A conductivity cell containing $5 \times 10^{-4} \mathrm{M} \mathrm{NaCl}$ solution develops resistance…

A conductivity cell containing $5 \times 10^{-4} \mathrm{M} \mathrm{NaCl}$ solution develops resistance $14000 \mathrm{ohms}$ at $25^{\circ} \mathrm{C}$. Calculate the conductivity of solution if the cell constant is $0.84 \mathrm{~cm}^{-1}$
  1. $6.0 \times 10^{-5} \Omega^{-1} \mathrm{~cm}^{-1}$
  2. $3.0 \times 10^{-5} \Omega^{-1} \mathrm{~cm}^{-1}$
  3. $9.0 \times 10^{-5} \Omega^{-1} \mathrm{~cm}^{-1}$
  4. $12.0 \times 10^{-5} \Omega^{-1} \mathrm{~cm}^{-1}$

Solution

$\begin{aligned} \mathrm{k} & =\frac{\text { cell constant }}{\mathrm{R}} \\ & =\frac{0.84 \mathrm{~cm}^{-1}}{14000 \Omega} \\ & =6.0 \times 10^{-5} \Omega^{-1} \mathrm{~cm}^{-1}\end{aligned}$

Asked in: MHT CET 2023 (12 May Shift 2)

Practice more Electrochemistry questions on Aicharya