A quantity of 2 g of dissolved in 25 g of benzene shows a depression in freezing point equal to 1.96 — Solutions and Colligative Properties Chemistry Question
Question
A quantity of 2 g of $\text{C}_6\text{H}_5\text{COOH}$ dissolved in 25 g of benzene shows a depression in freezing point equal to 1.96 K. Molar depression constant for benzene is $4.9 \text{ K-kg mol}^{-1}$. What is the percentage association of acid if it forms double molecules (dimer) in solution?
💡 Solution & Explanation
Molar mass of benzoic acid ($\text{C}_6\text{H}_5\text{COOH}$) is 122 g/mol. Molality $m = (2 / 122) / 0.025 = 0.6557 \text{ m}$. The expected depression is $\Delta T_f = K_f \times m = 4.9 \times 0.6557 \approx 3.213 \text{ K}$. The observed depression is 1.96 K. Thus, $i = 1.96 / 3.213 \approx 0.61$. For dimerization, $i = 1 - \alpha/2$, so $0.61 = 1 - \alpha/2 \Rightarrow \alpha/2 = 0.39 \Rightarrow \alpha = 0.78$. The percentage association is 78%. Therefore, correct answer is B.