Question

A conductivity cell filled with $0.01M$ $KCl$ gives at ${25}^{o}C$ the resistance of $604ohms$. The conductivity of $KCl$ at ${25}^{o}C$ is $0.00141{\Omega }^{-1}{cm}^{-1}$. The same cell filled with $0.001M$ $Ag{NO}_3$ gives a resistance of $6529ohms$. Calculate the molar conductivity of $0.001M$ $Ag{NO}_3$ solution at ${25}^{o}C$

Answer 1

We know, $K=K\times 1/R$

$\left(KCl\right)K_1{R}_1=K_2{R}_2\left(Ag{NO}_3\right)$

$0.00141\times 604=K_2\times 6529$

$K_2=1.30\times {10}^{-4}{r}^{-1}{cm}^{-1}$

molar conductivity, ${\lambda }_m=\frac{K}{C}\times 1000$

$=\frac{1.30\times {10}^{-4}}{0.001}\times 1000$

$=0.13\times 1000$

$=130{Scm}^2{mol}^{-1}$