Resistance of 0.2 M solution of an electrolyte is $50\Omega$.The specific conductance of the solution is $1.4S{m}^{-1}$. The resistance of 0.5 M solution of the same electrolyte is $280\Omega$. The molar conductivity of 0.5 M solution of the electrolyte in $S{m}^{2}mo{l}^{-1}$ is

The explanation for the correct option:

Option(B)$5\times {10}^{-4}$

1. In order to solve the problem, calculate the value of the cell constant of the first solution and then use this value of the cell constant to calculate the value of k of the second solution. Afterward, finally, calculate molar conductivity using values of k and m.

2. For the first solution,
$k=1.4S{m}^{-1},R=50\Omega ,M=0.2$
Specific conductance $\left(K\right)=\frac{1}{R}\times \frac{l}{A}$

$1.4=\frac{1}{50}\times \frac{1}{A}$

$\therefore \frac{1}{A}=50\times 1.4=70m^{-1}$

3. For second solution, $R=280,\frac{l}{A}=50\times 1.4m^{-1}$

$K=\frac{1}{280}\times 1.4\times 50=\frac{1}{4}$

Now, molar conductivity can be calculated as:

${\text{∧}}_m=\frac{k\times 1000}{M}\times {\left({10}^{-2}\right)}^3$

${\text{∧}}_m=\frac{1}{4}\times \frac{1000}{0.5}\times {10}^{-6}$

$=500\times {10}^{-6}=5\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

Hence, Option(B)

$5\times {10}^{-4}S{m}^{2}mo{l}^{-1}$

is the correct option The molar conductivity of 0.5 M solution of the electrolyte in

$S{m}^{2}mo{l}^{-1}$

is

$5\times {10}^{-4}S{m}^{2}mo{l}^{-1}$