The molar conductivity of $0.02 \ M$ $KCl$ solution is $410 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ at $25^{\circ} C$. Calculate its conductivity?

  • A
    $8.2 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
  • B
    $2.8 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
  • C
    $4.1 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$
  • D
    $5.4 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$

Explore More

Similar Questions

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 $.......... \, \times 10^{-3} \, cm^{-1}$.

The specific conductance of a saturated solution of silver bromide is $\kappa \ S \ cm^{-1}$. The limiting ionic conductances for $Ag^{+}$ and $Br^{-}$ are $x$ and $y$ $S \ cm^2 \ mol^{-1}$ respectively. Then the solubility of silver bromide (in $g/L$) will be $(Ag = 108, Br = 80)$.

Difficult
View Solution

The increase in equivalent conductance of an electrolyte solution with dilution is due to the increase in

What is the molar conductivity of a $0.05 \ M$ solution of sodium hydroxide,if its conductivity is $0.0118 \ S \ cm^{-1}$ at $298 \ K$?

The molar conductivities $(\lambda_{m}^0)$ at infinite dilution of $KBr$,$HBr$,and $KNH_2$ are $120.5$,$420.6$,and $90.48 \ S \ cm^2 \ mol^{-1}$ respectively. Find the value of $\lambda_{m}^0$ for $NH_3$.

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