Question

Each of the two platinum electrodes having area $64m{m}^2$ of a conductivity cell are separated by $8mm$. The resistance of the cell containing $7.5\times {10}^{-3}M$ $KCl$ solution at $298K$ is $1250ohm$. Calculate :
(i) Cell constant
(ii) Conductivity and
(iii) Molar conductivity

Answer 1

Area $=64m{m}^2=0.64sq.cm$

Distance $=8mm=0.8cm$

Resistance $=1250\Omega$

Concentration $=7.5\times {10}^{-3}M$

i) Cell constant $=\frac{Distance}{Area}=\frac{0.8}{0.64}=1.25c{m}^{-1}$

ii) Conductivity $K=\frac{1}{\rho }$

$\rho =R\frac{A}{l}=1250\times \frac{0.64}{0.8}=1000$

$\therefore k=\frac{1}{\rho }=\frac{1}{1000}={10}^{-3}Sc{m}^{-1}$

iii) Molar conductivity $=\frac{k\times 1000}{morality}=\frac{{10}^{-3}\times 1000}{7.5\times {10}^{-3}}$

$=133.33c{m}^{2}mo{l}^{-1}$