The conductivity of $0.012 \ M$ $NaBr$ solution is $2.67 \times 10^{-4} \ S \ cm^{-1}$. What is its molar conductivity?

  • A
    $26.7 \ S \ cm^2 \ mol^{-1}$
  • B
    $32.04 \ S \ cm^2 \ mol^{-1}$
  • C
    $12.2 \ S \ cm^2 \ mol^{-1}$
  • D
    $22.2 \ S \ cm^2 \ mol^{-1}$

Explore More

Similar Questions

The specific conductance $(k)$ of $0.02 \ M$ aqueous acetic acid solution at $298 \ K$ is $1.65 \times 10^{-4} \ S \ cm^{-1}$. The degree of dissociation $(\alpha)$ of acetic acid is: [Given: $\lambda_{H^{+}}^{\infty} = 349.1 \ S \ cm^2 \ mol^{-1}$ and $\lambda_{CH_3COO^{-}}^{\infty} = 40.9 \ S \ cm^2 \ mol^{-1}$]

Electrical conductance due to all the ions in $1 \ cm^3$ of a given solution is called as

The resistance of a conductivity cell of $0.1 \ M$ $KCl$ solution is $120 \ \Omega$ and conductivity is $1.64 \times 10^{-4} \ S \ cm^{-1}$. What is the value of cell constant (in $cm^{-1}$)?

Given below are two statements:
Assertion $(A)$: Conductivity of an electrolyte decreases on dilution.
Reason $(R)$: On dilution,the number of ions per unit volume increases.
The correct answer is:

How can the ionic conductivity of an unknown solution be determined?

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