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Solve the following.
When 2 g of benzoic acid (C₆H₅COOH) is dissolved in 25 g of benzene, it shows a depression in…
When $2\text{ g}$ of benzoic acid ($\text{C}_6\text{H}_5\text{COOH}$) is dissolved in $25\text{ g}$ of benzene, it shows a depression in freezing point equal to $1\cdot62\text{ K}$. Molal depression constant for benzene is $4\cdot7\text{ K kg mol}^{-1}$.
What is the percentage association of acid if it forms dimer in solution?
Answer
Answer
Formulas used: Depression of freezing point: dTf = i*Kf*m with Kf = 4.7 K kg mol^-1, w1 = 25 g, w2 = 2 g, dTf = 1.62 K, so M2 = 232.1 g mol^-1 van 't Hoff factor from molar masses: i = M_normal/M_observed with M_normal = 122 g mol^-1, M_observed = 232.1 g mol^-1, so i = 0.526 Degree of association from i: alpha = (1 - i)/(1 - 1/nu) with i = 0.526, nu = 2, so alpha = 0.949
Observed molar mass: $M_2 = \dfrac{1000\,K_f\,w_2}{\Delta T_f\,w_1} = \dfrac{1000\times4.7\times2}{1.62\times25} = 232.1\ \mathrm{g\,mol^{-1}}$
Normal molar mass of $\mathrm{C_6H_5COOH}$ = 122 g mol^-1, so $i = 122/232.1 = 0.526$.
For dimerisation $\alpha = \dfrac{1-i}{1-\frac12} = \dfrac{1-0.526}{0.5} = 0.949$
Final answer: 94.9 %