According to Kohlrausch's law of independent migration of ions: $\Lambda^{\circ}_m(\text{CH}_3\text{COOH}) = \Lambda^{\circ}_m(\text{CH}_3\text{COONa}) + \Lambda^{\circ}_m(\text{HCl}) - \Lambda^{\circ}_m(\text{NaCl})$ $= 91.05 + 425.9 - 126.4 = 390.55\text{ S cm}^2\text{ mol}^{-1}$

The molar conductivity at a given concentration $c$ is: $\Lambda_m = \frac{\kappa \times 1000}{c}$ $\Lambda_m = \frac{4.95 \times 10^{-5} \times 1000}{0.001028} = \frac{0.0495}{0.001028} \approx 48.15\text{ S cm}^2\text{ mol}^{-1}$

The degree of dissociation, $\alpha$ is: $\alpha = \frac{\Lambda_m}{\Lambda^{\circ}_m} = \frac{48.15}{390.55} \approx 0.1233$