∧ _ m ^ o for NaCl, HCl and NaA are 126.4, 425.9 and 100 .5 S cm ^...

Kohlrausch's law states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations.

${\lambda}_{m}^{o}\left(HA\right) = {\lambda}_{m}^{o}\left(HCl\right) + {\lambda}_{m}^{o}\left(NaA\right) - {\lambda}^{o}\left(NaCl\right)$

$= 425.9 + 100.5 - 126.4$

$= 400$

${\lambda}_{m}^{o} = \dfrac{K \times 1000}{M} = \dfrac{5 \times {10}^{- 5} \times {10}^{3}}{{10}^{- 3}} = 50$

$\alpha = \dfrac{50}{400} = 0 .125$