The freezing point of benzene decreases by 0.45°C on adding 0.2 g of acetic acid to 20 g of benzene. If acetic acid associates to form a dimer in benzene, then what is the percentage association of acetic acid in benzene?
K$$_{f}$$ for benzene = 5.12 K kg mol$$^{-1}$$

We start with the relation for depression in freezing point:

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

where $$\Delta T_f$$ is the observed fall in freezing point, $$i$$ is the van’t Hoff factor, $$K_f$$ is the cryoscopic constant of the solvent, and $$m$$ is the molality of the solution.

First we calculate the molality when no association is assumed (so that later we can compare with the observed value).

The molality $$m$$ is defined as

$$m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}.$$

Moles of acetic acid added:

$$\text{moles} = \frac{0.2\;{\rm g}}{60\;{\rm g\,mol^{-1}}} = \frac{1}{300} \;{\rm mol} = 0.003333\ldots{\rm \,mol}.$$

Mass of benzene (solvent) in kilograms:

$$20\;{\rm g} = 0.020\;{\rm kg}.$$

So the molality is

$$m = \frac{0.003333\ldots}{0.020} = 0.166666\ldots \;{\rm mol\,kg^{-1}}.$$

If there were no association ($$i=1$$), the theoretical depression would be

$$\Delta T_f^{\text{(theo)}} = K_f \, m = 5.12 \times 0.166666\ldots = 0.8533\;^\circ{\rm C}.$$

The experimental (observed) depression is given as $$\Delta T_f^{\text{(obs)}} = 0.45^\circ{\rm C}.$$

Hence the van’t Hoff factor actually present is

$$i = \frac{\Delta T_f^{\text{(obs)}}}{\Delta T_f^{\text{(theo)}}} = \frac{0.45}{0.8533} = 0.527 \;(\text{approximately}).$$

Now we relate $$i$$ to the degree of association $$\alpha$$ for dimerisation. For the association $$2 \,{\rm CH_3COOH} \rightleftharpoons ({\rm CH_3COOH})_2$$ we have $$n = 2$$ (two monomers form one dimer).

Let us take 1 mole of acetic acid initially. If the degree of association is $$\alpha$$, then

– moles remaining unassociated = $$1 - \alpha$$,

– moles of dimers formed = $$\frac{\alpha}{2}$$ (because two monomers give one dimer).

Total particles in solution after association:

$$N_{\text{total}} = (1 - \alpha) + \frac{\alpha}{2} = 1 - \frac{\alpha}{2}.$$

The van’t Hoff factor is the ratio of the number of particles present to the number of formula units taken initially, therefore

$$i = 1 - \frac{\alpha}{2}.$$

Solving for $$\alpha$$:

$$\alpha = 2(1 - i).$$

Substituting $$i = 0.527$$,

$$\alpha = 2(1 - 0.527) = 2 \times 0.473 = 0.946.$$

Expressed as a percentage,

$$\text{Percentage association} = 0.946 \times 100\% = 94.6\%.$$

Hence, the correct answer is Option C.