The resistance of a cell ( cell constant $= 1.1 {\mathrm{cm}}^{-1})$ containing $\frac{\mathrm{N}}{50}\mathrm{KCl}$ was found to be $400 \mathrm{ohms}.$ The equivalent conductivity of $\mathrm{KCl}$ at this dilution is :

Cell constant $=1.1{\mathrm{cm}}^{-}$ ;

$\mathrm{R}=400\mathrm{ohms};\wedge =?,$

normality $=\frac{\mathrm{N}}{50}.$

Specific conductance, $\mathrm{K}=\frac{1}{\mathrm{R}}\times$ cell constant$=\frac{1}{400\mathrm{ohm}}\times 1.1{\mathrm{cm}}^{-}$

But equivalent conductance,

$\begin{array}{l}\wedge =\frac{\mathrm{K}\times 1000{\mathrm{cm}}^3}{\text{ Normality }}\\ =\frac{1}{400\text{ ohm }}\times \frac{1.1{\mathrm{cm}}^{-}\times 1000{\mathrm{cm}}^3}{(1/50)\text{ g. equiv }}\end{array}$

$=137.5 {\mathrm{ohm}}^{-}{\mathrm{cm}}^2{\text{ (g. equiv) }}^{-}$