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 m2 mol-1 is :

Electrochemistry PYQ — JEE Main 2014 · Shift 0 · Q1 with Solution

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Correct answer: C

For conductivity measurements,

\kappa = G\left(\frac{l}{A}\right)=\frac{1}{R}\left(\frac{l}{A}\right)

where $\dfrac{l}{A}$ is the cell constant.

Given for the $0.2\,\text{M}$ solution:

So,

\frac{l}{A} = \kappa_1 R_1 = 1.4 \times 50 = 70\,\text{m}^{-1}

Given for the second solution:

Now,

\kappa_2 = \frac{1}{R_2}\left(\frac{l}{A}\right) = \frac{70}{280} = 0.25\,\text{S m}^{-1}

For molar conductivity,

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

with $c$ in $\text{mol m}^{-3}$ .

Given concentration:

0.5\,\text{M} = 0.5\,\text{mol L}^{-1} = 500\,\text{mol m}^{-3}

\Lambda_m = \frac{0.25}{500} = 5 \times 10^{-4}\,\text{S m}^2\text{ mol}^{-1}

\boxed{5 \times 10^{-4}\,\text{S m}^2\text{ mol}^{-1}}

This corresponds to Option C.

Stored correct answer: C

Our derived answer: C

So the answer agrees with the stored answer.