Explain Kohlrausch law of independent migration of ions.

(N/A) Law: The law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.
For example,if the limiting molar conductivity of positive and negative ions is $\lambda_{m^{+}}^{\circ}$ and $\lambda_{m^{-}}^{\circ}$,respectively,then the limiting molar conductivity of the solution $(\Lambda_{m}^{\circ})$ will be as follows:
$\Lambda_{m}^{\circ} = v_{+} \lambda_{m^{+}}^{\circ} + v_{-} \lambda_{m^{-}}^{\circ}$
Explanation:
If $\lambda_{m}^{\circ}$ of $K^{+} = 73.5$ and $\lambda_{m}^{\circ}$ of $Br^{-} = 78.1 \ S \ cm^{2} \ mol^{-1}$,then the limiting molar conductivity of $KBr$ solution at infinite dilution is as follows:
$\Lambda_{m}^{\circ}(KBr) = \lambda_{m}^{\circ}(K^{+}) + \lambda_{m}^{\circ}(Br^{-})$
$= 73.5 + 78.1$
$= 151.6 \ S \ cm^{2} \ mol^{-1}$
Limiting molar conductivity of some ions at $298 \ K$ temperature:

Ion | $\lambda^{\circ} / (S \ cm^{2} \ mol^{-1})$	Ion | $\lambda^{\circ} / (S \ cm^{2} \ mol^{-1})$
$H^{+} \ | \ 349.6$	$OH^{-} \ | \ 199.1$
$Na^{+} \ | \ 50.1$	$Cl^{-} \ | \ 76.3$
$K^{+} \ | \ 73.5$	$Br^{-} \ | \ 78.1$
$Ca^{2+} \ | \ 119.0$	$CH_{3}COO^{-} \ | \ 40.9$
$Mg^{2+} \ | \ 106.0$	$SO_{4}^{2-} \ | \ 160.0$