When 12.2 g of benzoic acid is dissolved in 100 g of water, the freezing point of solution was found to be $$-0.93$$ °C ($$K_f(H_2O) = 1.86$$ K kg mol$$^{-1}$$). The number (n) of benzoic acid molecules associated (assuming 100% association) is ______.

The molar mass of benzoic acid ($$C_6H_5COOH$$) is $$12 \times 7 + 6 + 16 \times 2 = 84 + 6 + 32 = 122$$ g/mol.

The number of moles of benzoic acid dissolved is $$\frac{12.2}{122} = 0.1$$ mol.

The molality if no association occurred would be $$\frac{0.1}{0.1} = 1$$ mol/kg (since the solvent mass is 100 g = 0.1 kg).

The depression in freezing point is given by $$\Delta T_f = i \times K_f \times m$$, where $$i$$ is the van't Hoff factor. Substituting: $$0.93 = i \times 1.86 \times 1$$.

Solving for $$i$$: $$i = \frac{0.93}{1.86} = 0.5$$.

When $$n$$ molecules of benzoic acid associate to form one associated molecule, the van't Hoff factor for 100% association is $$i = \frac{1}{n}$$.

Therefore $$\frac{1}{n} = 0.5$$, which gives $$n = 2$$.

This means benzoic acid forms dimers in aqueous solution (2 molecules associate together), which is consistent with the known behaviour of benzoic acid forming hydrogen-bonded dimers. The answer is $$2$$.