Question 11 Mark
The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm-1. Calculate its molar conductivity.
Answer
View full question & answer→Molar conductivity $\left(\Lambda_{ m }\right)=\frac{\kappa \times 1000}{\text { Molarity }}$
$\kappa=$ conductivity $=0.0248 S cm ^{-1}$, Molarity $=0.20 M$
Hence, $\Lambda_{ m }=\frac{0.0248\left(S cm ^{-1}\right) \times 1000 cm^3}{0.20 mol}$
$\Lambda_{ m }=124 S cm ^2 mol^{-1}$
$\kappa=$ conductivity $=0.0248 S cm ^{-1}$, Molarity $=0.20 M$
Hence, $\Lambda_{ m }=\frac{0.0248\left(S cm ^{-1}\right) \times 1000 cm^3}{0.20 mol}$
$\Lambda_{ m }=124 S cm ^2 mol^{-1}$