The resistance of $\frac{1}{10} \ M$ solution is $2.5 \times 10^3 \ \Omega$. What is the molar conductivity of the solution? (Cell constant $= 1.25 \ cm^{-1}$)

  • A
    $3.5 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
  • B
    $5.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
  • C
    $2.5 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
  • D
    $2.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$

Explore More

Similar Questions

Define $\Lambda _m = \frac{\kappa \times 1000}{C}$ where $C$ is concentration in $mol \ L^{-1}$ and $\kappa$ is conductivity in $S \ cm^{-1}$.

Which of the following graphs represents the variation of molar conductance $(\Lambda_m)$ of an electrolyte with $\sqrt{C}$ for weak and strong electrolytes,respectively?

What is the conductivity of $0.01 \ M$ $NaCl$ solution if the resistance and cell constant of the $NaCl$ solution are $375 \ \Omega$ and $0.5 \ cm^{-1}$ respectively at $298 \ K$?

Which of the following has maximum molar conductivity?

Difficult
View Solution

The resistance of a conductivity cell containing $0.01 \, M \, KCl$ solution at $298 \, K$ is $1750 \, \Omega$. If the conductivity of $0.01 \, M \, KCl$ solution at $298 \, K$ is $0.152 \times 10^{-3} \, S \, cm^{-1}$,then the cell constant of the conductivity cell is $.......... \, \times 10^{-3} \, cm^{-1}$.

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