If 0.01 M solution of an electrolyte has a resistance of 40 ohms in a cell having a cell constant of 0.4 cm^–1, then its molar conductance in ohm^–1 cm² mol^–1 is

Molarity = 0.01 M ; Resistance = 40 ohm;

Cell constant $\frac{l}{A}=0.4c{m}^{-1}$

Specific conductivity $\left(\kappa \right)$ 

$=\frac{\mathrm{cell} constant}{\mathrm{resistance}}=\frac{0.4}{40}=0.01 oh{m}^{-1}c{m}^{-1}$

Molar conductance $\left({\wedge }_m\right)=\frac{1000\kappa }{Molarity}$

$=\frac{1000\times 0.01}{0.01}={10}^{3}oh{m}^{-1}c{m}^{2}mo{l}^{-1}$