Question

The resistance of $0.5 \, M$ solution of an electrolyte in a cell was found to be $50 \, \Omega$. If the electrodes in the cell are $2.2 \, cm$ apart $\&$ having an area of $4.4\, cm ^{2}$ then the molar conductivity (in $Sm ^{2} mol ^{-1}$ ) of solution is:

• A. 0.2
• B. 0.02
• C. 0.002
• D. None of these

Answer 1

Answer: (C)

Apply, $K=\frac{1}{R} \frac{\ell}{a}$

$\Rightarrow 50 \Omega=\frac{1}{K} \times \frac{2.2 \times 10^{-2} m }{4.4 \times\left(10^{-2} m \right)^{2}}$

$K=1 \Omega^{-1} m^{-1} \quad$ or

Then, use $\Lambda_{ m }= K \times v ( ml ) \rightarrow$ this is the volume which

contains 1 mole of electrolyte

$\Lambda_{ m }=1 m ^{-1} \times 2000 cm ^{3} mol ^{-1}$

$\Lambda_{ m }=1 m ^{-1} \times 2000 \times 10^{-6} m ^{3} mol ^{-1}(0.5$ moles are in $11 tr$

so 1 mole is present in $2 L$ )

$\Lambda_{ m }=2 \times 10^{-3} S m ^{2} mol ^{-1}$