1. Find moles of acetic acid added

Given volume of acetic acid = $1.2\,\text{mL}$
Density = $1.02\,\text{g mL}^{-1}$

So mass of acetic acid is $m = 1.2 \times 1.02 = 1.224\,\text{g}$

Molar mass of acetic acid = $60\,\text{g mol}^{-1}$

Hence moles of acetic acid, $n = \frac{1.224}{60} = 0.0204\,\text{mol}$

2. Mass of solvent (water)

Final solution volume = $2.0\,\text{L}$. For such a dilute aqueous solution, take mass of solvent approximately as $2.0\,\text{kg}$.

So, $\text{mass of water} \approx 2.0\,\text{kg}$

3. Calculate molality of acetic acid

$m = \frac{0.0204}{2.0} = 0.0102\,\text{mol kg}^{-1}$

4. Use depression in freezing point relation

$\Delta T_f = i K_f m$

Given:

• $\Delta T_f = 0.0198^\circ \text{C}$
• $K_f = 1.85\,\text{K kg mol}^{-1}$
• $m = 0.0102\,\text{mol kg}^{-1}$

So, $i = \frac{\Delta T_f}{K_f m} = \frac{0.0198}{1.85 \times 0.0102}$

$1.85 \times 0.0102 = 0.01887$

$i = \frac{0.0198}{0.01887} \approx 1.049$

5. Relate van't Hoff factor to dissociation

Acetic acid dissociates as $\text{CH}_3\text{COOH} \rightleftharpoons \text{H}^+ + \text{CH}_3\text{COO}^-$

If degree of dissociation is $\alpha$, then for a solute producing 2 particles from 1 particle, $i = 1 + \alpha$

Thus, $\alpha = i - 1 = 1.049 - 1 = 0.049$

Percentage dissociation, $\%\text{ dissociation} = 0.049 \times 100 = 4.9\%$

Nearest integer: $\boxed{5}$

6. Comparison with stored answer

Stored correct answer = $5$
Derived answer = $5$

So the derived answer agrees with the stored correct answer.