Correct order of limiting molar conductivity for cations in water at $298 \ K$ is $:$

  • A
    $H^{+} > Na^{+} > K^{+} > Ca^{2+} > Mg^{2+}$
  • B
    $H^{+} > Ca^{2+} > Mg^{2+} > K^{+} > Na^{+}$
  • C
    $Mg^{2+} > H^{+} > Ca^{2+} > K^{+} > Na^{+}$
  • D
    $H^{+} > Na^{+} > Ca^{2+} > Mg^{2+} > K^{+}$

Explore More

Similar Questions

When a certain conductivity cell was filled with $0.1 \ M \ KCl$,it had a resistance of $85 \ \Omega$ at $25 \ ^oC$. When the same cell was filled with an aqueous solution of $0.052 \ M$ unknown electrolyte,the resistance was $96 \ \Omega$. Calculate the molar conductivity of the unknown electrolyte at this concentration ............. $\Omega^{-1} \ cm^2 \ mol^{-1}$ (Given: Specific conductance of $0.1 \ M \ KCl = 1.29 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$)

The conductivity of $0.3 \ M$ solution of $KCl$ at $298 \ K$ is $0.0627 \ S \ cm^{-1}$. What is its molar conductivity?

Electrolytic conduction differs from metallic conduction in that in the case of electrolytic conduction:

The equivalent conductances of $CH_3COONa$,$HCl$,and $CH_3COOH$ at infinite dilution are $91$,$426$,and $391 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ respectively. What is the equivalent conductance of $NaCl$ at infinite dilution?

What is the molar conductivity at zero concentration in $\Omega^{-1} \ cm^2 \ mol^{-1}$ for aluminium sulphate,if the molar ionic conductivities at zero concentration of $Al^{3+}$ and $SO_4^{2-}$ are $189 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $50.1 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively?

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