When 9.45 g of $$ClCH_2COOH$$ is added to 500 mL of water, its freezing point drops by 0.5°C. The dissociation constant of $$ClCH_2COOH$$ is $$x \times 10^{-3}$$. The value of $$x$$ is ______ (off to the nearest integer)
$$K_{f_{H_2O}} = 1.86$$ K kg mol$$^{-1}$$

We are given that 9.45 g of $$ClCH_2COOH$$ is added to 500 mL of water and the freezing point drops by 0.5°C. We need to find the dissociation constant.

The molar mass of $$ClCH_2COOH$$ is: $$35.5 + 12 + 2(1) + 12 + 2(16) + 1 = 94.5 \text{ g/mol}$$.

The number of moles of solute is:

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

The molality of the solution (taking 500 mL of water as approximately 500 g = 0.5 kg):

$$m = \frac{0.1}{0.5} = 0.2 \text{ mol/kg}$$

Using the freezing point depression formula: $$\Delta T_f = i \cdot K_f \cdot m$$

$$0.5 = i \times 1.86 \times 0.2$$ $$i = \frac{0.5}{0.372} = 1.344$$

For a weak monoprotic acid $$ClCH_2COOH \rightleftharpoons ClCH_2COO^- + H^+$$, the van't Hoff factor is $$i = 1 + \alpha$$, where $$\alpha$$ is the degree of dissociation.

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

The dissociation constant $$K_a$$ is given by:

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

where $$C$$ is the initial concentration. Since the solution is dilute, we use molality as the concentration:

$$K_a = \frac{0.2 \times (0.344)^2}{1 - 0.344} = \frac{0.2 \times 0.11834}{0.656}$$ $$= \frac{0.023668}{0.656} = 0.03608$$ $$= 36.08 \times 10^{-3}$$

Rounding to the nearest integer: $$x = 36$$.

So, the answer is $$36$$.