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)

Given $W_A$, benzoic acid $= 2 \,g$
$W_B$, Benzene $= 25 \,g$
$K_f = 4.9 \,K \,kg\, mol^{-1}$
$T_f = 1.62 \,K$
$M_B =241.98$
$2C_6H_5COOH \rightleftharpoons (C_6H_5COOH)_2$
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 $99.2\%$.