Question

$2g$ of benzoic acid $(C_6{H}_{6}COOH)$ is dissolved in $25g$ of benzene. The observed molar mass of benzoic acid is found to be $241.98$ . What is the percentage association of acid if it forms in solution?

• A. $85.2\%$
• B. $89.2\%$
• C. $95.2\%$
• D. $99.2\%$

Answer 1

Answer: (D)

Given $W_A$, benzoic acid $=2g$
$W_B$, Benzene $=25g$
$K_f=4.9Kkgmo{l}^{-1}$
$T_f=1.62K$
$M_B=241.98$
$2C_6{H}_{5}COOH\rightleftharpoons (C_6{H}_{5}COOH{)}_2$
Let, $x=$ degree of association of solute
$\therefore (1-x)=$ moles of benzene left in unassociated form
$\frac{x}{2}=$ moles of benzoic acid in equilibrium
Total number of moles at equilibrium $=1-x+\frac{x}{2}$
van,t Hoff factor, $i=1-\frac{x}{2}$
$i=\frac{\text{Total number of moles of particles after association }}{\text{Number of moles of particle before association}}$
$i=\frac{122gmol}{241.98gmol}$
or, $1-\frac{x}{2}=\frac{122}{241.98}$
$\therefore \frac{x}{2}=1-\frac{122}{241.98}$
$\frac{x}{2}=0.49$
$x=2\times 0.49$
$x=0.9916$
$\therefore$ The degree of association of benzoic acid in benzene is $99.2\%$.