Calculate the molar conductivity of a $0.02 \, M$ solution if its conductivity is $2.06 \times 10^{-3} \, S \, cm^{-1}$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The formula for molar conductivity is $\Lambda_m = \frac{\kappa \times 1000}{C}$.
Given conductivity $\kappa = 2.06 \times 10^{-3} \, S \, cm^{-1}$ and concentration $C = 0.02 \, M$.
Substituting the values: $\Lambda_m = \frac{2.06 \times 10^{-3} \times 1000}{0.02} = \frac{2.06}{0.02} = 103 \, S \, cm^2 \, mol^{-1}$.

Explore More

Similar Questions

Calculate the cell constant of a conductivity cell containing $0.01 \ M \ AgNO_3$ solution having a resistance of $1440 \ \Omega$ and a conductivity of $0.001262 \ \Omega^{-1} \ cm^{-1}$. (in $cm^{-1}$)

Write a note on Kohlrausch law of independent migration of ions and limiting molar conductivity $\Lambda_{m}^{o}$ of strong electrolyte.

Difficult
View Solution

What is the conductivity of $0.02 \ M$ $AgNO_3$ solution having cell constant $1.1 \ cm^{-1}$ and resistance $94.5 \ \Omega$?

The resistance of a decimolar solution of $NaCl$ is $30 \ \Omega$. Calculate the conductivity of the solution if the cell constant is $0.33 \ cm^{-1}$.

The specific conductivity of $N/10$ $KCl$ solution at $20 \, ^oC$ is $0.012 \, \Omega^{-1} \, cm^{-1}$ and the resistance of the solution in the cell at $20 \, ^oC$ is $56 \, \Omega$. The cell constant is ........... $cm^{-1}$.

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