Question

The specific conductivity of $0.1NKCI$ solution at ${20}^{o}C$ is $0.212{oho}^{-1}{cm}^{-1}$. This solution was found to offer resistance of $55$ ohm. Find cell constant of the conductivity cell. If a solution of $1.0MAB,$ a uni-univalent electrolyte offers a resistance of $25$ ohm, using the same conductivity cell, find its equivalent conductance?

Answer 1

(i) $\because x=\frac{l}{A}=\frac{k}{c}=\frac{0.212oh{m}^{-1}c{m}^{-1}}{1/55}=11.66c{m}^{-1}\left(cellcontant\right)$

(ii) $C=\frac{k}{x}\Rightarrow k=xc(11.66c{m}^{-1}\times \frac{1}{25})oh{m}^{-1}\therefore {\Lambda }_{eq}=(k\times \frac{1000}{m\times n_f})=(11.66\times \frac{1}{25}\times \frac{1000}{1\times 1})=466.4{\Omega }^{-1}c{m}^{2}e{q}^{-1}$