Question

The resistance of solution of a salt occupying a volume between two platinum electrodes 1.8 cm apart and \[5.4c{m^2}\] in area was found to be 32 ohms. Calculate the conductivity of the solution.

Answer 1

Hint: Since, we have the area of cross section ‘A’ and the distance between the two platinum electrodes ‘l’ which are given in the question, and also, we have the resistance of the solution ‘R’. We know that the conductivity of the solution is given by the following equation, $K = \dfrac{1}{R} \times \dfrac{l}{A}$, so, substituting the value in then equation, we can calculate the conductivity of the solution, κ.

Complete step by step answer:
Given in the question are,
Resistance, \[R = 32\] ohms
Length, \[l = 1.8cm\]
Area, \[a = 5.4c{m^2}\]
A conductivity cell which is composed of two platinum electrodes and the electrodes having area of cross section equal to A and are separated by distance ‘l’. Hence, the solution confined between the two electrodes is a column of length l and area of cross section A. The resistance for this column of solution is given by the following equation.
 $R = \rho \dfrac{ l }{A} = lKA$
The quantity $\dfrac{l}{A}$ is called cell constant and is denoted by the symbol \[{G^*}\].
 $G = \dfrac{l}{A}R\,K$
Using the values of cell constant and resistance of solution in the cell, we can find the conductivity of the solution by using the following equation.
 $K = \dfrac{{cell\,cons \tan t}}{{R = \dfrac{G}{R}}}$
Conductivity, $K = \dfrac{1}{R} \times \dfrac{ l }{A}$
 $ = \dfrac{1}{{32}} \times \dfrac{{1.8}}{{5.4}}$
 \[ = 0.0104166c{m^{ - 1}}oh{m^{ - 1}}\]

Therefore, the conductivity of the solution is \[0.0104166c{m^{ - 1}}oh{m^{ - 1}}\].

Note: The conductance of electricity by ions present in solutions is called electrolytic conductance or ionic conductance and this conductivity of electrolytic solution depends on the nature of electrolyte, size of ions produced and their solvation, nature of solvent and its viscosity, concentration of electrolyte and temperature (conductance increases with increase in temperature).