Resistance of 0.2 M solution of an electrolyte is 50 $$\Omega$$. 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 $$\Omega$$. The molar conductivity of 0.5 M solution of the electrolyte in S m$$^2$$ mol$$^{-1}$$ is:

We start with the information for the first solution. Its resistance is given as $$R_1 = 50\; \Omega$$ and its specific conductance (also called conductivity) is $$\kappa_1 = 1.4\; \text{S m}^{-1}$$.

The relation that links the conductivity $$\kappa$$, the conductance $$G$$ and the cell constant $$\frac{\ell}{A}$$ is first stated:

$$\kappa = G \left(\frac{\ell}{A}\right)$$ and $$G = \frac{1}{R}$$.

So, for the first solution we have

$$G_1 = \frac{1}{R_1} = \frac{1}{50}\; \text{S}$$.

Substituting this value in the conductivity formula, we get the cell constant:

$$\frac{\ell}{A} = \frac{\kappa_1}{G_1} = \kappa_1 \times R_1 = 1.4 \times 50 = 70\; \text{m}^{-1}.$$

The cell constant is a property of the conductivity cell, so it remains the same when we measure the second solution.

Now we turn to the second solution whose resistance is $$R_2 = 280\; \Omega.$$

First we calculate its conductance:

$$G_2 = \frac{1}{R_2} = \frac{1}{280}\; \text{S}.$$

Using the same formula $$\kappa = G \left(\frac{\ell}{A}\right)$$, we find its conductivity $$\kappa_2$$:

$$\kappa_2 = G_2 \left(\frac{\ell}{A}\right) = \frac{1}{280} \times 70 = \frac{70}{280} = 0.25\; \text{S m}^{-1}.$$

The concentration of the second solution is given as $$0.5\; \text{M}$$. Because $$1\; \text{M} = 1\; \text{mol L}^{-1} = 1000\; \text{mol m}^{-3}$$, we convert:

$$c = 0.5\; \text{M} = 0.5 \times 1000 = 500\; \text{mol m}^{-3}.$$

The formula that connects molar conductivity $$\Lambda_m$$ with conductivity $$\kappa$$ and concentration $$c$$ (in $$\text{mol m}^{-3}$$) is

$$\Lambda_m = \frac{\kappa}{c}.$$

Substituting the values we have just found,

$$\Lambda_m = \frac{0.25}{500}\; \text{S m}^2 \text{ mol}^{-1} = 0.0005\; \text{S m}^2 \text{ mol}^{-1}.$$

Writing the answer in scientific notation,

$$\Lambda_m = 5 \times 10^{-4}\; \text{S m}^2 \text{ mol}^{-1}.$$

Hence, the correct answer is Option A.