The electrical resistance of a column of $0.05 \ mol \ L^{-1}$ $NaOH$ solution of diameter $1 \ cm$ and length $50 \ cm$ is $5.55 \times 10^{3} \ \Omega$. Calculate its resistivity,conductivity and molar conductivity.

$A = \pi r^{2} = 3.14 \times (0.5 \ cm)^{2} = 0.785 \ cm^{2} = 0.785 \times 10^{-4} \ m^{2}$
$l = 50 \ cm = 0.5 \ m$
$\rho = \frac{R A}{l} = \frac{5.55 \times 10^{3} \ \Omega \times 0.785 \ cm^{2}}{50 \ cm} = 87.135 \ \Omega \ cm$
$\kappa = \frac{1}{\rho} = \frac{1}{87.135} \ S \ cm^{-1} = 0.01148 \ S \ cm^{-1}$
$\Lambda_{m} = \frac{\kappa \times 1000}{c} = \frac{0.01148 \ S \ cm^{-1} \times 1000 \ cm^{3} \ L^{-1}}{0.05 \ mol \ L^{-1}} = 229.6 \ S \ cm^{2} \ mol^{-1}$