1.2 mL of acetic acid is dissolved in water to make 2.0 L of solution. The depression in freezing point observed for this strength of acid is 0.0198°C. The percentage of dissociation of the acid is ______ (Nearest integer)
[Given: Density of acetic acid is 1.02 g mL$$^{-1}$$ Molar mass of acetic acid is 60 g mol$$^{-1}$$, K$$_f$$(H$$_2$$O) = 1.85 K kg mol$$^{-1}$$]

We need to find the percentage of dissociation of acetic acid.

The volume of acetic acid is 1.2 mL, its density is 1.02 g mL$$^{-1}$$, its molar mass is 60 g mol$$^{-1}$$, and the total volume of solution is 2.0 L. The depression in freezing point ($$\Delta T_f$$) is 0.0198 °C and K$$_f$$(H$$_2$$O) = 1.85 K kg mol$$^{-1}$$.

The mass of acetic acid is 1.2 × 1.02 = 1.224 g, and the moles of acetic acid are $$\frac{1.224}{60}$$ = 0.0204 mol.

Assuming the solution is very dilute so that 2.0 L of solution corresponds to approximately 2.0 kg of solvent, the molality (m) is $$\frac{0.0204}{2.0}$$ = 0.0102 mol kg$$^{-1}$$.

The van't Hoff factor (i) is determined from the relationship $$\Delta T_f = i \times K_f \times m$$. Substituting the known values gives $$0.0198 = i \times 1.85 \times 0.0102$$, which simplifies to $$0.0198 = i \times 0.01887$$, and hence $$i = \frac{0.0198}{0.01887} = 1.0493$$.

For acetic acid: CH$$_3$$COOH $$\rightleftharpoons$$ CH$$_3$$COO$$^-$$ + H$$^+$$, the van't Hoff factor is given by $$i = 1 + \alpha(n - 1)$$ where n = number of particles formed = 2 (one CH$$_3$$COO$$^-$$ and one H$$^+$$). Substituting gives $$1.0493 = 1 + \alpha(2 - 1)$$, and therefore $$\alpha = 0.0493$$.

The percentage dissociation is calculated as $$\alpha \times 100 = 0.0493 \times 100 \approx 5\%$$.

Therefore, the percentage of dissociation of acetic acid is 5%.