What is the conductivity of $0.05 \ M$ $NaOH$ solution having resistance $31.5 \ \Omega$ and cell constant $0.315 \ cm^{-1}$?

Molar conductivity of a weak acid $HQ$ of concentration $0.18 \ M$ was found to be $1/30$ of the molar conductivity of another weak acid $HZ$ with concentration of $0.02 \ M$. If $\lambda_{Q^{-}}^0 = \lambda_{Z^{-}}^0$, then the difference of the $pK_a$ values of the two weak acids $(pK_a(HQ) - pK_a(HZ))$ is . . . . . . (Nearest integer).
[Given: degree of dissociation $(\alpha)$ $\ll 1$ for both weak acids, $\lambda^0$: limiting molar conductivity of ions]

The cell constant of a cell is $0.5 \ cm^{-1}$ and the resistance $R = 50 \ \Omega$. If the normality is $1 \ N$,then the equivalent conductivity of the electrochemical cell is ...... $\Omega^{-1} \ cm^2 \ (g \ eq)^{-1}$.

The conductivity of a strong electrolyte is:

The ionic conductances of $Ba^{2+}$ and $Cl^{-}$ at infinite dilution are $127$ and $76 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ respectively. The equivalent conductance of $BaCl_2$ at infinite dilution is ..... (in $.5$)

Explain Kohlrausch law of independent migration of ions.