Q. Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100Ω. The conductivity of this solution is 1.29 S $m^{- 1}.$ Resistance of the same cell when filled with 0.2 M of the same solution is 520Ω. The molar conductivity of 0.02 M solution of the electrolyte will be
A) $124 \times 10^{- 4}\ S\ m^{2}\ {mol}^{- 1}$
B) $1240 \times 10^{- 4}\ S\ m^{2}\ {mol}^{- 1}$
C) $1.24 \times 10^{- 4}\ S\ m^{2}\ {mol}^{- 1}$
D) $12.4 \times 10^{- 4}\ S\ m^{2}\ {mol}^{- 1}$
Explanation
(d)
$R\ = \ 100\mathrm{\Omega}$
$\kappa\ = \ \frac{1}{R}\left( \frac{l}{a} \right)$
$\frac{l}{a}\$(cell constant) $= \ 1.29\ \times \ 100m^{- 1}\$
Given,$\ R\ = \ 520\mathrm{\Omega};\ C\ = \ 0.2\ M\$
$µ$ (molar conductivity) =?
$\mu\ = \ \kappa\ \times \ V$
($\kappa$ can be calculated as $\kappa = \ \frac{1}{R}\left( \frac{l}{a} \right)$ now cell constant is known)
Hence, $\mu\ = \ \frac{1}{520}\ \times 129\ \times \ \frac{1000}{0.2}\ \times \ 10^{- 6}\ m^{3}$
$= \ 12.4 \times 10^{- 4}\ S\ m^{2}\ {mol}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.