Resistance of 0.2 M solution of an electrolyte 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 m²mol^-1 is:

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$5 \times 10^{- 3}$

$5 \times 10^{3}$

$5 \times 10^{2}$

$5 \times 10^{- 4}$

answer is A.

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Detailed Solution

Given for 0.2 M solution
$\begin{array}{l} R = 50 Ω \\ κ = 1 .4 {Sm}^{- 1} = 1 .4 \times 10^{- 2} {Scm}^{- 1} \end{array}$
$Now, R = ρ \frac{l}{a} = \frac{l}{κ} \times \frac{1}{a}$

$\Rightarrow \frac{l}{a} = R \times K = 50 \times 1 .4 \times 10^{- 2}$
For 0.5 M solution
$\begin{array}{l} R = 280 Ω \\ κ = ? \\ \frac{l}{a} = 50 \times 1 .4 \times 10^{- 2} \\ \Rightarrow R = ρ \frac{l}{a} = \frac{1}{κ} \times \frac{l}{a} \\ \Rightarrow κ = \frac{l}{280} \times 50 \times 1 .4 \times 10^{- 2} \\ = \frac{l}{280} \times 70 \times 10^{- 2} \\ = 2 .5 \times 10^{- 3} {Scm}^{- 1} \end{array}$
Now, $Λ_{m} = \frac{κ \times 1000}{M} = \frac{2 .5 \times 10^{- 3} \times 1000}{0 .5}$
$= 5 S {cm}^{2} {mol}^{- 1} = 5 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$