The values of $\mu^{\infty}$ for $NH_4Cl$,$NaOH$,and $NaCl$ are $129.8$,$248.1$,and $126.4 \ \Omega^{-1} \ cm^{2} \ mol^{-1}$ respectively. Calculate $\mu^{\infty}$ for $NH_4OH$ solution.

  • A
    $285.3$
  • B
    $278.6$
  • C
    $251.5$
  • D
    $243.9$

Explore More

Similar Questions

Molar conductivities at infinite dilution $\wedge_{m}^{\circ}$ for $Ba(OH)_2$,$BaCl_2$ and $NH_4Cl$ are $457.0$,$240.6$ and $213.0 \ S \ cm^2 \ mol^{-1}$ respectively. The $\wedge_{m}^{\circ}$ for ammonium hydroxide (in $S \ cm^2 \ mol^{-1}$) is (in $.2$)

Write the formula to calculate conductivity by using a conductivity cell.

If the values of $\Lambda_{\infty}$ of $NH_4Cl$,$NaOH$ and $NaCl$ are $130$,$217$ and $109 \ ohm^{-1} \ cm^2 \ equiv^{-1}$ respectively,the $\Lambda_{\infty}$ of $NH_4OH$ in $ohm^{-1} \ cm^2 \ equiv^{-1}$ is:

State Kohlrausch's law of independent migration of ions and explain its applications.

The property which decreases with dilution is:

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