Resistance of a conductivity cell filled with a solution of an electrolyte of concentration $$0.1\,M$$ is $$100\,\Omega$$. The conductivity of this solution is $$1.29\,S\,m^{-1}$$. Resistance of the same cell when filled with $$0.2\,M$$ of the same solution is $$520\,\Omega$$. The molar conductivity of $$0.02\,M$$ solution of the electrolyte will be

The conductivity cell is first calibrated with the $$0.1\,M$$ solution.

Cell constant, $$G^{\ast}$$ Using $$\kappa = G^{\ast} R$$, $$G^{\ast}= \kappa R = 1.29\,S\,m^{-1}\times 100\,\Omega = 129\,m^{-1}$$

For the second filling (resistance $$R_2 = 520\,\Omega$$):

Conductivity $$\kappa_2 = \frac{G^{\ast}}{R_2}= \frac{129}{520}=0.248\,S\,m^{-1}$$

Molar conductivity of the solution at concentration $$C_2 = 0.20\,M$$ (remember $$0.20\,M = 0.20\,mol\,L^{-1}=200\,mol\,m^{-3}$$): $$\Lambda_m = \frac{\kappa_2}{C_2\,(mol\,m^{-3})}= \frac{0.248}{200}=1.24\times 10^{-3}\,S\,m^{2}\,mol^{-1}$$

Expressing in the form required by the options: $$1.24\times 10^{-3}\,S\,m^{2}\,mol^{-1}=12.4\times 10^{-4}\,S\,m^{2}\,mol^{-1}$$

Hence the molar conductivity is Option D which is: $$12.4 \times 10^{-4}\,S\,m^{2}\,mol^{-1}$$