Question

A conductivity cell was filled with a 0.02 M KCl solution which has a specific conductance of $2.768\times {{10}^{-3}}\text{ }oh{{m}^{-1}}c{{m}^{-1}}$. If its resistance is 82.4 ohm at ${{25}^{\circ }}C$ , the cell constant is

Answer 1

Hint In order to solve this question, we will first find out the relation between cell constant and specific conductivity by using the formula $R=\rho \dfrac{l}{a}$ . Then from relation we can find the cell constant value.

Complete Step by step solution:
- Let us first find out the relation between cell constant and conductivity or we specific conductivity.
- We know that $R=\rho \dfrac{l}{a}$
And l/a is a cell constant.
- Hence, we can write it as l/a = $R/\rho$
As we also know that $l/\rho =G$ , that is the reciprocal of resistivity is conductivity.
- Hence, we can write the final relation as:
l/a= G X R
= $2.768\times {{10}^{-3}}\text{ }oh{{m}^{-1}}c{{m}^{-1}}$ $\times 82.4ohm=0.2281c{{m}^{-1}}$

- Hence, we can conclude that the value of cell constant is 0.2281 per cm.

Additional information:
- As we know that cell constant is the ratio of the distance between the cell electrodes to the area that the electrodes occupy.
- It is basically measured for a cell containing a solution whose conductivity is already known to us. This is the reason due to which we use KCl solution as its conductivity can be known to accurately measure at various temperatures and concentrations.

Note:
- As we know that for a particular cell the value of l/a is basically constant. This value can be also determined experimentally. Mainly, cell constant when multiplied with the measured current gives the electrical conductivity of the solution.