Resistance of 0.2 M solution of an electrolytes is 50 $Ω$ .The specific conductance of the solution is 1.4 S m^-1.The resistance of 0.5 M solution of the same electrolyte is 280 $Ω$ .The molar conductivity of 0.5 M solution of the electrolyte in S m² mol^-1 is:

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5 x 10^-3

5 x 10^-4

5 x 10³

5 x 10²

answer is A.

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Detailed Solution

$\begin{array}{l} R = 50 Ω \\ k = 1.4 \times 10^{- 2} S {cm}^{- 1} \\ R = p \cdot \frac{l}{a} = \frac{1}{k} \cdot \frac{l}{a} \\ \frac{l}{a} = R \times k = 50 \times 1.4 \times 10^{- 2} \\ R = 280 Ω for 0.5 M \\ R = p \cdot \frac{l}{a} = \frac{1}{280} \times 50 \times 1.4 \times 10^{- 2} \\ = \frac{1}{280} \times 70 \times 10^{- 2} \\ = 2.5 \times 10^{- 3} S {cm}^{- 1} \\ λ_{m} = \frac{k \times 1000}{M} = \frac{2.5 \times 10^{- 3} \times 1000}{0.5} = 5 \times 10^{- 4} {Sm}^{2} {mol}^{- 1} \end{array}$