Question

A conductivity cell filled with $0.01MKCl$ gives at ${25}^{o}C$ the resistance of $604ohms$. The conductivity of $KCl$ at ${25}^{o}C$ is $0.00141{\Omega }^{-1}c{m}^{-1}$. The same cell filled with $0.001MAgN{O}_3$ gives a resistance of $6529ohms$. Calculate the molar conductivity of $0.001MAgN{O}_3$ solution at ${25}^{o}C$.

Answer 1

Cell constant$\left(K\right)$ would be same for both cell as the both cells are same,

For $1st$ cell-

$\lambda =\frac{1000\times K}{M}$ or $K=\frac{\lambda \times M}{1000}=1.41\times {10}^{-8}$

Put the value of $K$ for the $2nd$ cell-

So ${\lambda }_M=\frac{1000\times K}{M}=\frac{{10}^3\times 1.41\times {10}^{-8}}{{10}^{-3}}=1.41\times {10}^{-2}$