Question

The conductivity of $0.02\mathrm{N}$ solution of a cell of $\mathrm{KCl}$ at ${25}^{\circ }\mathrm{C}$ is $2.765\times {10}^{-3}S{\mathrm{cm}}^{-1}$. If the resistance of a cell containing this solution is $4\times {10}^3$ ohm, what will be the cell constant?

Answer 1

Correct Answer: 11.06

Explanation:
Given:Normality concentration of $KCl$ solution $=0.02N$Conductivity of $KCl$ solution $K=2.765\times {10}^{-3}{\mathrm{Scm}}^{-1}$Temperature, $T={25}^{\circ }\mathrm{C}$Resistance of the cell, $R=4000\mathrm{Ohm}$We have to find the value of the cell constant.Conductivity, $k$ is given as:$\kappa =\frac{1}{R}\left[\frac{1}{A}\right]$where, $\overline{K}=$ Conductivity$R=$ Resistance$I=$ Distance$A=$ Cross section Area$\frac{1}{A}=$ Cell constantThus, from equation (i) $\frac{1}{A}=2.765\times {10}^{-3}\times 4\times {10}^3$$=2.765\times {10}^{-3}{\mathrm{Scm}}^{-1}\times 4\times {10}^3{S}^{-1}$$=2.765\times 4{\mathrm{cm}}^{-1}$$=11.06{\mathrm{cm}}^{-1}$Therefore cell constant is $11.06{\mathrm{cm}}^{-1}$.Hence, the correct answer is $11.06{\mathrm{cm}}^{-1}$.