Question

The freezing point of a solution containing 5g of benzoic acid (M = 122g mol−1 ) in 35 g of benzene is depressed by 2.94 K. What is the percentage association of benzoic acid, if it forms a dimer in solution? (Kf for benzene = 4.9 K kg mol−1 )

Answer 1

97.52%

Answer 2

The problem involves the colligative property of freezing point depression and the association of solute molecules. Benzoic acid forms a dimer in benzene, meaning two molecules associate to form one.

1. Calculate the molality (m) of the solution:

   • Molar mass of benzoic acid (M) = 122 g/mol
   • Mass of benzoic acid ($w_2$) = 5 g
   • Moles of benzoic acid ($n_2$) = $\frac{w_2}{M} = \frac{5 \, \text{g}}{122 \, \text{g/mol}} = 0.04098 \, \text{mol}$
   • Mass of benzene ($w_1$) = 35 g = 0.035 kg
   • Molality ($m$) = $\frac{n_2}{w_1(\text{in kg})} = \frac{0.04098 \, \text{mol}}{0.035 \, \text{kg}} = 1.1709 \, \text{mol/kg}$

2. Calculate the Van't Hoff factor (i) using the freezing point depression formula: The formula for freezing point depression is: $\Delta T_f = i \cdot K_f \cdot m$ Given: $\Delta T_f = 2.94 \, \text{K}$, $K_f = 4.9 \, \text{K kg mol}^{-1}$ $2.94 \, \text{K} = i \cdot (4.9 \, \text{K kg mol}^{-1}) \cdot (1.1709 \, \text{mol/kg})$ $2.94 = i \cdot 5.73741$ $i = \frac{2.94}{5.73741} \approx 0.5124$

3. Relate the Van't Hoff factor (i) to the degree of association ($\alpha$): For dimerization, the association reaction is $2A \rightleftharpoons A_2$. If $\alpha$ is the degree of association, then for every 1 mole of solute initially, at equilibrium we have:

   • Moles of unassociated A = $(1 - \alpha)$
   • Moles of associated $A_2$ = $\frac{\alpha}{2}$
   • Total moles at equilibrium = $(1 - \alpha) + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$ The Van't Hoff factor ($i$) is the ratio of observed moles to initial moles: $i = \frac{1 - \frac{\alpha}{2}}{1} = 1 - \frac{\alpha}{2}$ Substitute the calculated value of $i$: $0.5124 = 1 - \frac{\alpha}{2}$ $\frac{\alpha}{2} = 1 - 0.5124$ $\frac{\alpha}{2} = 0.4876$ $\alpha = 2 \times 0.4876 = 0.9752$

4. Calculate the percentage association: Percentage association = $\alpha \times 100\%$ Percentage association = $0.9752 \times 100\% = 97.52\%$

The percentage association of benzoic acid is approximately 97.52%.