Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Conductivity of a solution is defined as the conductance of a solution of $1 \, cm$ in length and area of cross-section $1 \, cm^2$. The inverse of resistivity is called conductivity or specific conductance. It is represented by the symbol $\kappa$. If $\rho$ is resistivity,then we can write:
$\kappa = \frac{1}{\rho}$
The conductivity of a solution at any given concentration is the conductance $(G)$ of one unit volume of solution kept between two platinum electrodes with the unit area of cross-section and at a distance of unit length.
i.e.,$G = \kappa \frac{a}{l} = \kappa \cdot 1 = \kappa$
(since $a = 1, l = 1$)
Conductivity always decreases with a decrease in concentration,both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.
Molar conductivity:
Molar conductivity of a solution at a given concentration is the conductance of volume $V$ of a solution containing $1 \, mole$ of the electrolyte kept between two electrodes with the area of cross-section $A$ and distance of unit length.
$\Lambda_m = \kappa \cdot \frac{A}{l}$
Now,$l = 1$ and $A = V$ (volume containing $1 \, mole$ of the electrolyte).
$\therefore \Lambda_m = \kappa \cdot V$
Molar conductivity increases with a decrease in concentration. This is because the total volume $V$ of the solution containing one mole of the electrolyte increases on dilution.
The variation of $\Lambda_m$ with $\sqrt{c}$ for strong and weak electrolytes is shown in the following plot:

Explore More

Similar Questions

If the distance between two $Pt$ electrodes is $2 \, cm$,the cross-sectional area is $4.0 \, cm^2$,and the resistance is $25 \, \Omega$,find the molar conductivity of a $0.5 \, M$ solution.

If the specific conductance of $1 \, M \, H_2SO_4$ solution is $26 \times 10^{-2} \, S \, cm^{-1}$,then the equivalent conductivity would be

What is the cell constant of $\frac{N}{10}$ $KCl$ solution at $25^{\circ} C$,if conductivity and resistance of a solution is $0.0112 \ \Omega^{-1} cm^{-1}$ and $55.0 \ \Omega$ respectively (in $cm^{-1}$)?

Which of the following statements is incorrect?

What is the conductivity of $0.02 \text{ M}$ $AgNO_3$ solution having cell constant $1.2 \text{ cm}^{-1}$ and resistance $95.0 \text{ }\Omega$?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo