Question

Resistance of a conductivity cell filled with 0.15 mol $L^{-1}$ NaCl solution is 50 $\Omega$. If resistance of the same cell when filled with 0.02 mol $L^{-1}$ NaCl solution is 500 $\Omega$, calculate the conductivity of 0.02 mol $L^{-1}$ NaCl solution. (The conductivity of 0.15 mol $L^{-1}$ NaCl solution is 1.5 s/m).

Answer 1

$\kappa =\frac{1}{R}\frac{l}{a}$
Here $\kappa$ is conductivity, R is resistance and $\frac{l}{a}$ is the cell constant.
For $0.15M$ solution
$\kappa =\frac{1}{R}\frac{l}{a}$
$1.5s/m=\frac{1}{50\Omega }\frac{l}{a}$
$\frac{l}{a}=75/m$
For $0.02M$ solution
$\kappa =\frac{1}{R}\frac{l}{a}$

$\kappa =\frac{1}{500\Omega }\times 75/m=0.150S/m$