Question

Benzoic acid molecules undergo dimerisation in benzene. 2.44 g of benzoic acid when dissolved in 30 g of benzene caused depression in freezing point of 2 K. What is the percentage of association of it ? (Given $K_f (C_6H_6) = 5\operatorname{K kg mol^{-1}}$; molar mass of benzoic acid $= 122\operatorname{g mol^{-1}}$)

• A. 20
• B. 70
• C. 60
• D. 90

Hint:

For association of 'n' moles into 1 mole (e.g., dimerization where n=2), the van't Hoff factor 'i' is related to the degree of association ($\alpha$) by the formula: $i = 1 - \alpha + \frac{\alpha}{n}$. In the case of dimerization (n=2): $i = 1 - \alpha + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}$. This relationship is crucial for solving problems involving colligative properties of associating solutes. Remember that for association, the observed molar mass will be greater than the theoretical molar mass, leading to $i<1$.

Answer 1

The Correct Option is A

The problem involves calculating the percentage association of benzoic acid in benzene. Given the molecular information and the depression in freezing point, we use the colligative properties to find the extent of dimerisation.

1. Determine the molality: Molality \(m\) is defined as the moles of solute per kilogram of solvent. Calculate the moles of benzoic acid:

Moles of benzoic acid = \( \frac{2.44 \text{ g}}{122 \text{ g/mol}} = 0.02 \text{ mol} \)

Now, find the molality:

\( m = \frac{0.02 \text{ mol}}{0.03 \text{ kg}} = 0.6667 \text{ mol/kg} \)

2. Calculate the expected freezing point depression if no association occurs:

The formula for depression in freezing point is:

\( \Delta T_f = i \cdot K_f \cdot m \)

For non-associated solute, van't Hoff factor \(i = 1\).

\( \Delta T_f = 1 \cdot 5 \text{ K kg mol}^{-1} \cdot 0.6667 \text{ mol/kg} = 3.3335 \text{ K} \)

3. Calculate the observed van’t Hoff factor using observed depression:

Given that the observed depression is \(2 \text{ K}\), calculate the van’t Hoff factor:

\( i = \frac{\text{observed } \Delta T_f}{\text{expected } \Delta T_f} = \frac{2}{3.3335} = 0.6 \)

4. Relate van’t Hoff factor to association:

In dimerization, 2 moles of benzoic acid associate to form 1 mole of dimer, thus \(i = \frac{1+\alpha}{2}\), where \(\alpha\) is the degree of association.

Solving for \(\alpha\): \( 0.6 = \frac{1+\alpha}{2} \)

\( 1+\alpha = 1.2 \Rightarrow \alpha = 0.2 \)

5. Calculate the percentage association:

Percentage association = \(\alpha \times 100 = 0.2 \times 100 = 20\% \)

Therefore, the percentage of association of benzoic acid molecules in benzene is 20%.