Question

The conductivity (in ${\Omega }^{-1}{m}^{-1}$) of $K_{2}S{O}_4$ solution is: ( write your answer to nearest integer)

Answer 1

For KCl solution $\kappa =\frac{1}{R}\frac{l}{a}=$$0.2768oh{m}^{-1}{m}^{-1}=\frac{1}{82.40ohm}\frac{l}{a}$
So,$\frac{l}{a}=22.8m^{-1}$
For $K_{2}S{O}_4$
$\kappa =\frac{1}{326.0ohm}\times 22.8m^{-1}=0.07oh{m}^{-1}{m}^{-1}$