1. Find molar mass of $\mathrm{ClCH_2COOH}$

For monochloroacetic acid, $\mathrm{ClCH_2COOH}$:

$M = 2(12) + 3(1) + 35.5 + 2(16) = 94.5\ \text{g mol}^{-1}$

2. Find moles of solute added

Given mass $= 9.45\ \text{g}$,

$n = \frac{9.45}{94.5} = 0.1\ \text{mol}$

3. Find molality of the solution

Mass of water $= 500\ \text{mL} \approx 500\ \text{g} = 0.5\ \text{kg}$

$m = \frac{0.1}{0.5} = 0.2\ \text{mol kg}^{-1}$

4. Use depression in freezing point relation

$\Delta T_f = iK_f m$

Given:

• $\Delta T_f = 0.5^\circ \mathrm{C}$
• $K_f = 1.86\ \mathrm{K\,kg\,mol^{-1}}$
• $m = 0.2$

So,

$0.5 = i(1.86)(0.2)$

$i = \frac{0.5}{0.372} \approx 1.344$

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

Monochloroacetic acid dissociates as:

$\mathrm{ClCH_2COOH \rightleftharpoons H^+ + ClCH_2COO^-}$

One molecule gives $2$ particles, so

$i = 1 + \alpha$

Thus,

$\alpha = i-1 = 1.344 - 1 = 0.344$

6. Find concentration for dissociation constant

Initial concentration:

$C = \frac{0.1}{0.5} = 0.2\ \text{M}$

For a weak acid,

$K_a = \frac{C\alpha^2}{1-\alpha}$

Substitute values:

$K_a = \frac{(0.2)(0.344)^2}{1-0.344}$

$(0.344)^2 = 0.118336$

$K_a = \frac{0.2 \times 0.118336}{0.656}$

$K_a = \frac{0.0236672}{0.656} \approx 0.03608$

$K_a \approx 3.61 \times 10^{-2} = 36.1 \times 10^{-3}$

So,

$x \approx 36$

7. Comparison with stored answer

Stored correct answer is $34.4$.

My calculated value is about $36.1$, so rounded to nearest integer it is $36$.

The likely reason for mismatch is that the exact molar mass may have been approximated differently (for example using $\mathrm{Cl}=35$ instead of $35.5$), but with the given data the value comes closest to $36$.