Question

$1.0N$ solution of a salt surrounding two platinum electrodes $2.1cm$ apart and $4.2$ sq cm in area was found to offer a resistance of $50ohm$. Calculate the equivalent conductivity of the solution.

Answer 1

Given, $l-2.1cm,a=4.2sqcm,R=50ohm$
Specific conductance, $k=\frac{1}{a}\cdot \frac{1}{R}$
or $k=\frac{2.1}{4.2}\times \frac{1}{50}=0.01oh{m}^{-1}c{m}^{-1}$
Equivalent conductivity $=k\times V$
$V=$ the volume containing $1g-equivalent=1000mL$
So, Equivalent conductivity $=0.01\times 1000$
$=10oh{m}^{-1}c{m}^{2}e{q}^{-1}$.