Question

The resistance of $0.5\text{M}$ solution of an electrolyte in a cell was found to be $50\Omega$. If the electrodes in the cell are $2.2\text{cm}$ apart and have an area of $4.4{\text{cm}}^2$ then the molar conductivity (in ${\Omega }^{-1}{\text{cm}}^2{\text{mol}}^{-1}$) of the solution is:

Answer 1

Specific conductivity is
$K=\frac{1}{R}\times \frac{l}{a}=\frac{1}{50}\times \frac{2.2}{4.4}=0.01{\Omega }^{-1}c{m}^{-1}$
Molar conductivity,
${\wedge }_m=K\times \frac{1000}{C}=0.01\times \frac{1000}{0.5}=20{\Omega }^{-1}{\text{cm}}^2{\text{mol}}^{-1}$