The resistance of conductivity cell filled with 0.1 M KCl solution is 100 ohm and conductivity is $1.70…
The resistance of conductivity cell filled with 0.1 M KCl solution is 100 ohm and conductivity is $1.70 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}$. What is the cell constant of the cell?
$1.70 \mathrm{~cm}^{-1}$
$0.017 \mathrm{~cm}^{-1}$
$0.058 \mathrm{~cm}^{-1}$
$0.012 \mathrm{~cm}^{-1}$
Solution
$G^*=\kappa \times R$
Where:
- $G^*$ is the cell constant $\left(\mathrm{cm}^{-1}\right)$,
- $\kappa$ is the conductivity $\left(\mathrm{S} \mathrm{cm}^{-1}\right)$,
- $R$ is the resistance $(\Omega)$.
Given:
- $\kappa=1.70 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}$,
- $R=100 \Omega$.
Calculation:
$G^*=\left(1.70 \times 10^{-4}\right) \times 100=0.017 \mathrm{~cm}^{-1}$
Final Answer:
The cell constant is:
$0.017 \mathrm{~cm}^{-1}$
Hence, the correct option is (2) $0.017 \mathrm{~cm}^{-1}$.