What is the cell constant of $\frac{\mathrm{N}}{10} \mathrm{KCl}$ solution at $25^{\circ} \mathrm{C}$, if…

What is the cell constant of $\frac{\mathrm{N}}{10} \mathrm{KCl}$ solution at $25^{\circ} \mathrm{C}$, if conductivity and resistance of a solution is $0.0112 \Omega^{-1} \mathrm{~cm}^{-1}$ and $55 \cdot 0 \Omega$ respectively?
  1. $0.616 \mathrm{~cm}^{-1}$
  2. $0 \cdot 491 \mathrm{~cm}^{-1}$
  3. $2 \cdot 0 \mathrm{~cm}^{-1}$
  4. $0 \cdot 2 \mathrm{~cm}^{-1}$

Solution

$\begin{aligned} \mathrm{k}=0.0112 \Omega^{-1} \mathrm{~cm}^{-1} &, \mathrm{R}=55.0 \Omega \\ \text { Cell constant, } \mathrm{b} &=\mathrm{k} \times \mathrm{R} \\ &=0.0112 \Omega^{-1} \mathrm{~cm}^{-1} \times 55.0 \Omega \\ \therefore \mathrm{b} &=0.616 \mathrm{~cm}^{-1} \end{aligned}$

Asked in: MHT CET 2020 (14 Oct Shift 1)

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