Chemistry - Electrochemistry Question with Solution | TestHub

Conductivity $\kappa$ is the reciprocal of resistivity $\varrho$, so $\kappa =\frac{1}{\varrho }$.

Resistance $R$ is given by $R=\varrho \frac{l}{a}$.

Taking the reciprocal, $\frac{1}{R}=\frac{1}{\varrho }\frac{a}{l}$.

Since conductance $G=\frac{1}{R}$ and $\kappa =\frac{1}{\varrho }$, we get $G=\kappa \frac{a}{l}$.

Rearranging for $\kappa$, we get $\kappa =G\frac{l}{a}=\frac{1}{R}\frac{l}{a}$.

Molar conductivity $\Lambda$ is defined as $\Lambda =\frac{\kappa }{c}$, which gives $\kappa =\Lambda \cdot c$.

The expression $\kappa =\frac{1}{R}\frac{a}{l}$ is incorrect.

Answer: $\kappa =\frac{1}{\mathrm{R}}\frac{\mathrm{a}}{l}$