Q. Resistance of 0.2 M solution of an electrolyte is$\ 50\ \mathrm{\Omega}\$. The specific conductance of the solution is 1.3 S $m^{- 1}$. If resistance of the 0.4M solution of the same electrolyte is 260Ω , its molar conductivity is
A) $6250\ Sm^{2}\ mol^{- 1}$
B) $6.25\ \times \ 10^{- 4}\ Sm^{2}\ mol^{- 1}$
C) $625\ \times \ 10^{- 4}\ Sm^{2}\ mol^{- 1}$
D) $62.5\ Sm^{2}\ mol^{- 1}$
Explanation
(b)
Specific conductance = conductance × cell constant
${1.3\ Sm}^{- 1} = \frac{1}{50}\ S\ \times cell\ constant$
∴ Cell constant
$= \ 1.3\ \times \ {50\ m}^{- 1} = \ {65\ m}^{- 1} = (65/100){cm}^{- 1}\$
Molar conductivity = $\frac{1000\ \times conductance\ \times cell\ constant}{molarity}\$
$= \ \frac{1000}{0.4}\ \times \ \frac{1}{260}\ \times \ \frac{65}{100} = 6.25\ S{cm}^{2}\ {mol}^{- 1}$
$= \ 6.25\ \times \ 10^{- 4}\ {Sm}^{2}\ {mol}^{- 1}\$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.