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

Metallic and ionic conductivity depends on which factors?

The molar conductivity of $0.4 \ M \ KCl$ solution is $2.5 \times 10^2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$. What is the resistivity of the solution?

At $291 \ K$,a saturated solution of $BaSO_4$ was found to have a specific conductivity of $3.648 \times 10^{-6} \ \Omega^{-1} \ cm^{-1}$ and that of the water used is $1.25 \times 10^{-6} \ \Omega^{-1} \ cm^{-1}$. If the ionic conductances of $Ba^{2+}$ and $SO_4^{2-}$ are $110$ and $136.6 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively,the solubility of $BaSO_4$ at $291 \ K$ will be (Atomic masses: $Ba=137, S=32, O=16$)

What is the mathematical relationship between the radius $(r)$ of a cylindrical conductor and its cross-sectional area $(A)$?

At $298 \ K$,the equivalent conductance of $0.1 \ M$ acetic acid is $5.20 \ S \ cm^{2} \ eq.^{-1}$. Calculate the degree of dissociation of acetic acid at this concentration in $\%$. Given: $\lambda^{\infty} (H^{+}) = 349.8 \ S \ cm^{2} \ mol^{-1}$ and $\lambda^{\infty} (CH_{3}COO^{-}) = 40.9 \ S \ cm^{2} \ mol^{-1}$.

Difficult
View Solution

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