1. Use freezing point depression formula

For a solution, $\Delta T_f = i K_f m$ where:

• $\Delta T_f$ = depression in freezing point
• $i$ = van't Hoff factor
• $K_f = 1.85\ \text{K kg mol}^{-1}$
• $m = 2$

Since pure water freezes at $0^\circ C$ and solution freezes at $-3.885^\circ C$, $\Delta T_f = 3.885^\circ C$

So, $3.885 = i \times 1.85 \times 2$ $3.885 = 3.70 i$ $i = \frac{3.885}{3.70} = 1.05$

2. Relate van't Hoff factor to degree of dissociation

For weak acid: $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$

If degree of dissociation is $\alpha$, then starting from 1 molecule, total particles become: $1-\alpha + \alpha + \alpha = 1+\alpha$ Hence, $i = 1 + \alpha$

So, $1.05 = 1 + \alpha$ $\alpha = 0.05$

3. Express in the required form

$0.05 = 50 \times 10^{-3}$

Therefore, the required integer is: $\boxed{50}$

4. Comparison with stored answer

Stored correct answer = 50

Our derived answer also = 50, so they agree.