W_A}$

$M_B=\frac{4.9\times 2\times 1000}{1.62\times 25}=241.98g/mol$

$2C_6{H}_{5}COOH\rightleftharpoons (C_6{H}_{5}COOH{)}_2$

If $x$ is the degree of association, $(1-x)$ mole of benzoic acid left undissociated & corresponding $x/2$ as associated moles of $C_6{H}_{5}COOH$ at equilibrium.

$\therefore$ Total number of moles of particles at eqm,

$\Rightarrow 1-x+\frac{x}{2}=1-\frac{x}{2}=i$

$\Rightarrow i=\frac{Normalmolecularmass}{Abnormalmolecularmass}$

$\Rightarrow 1-\frac{x}{2}=\frac{122}{241.98}$

$\Rightarrow \frac{x}{2}=1-\frac{122}{241.98}\Rightarrow x=0.992=99.2$%

$\therefore$ Degree of association of benzene $=99.2$%