The resistance of a conductivity cell containing $0.001MKCl$ solution at $298K$ is $1500\Omega$. What is the cell constant if conductivity of $0.001MKCl$ solution at $298K$ is $0.146\times {10}^{-3}Sc{m}^{-1}$.

The correct option is A $0.219c{m}^{-1}$
Given:
$\text{Conductivity,}\kappa =0.146\times {10}^{-3}Sc{m}^{-1}$
$\text{Resistance, R}=1500\Omega$

$\text{Conductance, G}=\frac{1}{\text{Resistance}}$
$\text{Resistance, R}=\rho \frac{l}{A}$
where,
$\rho$ is resistivity/specific resistance
$l$ is lemgth of electrolytic cell
$A$ is cross section of cell
Thus,
$G=\frac{1}{\rho }\times \frac{A}{l}$

The inverse of​ resistivity is called conductivity / specific conductance $\left(\kappa \right)$.​

$G=\kappa \times \frac{A}{l}$

$\kappa =G\times \frac{l}{A}$

$\frac{l}{a}$ is the cell constant $\left(G^{\ast }\right)$ of the cell.
$\therefore$

$\kappa =G\times G^{\ast }$
$\kappa =\frac{1}{R}\times G^{\ast }$
$G^{\ast }=\kappa \times R$

$\text{Cell constant,}{G}^{\ast }=0.146\times {10}^{-3}{\Omega }^{-1}c{m}^{-1}\times 1500\Omega$

$\text{Cell constant,}{G}^{\ast }=0.219{cm}^{-1}$