The molar conductivity of $0.04\text{M}$ solution of $MgC{l}_2$​ is $200{\text{S cm}}^3{\text{mol}}^{-1}$ at $298\text{K}$. A cell with electrodes that are $2.0{\text{cm}}^2$ in surface area and $0.50\text{cm}$ apart is filled with $MgC{l}_2$​ solution. How much current will flow when the potential difference between the two electrodes is $5.0\text{V}$?

The correct option is B $0.16\text{A}$
Given,
$\text{Molar conductivity}=200{\text{S cm}}^3{\text{mol}}^{-1}$
$\text{Molar concentration}=0.04\text{M}$
${\Lambda }_m=\frac{k\times 1000}{M}$
Where, ${\Lambda }_m$ = Molar conductivity
$k$ = specific conductance
$M$ = Molar concentration of electrolyte
$\Rightarrow k=\frac{{\Lambda }_m\times M}{1000}$
$\Rightarrow k=\frac{200\times 0.04}{1000}$
$\Rightarrow k=8\times {10}^{-3}\text{S}$
We know,
$k=C\times \text{cell constant}...\left(1\right)$
$\text{cell constant}=\frac{l}{A}$
$\Rightarrow \text{cell constant}=\frac{0.5\text{cm}}{2{\text{cm}}^2}$
$\Rightarrow \text{cell constant}=0.25{\text{cm}}^{-1}$
From equation $\left(1\right)$
$\Rightarrow C=\frac{k}{\text{cell constant}}$
$\Rightarrow C=\frac{8\times {10}^{-3}}{0.25}$
$\Rightarrow C=0.032\text{S cm}$
According to Ohm's law,
$I=\frac{V}{R}$
$\Rightarrow I=V\times C(\frac{1}{R}=C)$
$\Rightarrow I=5\times 0.032=0.16\text{A}$