The resistance of 0.5 N solution of an electro-lyte in a conductivity cell was found to be 45 ohms. If the electrodes in the cell are 2.2 cm apart and have an area of 3.8 cm² then the equivalent conductance (in Scm² eq^–1) of a solution is 

$$\begin{gathered} Specific conduc\tan ce \left(k\right)\hspace{0.17em}= \frac{1}{R}\times \left(\frac{l}{A}\right) \\ = \frac{1}{45}\times \frac{2.2}{3.8} \\ = 0.0128\hspace{0.17em}S c{m}^{-1}. \end{gathered}$$

$$\begin{gathered} Molar conduc\tan ce \left({\wedge }_m\right) =\frac{k\times 1000}{molarity } \\ = \frac{0.0128 \times 1000}{0.5} \\ = 25.73 S c{m}^2 e{q}^{-1.} \end{gathered}$$