K L Chugh Solutions for Chapter: Electrochemistry, Exercise 12: Follow Up Problems

The resistance of a conductivity cell containing $0.0001 \mathrm{M} \mathrm{KCl}$ solution at $298 \mathrm{K}$ is $1500 \mathrm{\Omega }.$ What is the cell constant if conductivity of $0.001 \mathrm{M} \mathrm{KCl}$ solution at $298 \mathrm{K}$ is $0.146\times {10}^{-3} \mathrm{S}{ \mathrm{cm}}^{-1}.$

The conductivity of $0.20 \mathrm{M}$ solution of $\mathrm{KCl}$ at $298 \mathrm{K}$ is $0.0248 \mathrm{S}{ \mathrm{cm}}^{-1}.$ Calculate its molar conductivity.

$0.5 \mathrm{N}$ solution of a salt placed between two $\mathrm{Pt}$ electrodes $2.0 \mathrm{cm}$ apart and having area of cross-section $2.5 {\mathrm{cm}}^2$ has resistance of $25$ ohms. Calculate the conductance and cell constant.

Calculate the resistivity, conductivity and molar conductivity of a $0.05 \mathrm{M} \mathrm{NaOH}$ solution if the resistance of its column having diameter $1 \mathrm{cm}$ and length $50 \mathrm{cm}$ is $5.55\times {10}^3 \mathrm{ohm}.$

Hint. $\mathrm{A}={\mathrm{\pi r}}^2=3.14\times {\left(0.5\right)}^2{ \mathrm{cm}}^2=0.785{ \mathrm{cm}}^2$ or $0.785\times {10}^{-4}{\mathrm{m}}^2; \mathrm{l}=50 \mathrm{cm}$ or $0.5\mathrm{m}, \mathrm{\rho }=\frac{\mathrm{R}·\mathrm{A}}{\mathrm{l}}$