Resistance of a conductivity cell filled with $0.1 \, mol \, L^{-1}$ $KCl$ solution is $100 \, \Omega$. If the resistance of the same cell when filled with $0.02 \, mol \, L^{-1}$ $KCl$ solution is $520 \, \Omega$,calculate the conductivity and molar conductivity of $0.02 \, mol \, L^{-1} \, KCl$ solution. The conductivity of $0.1 \, mol \, L^{-1} \, KCl$ solution is $1.29 \, S / m$.

(N/A) The cell constant $(G^*)$ is calculated as:
$G^* = \text{conductivity} \times \text{resistance} = 1.29 \, S \, m^{-1} \times 100 \, \Omega = 129 \, m^{-1}$.
Conductivity $(\kappa)$ of $0.02 \, mol \, L^{-1}$ $KCl$ solution:
$\kappa = \frac{G^*}{R} = \frac{129 \, m^{-1}}{520 \, \Omega} = 0.248 \, S \, m^{-1}$.
Molar conductivity $(\Lambda_m)$ is given by $\Lambda_m = \frac{\kappa}{c}$.
Concentration $c = 0.02 \, mol \, L^{-1} = 20 \, mol \, m^{-3}$.
$\Lambda_m = \frac{0.248 \, S \, m^{-1}}{20 \, mol \, m^{-3}} = 0.0124 \, S \, m^2 \, mol^{-1} = 124 \times 10^{-4} \, S \, m^2 \, mol^{-1}$.