Molecules of benzoic acid ($$C_6H_5COOH$$) dimerise in 30 g of benzene. 'w' g of benzoic acid shows a depression in freezing point equal to 2 K. If the percentage association of the acid to form dimer in the solution is 80, then w is: (Given that $$K_f = 5$$ K mol$$^{-1}$$, molar mass of benzoic acid = 122 g mol$$^{-1}$$)

We have a solution containing w grams of benzoic acid dissolved in 30 g = 0.03 kg of benzene. The observed depression in freezing point is $$\Delta T_f = 2\ \text{K}$$.

For any dilute solution the freezing-point depression is given by the formula

$$\Delta T_f = i\,K_f\,m,$$

where $$i$$ is the van’t Hoff factor that counts the effective number of solute particles, $$K_f$$ is the cryoscopic constant of the solvent, and $$m$$ is the molality of the solute.

The cryoscopic constant is supplied as $$K_f = 5\ \text{K\,kg\,mol}^{-1}$$. Now we express the molality in terms of the unknown mass w of benzoic acid.

The molar mass of benzoic acid is $$M = 122\ \text{g\,mol}^{-1}$$, so the number of moles present is

$$\text{moles of acid} = \frac{w}{122}\ \text{mol}.$$

The solvent mass is 0.03 kg, therefore the molality is

$$m = \frac{\dfrac{w}{122}}{0.03} = \frac{w}{122 \times 0.03} = \frac{w}{3.66}\ \text{mol\,kg}^{-1}.$$

Benzoic acid molecules associate to form dimers, and the question states that the percentage association is 80 %. Let $$\alpha$$ be the degree of association; hence $$\alpha = 0.80$$.

For dimerisation ($$2\text{A}\;\rightleftharpoons\;\text{A}_2$$) the van’t Hoff factor is obtained from

$$i = 1 - \tfrac{\alpha}{2},$$

because every two monomers combine to give one dimer, reducing the total number of solute particles. Substituting $$\alpha = 0.80$$ we get

$$i = 1 - \frac{0.80}{2} = 1 - 0.40 = 0.60.$$

Now we substitute all the known quantities into the freezing-point depression equation:

$$\Delta T_f = i\,K_f\,m \quad\Longrightarrow\quad 2 = 0.60 \times 5 \times \frac{w}{3.66}.$$

First evaluate the product $$0.60 \times 5 = 3$$, so

$$2 = 3 \times \frac{w}{3.66}.$$

Rearranging for w gives

$$w = 2 \times \frac{3.66}{3} = 2 \times 1.22 = 2.44\ \text{g}.$$

On rounding to the significant digits provided in the options, we report $$w \approx 2.4\ \text{g}.$$

Hence, the correct answer is Option B.