1. Given data

• Concentration of acetic acid: $C = 0.0025~\text{M}$
• Specific conductance: $\kappa = 5 \times 10^{-5}~\text{S cm}^{-1}$
• Limiting molar conductivity: $\Lambda_m^\circ = 400~\text{S cm}^2\text{ mol}^{-1}$

We need to find the dissociation constant $K_a$ in the form: $\boxed{\_\_\_\_\times 10^{-7}}$

2. Calculate molar conductivity

Molar conductivity is: $\Lambda_m = \frac{\kappa \times 1000}{C}$

Substitute the values: $\Lambda_m = \frac{5 \times 10^{-5} \times 1000}{0.0025}$

$\Lambda_m = \frac{5 \times 10^{-2}}{0.0025} = 20~\text{S cm}^2\text{ mol}^{-1}$

3. Calculate degree of dissociation

For a weak electrolyte: $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$

So, $\alpha = \frac{20}{400} = 0.05$

4. Use Ostwald's dilution law

For weak acid $\text{CH}_3\text{COOH}$: $K_a = \frac{C\alpha^2}{1-\alpha}$

Substitute values: $K_a = \frac{0.0025 \times (0.05)^2}{1-0.05}$

First, $(0.05)^2 = 0.0025$

Thus, $K_a = \frac{0.0025 \times 0.0025}{0.95}$

$K_a = \frac{6.25 \times 10^{-6}}{0.95}$

$K_a \approx 6.58 \times 10^{-6}$

Now write it as: $K_a = 65.8 \times 10^{-7}$

Nearest integer: $\boxed{66}$

5. Comparison with stored answer

Derived answer = $66$

Stored correct answer = $66$

So they agree.