The specific conductivity of $\mathrm{N} / 10 \mathrm{KCl}$ solution at $20^{\circ} \mathrm{C}$ is $0.212…

The specific conductivity of $\mathrm{N} / 10 \mathrm{KCl}$ solution at $20^{\circ} \mathrm{C}$ is $0.212 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$ and the resistance of the cell containing this solution at $20^{\circ} \mathrm{C}$ is $55 \mathrm{ohm}$. The cell constant is
  1. $4.616 \mathrm{~cm}^{-1}$
  2. $11.66 \mathrm{~cm}^{-1}$
  3. $\quad 2.173 \mathrm{~cm}^{-1}$
  4. $3.324 \mathrm{~cm}^{-1}$

Solution

Cell constant
$=\kappa \times \mathrm{R}$
$=0.212 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1} \times 55 \mathrm{ohm}$
$=11.66 \mathrm{~cm}^{-1}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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