Question

The specific conductance of $0.1NKCl$ solution at ${23}^{o}C$ is $0.012oh{m}^{-1}c{m}^{-1}$. The resistane of cell containing the solution at the same temperature was found to be $55ohm.$ The cell constant will be:

• A. $0.142{cm}^{-1}$
• B. $0.66{cm}^{-1}$
• C. $0.198{cm}^{-1}$
• D. $1.12{cm}^{-1}$

Answer 1

The correct option is B $0.66{cm}^{-1}$

Given:
$\text{Specific conductance,}\kappa =0.012oh{m}^{-1}c{m}^{-1}$
$\text{Resistance, R}=55ohm$

$\text{Conductance, G}=\frac{1}{\text{Resistance}}$
$\text{Resistance, R}=\rho \frac{l}{A}$
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.

For a electrolytic conductor, cell constant depends on the distance between the electrodes (l) and their area of cross-section (A),
The cell constant is usually determined by measuring the resistance of a cell containing the solution of known conductivity.
$\therefore$
$\kappa =G\times G^{\ast }$
$\kappa =\frac{1}{R}\times G^{\ast }$
$G^{\ast }=\kappa \times R$

$\text{Cell constant,}{G}^{\ast }=0.012oh{m}^{-1}c{m}^{-1}\times 55ohm$
$\text{Cell constant,}{G}^{\ast }=0.66{cm}^{-1}$