Question

A $0.05N$ solution of a salt occupying a volume between two platinum electrodes separated by a distance of $1.72cm$ and having an area of $4.5c{m}^2$ has a resistance of $250ohm$.
Calculate the equivalent conductance of the solution.

Answer 1

Specific conductance $=Conductance\times Cellconstant$
$K=C\times \frac{l}{A}$
$=\frac{1}{R}\times \frac{l}{A}$
$=\frac{1}{250}\times \frac{1.72}{4.5}$
$=1.5288\times {10}^{-3}oh{m}^{-1}c{m}^{-1}$
${\wedge }_e=K\times \frac{1000}{N}$
$=1.5288\times {10}^{-3}\times \frac{1000}{0.05}$
$=30.56oh{m}^{-1}c{m}^{2}e{q}^{-1}$.