Question

Specific conductivity of $0.20M$ solution of $KCl$ at $298K$ is $0.025$ $S$ ${cm}^{-1}$. Calculate its molar conductivity.

Answer 1

The molar conductivity ${\Lambda }_m=\frac{1000\kappa }{C}$
${\Lambda }_m=\frac{1000c{m}^3/d{m}^{3}0.025S/cm}{0.20mold{m}^{-3}}$
${\Lambda }_m=125Smo{l}^{-1}c{m}^2$
The molar conductivity is $125Smo{l}^{-1}c{m}^2$