The molar conductance of 0.05 M solution of $MgC{l}_2$ is $194.5{\Omega }^{-1}c{m}^{2}mo{l}^{-1}$ at ${25}^{\circ }C,$ A cell with electrods having $1.50c{m}^2$ surface area and 0.50 cm apart is filled with 0.05 M solution $mgC{l}_2$. How much current will flow when the potential difference between the electrodes is 5.0 V ?

Molar conductivity= $\frac{\text{Specific conductance (K)}}{\text{Concentration of electrolyte(C)}}$

$\Rightarrow K=\wedge m\times C=194.5{\Omega }^{-1}c{m}^{2}mo{l}^{-1}\times 0.05mol{L}^{-1}$

$\Rightarrow K=9.725{\Omega }^{-1}c{m}^2{L}^{-1}$ $1L=1000c{m}^3;1L^{-1}={10}^{-3}c{m}^{-3}$

as $\frac{1}{R}=K\frac{A}{l}$ $A⟶$ Area of cross-section of cell; $l⟶$ length of electrode cell; $R⟶$ Resistance

$\Rightarrow \frac{1}{R}=9.725{\Omega }^{-1}c{m}^2\times {10}^{-3}c{m}^{-3}\frac{1.50c{m}^2}{0.50cm}=0.029{\Omega }^{-1}$

$\Rightarrow R=34.27\Omega$

as $V=IR$ $V⟶$ Electric potential; $I⟶$ Current

$\Rightarrow I=\frac{V}{R}=\frac{5.0V}{34.27\Omega }=0.145Ampere$

Thus, $0.145Ampere$ of current is required.