Molar conductivity of 0.04 M ${MgCl}_{2}$ solution at 298K is 200 S cm²mole^-1.A conductivity cell which is filled with ${MgCl}_{2}$ have area of cross-section of electrode $4 {cm}^{2}$ distance electrode is 8cm. If potential difference between electrode is 10V then find current flow in mA. (Report your answer after dividing by 10)

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Detailed Solution

$λ_{m} = \frac{k \times 1000}{M}$
$\begin{array}{l} 200 = \frac{1}{R} \times \frac{ℓ}{A} \times \frac{1000}{M} \Rightarrow 200 = \frac{I}{V} \times \frac{ℓ}{A} \times \frac{1000}{M} \\ 200 = \frac{I}{10} \times \frac{8}{4} \times \frac{1000}{0.04} \Rightarrow I = \frac{40}{1000} = 0.04 A = 40 mA \end{array}$