Question

The resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is $100\Omega$. The conductivity of this solution is $1.29S{m}^{-1}$. The resistance of the same cell, when filled with 0.02 M of the same solution, is $520\Omega$. The molar conductivity of 0.02 M solution of the electrolyte will be: (Take $\frac{129}{520}=0.248$)

Answer 1

Cell constant can be calculated using 0.1 M KCl.
$=K\times R$
$=0.0129\times 100$
$=1.29c{m}^{-1}$
Calculating conductivity of 0.02 M KCl solution.
K $=\frac{Cellconstant}{Resistance}$
$=\frac{1.29}{520}$
$=2.48\times {10}^{-3}Sc{m}^{-1}$
Therefore, molar conductivity:
$\wedge m={10}^{3}k/M$
$={10}^3\times 2.48\times \frac{{10}^{-3}}{0.02}$
$=124Sc{m}^{2}mo{l}^{-1}$