Question

$2 \,g$ of benzoic acid $(C_6H_6COOH)$ is dissolved in $25\, g$ 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)

$99.2\%$

Answer 2

Given $W_A$, benzoic acid $= 2 \,g$
$W_B$, Benzene $= 25 \,g$
$K_f = 4.9 \,K \,kg\, mol^{-1}$
$T_f = 1.62 \,K$
$M_B =241.98$
$2C_6H_5COOH {<=>} (C_6H_5COOH)_2$
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, $i = 1 - \frac{x}{2}$
$i = \frac{\text{Total number of moles of particles after association }}{\text{Number of moles of particle before association}}$
$i = \frac{122\,g\,mol}{241.98\,g\,mol}$
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 $99.2\%$.