Resistance of a decimolar solution between two electrodes $0.02 \ m$ apart and $0.0004 \ m^2$ in area was found to be $50 \ \Omega.$ Specific conductance $(\kappa)$ is .......... $S \ m^{-1}$

  • A
    $0.1$
  • B
    $1$
  • C
    $10$
  • D
    $4 \times 10^{-4}$

Explore More

Similar Questions

The resistance of a $0.1 \, N$ solution of a salt is $2.5 \times 10^{3} \, \Omega$. If the cell constant is $1.15 \, cm^{-1}$,what will be the equivalent conductance of the solution in $\Omega^{-1} \, cm^{2} \, eq^{-1}$ (in $.6$)?

The conductivity of a solution of concentration $0.1 \ mol \ L^{-1}$ of a weak monobasic acid $(HA)$ (in $S \ cm^{-1}$) is (Given $\Lambda_{HA}^{\circ}=400 \ S \ cm^2 \ mol^{-1}$ and degree of dissociation $(\alpha)$ of $HA=0.02$)

At $25\,^{\circ}C$,the specific conductivity of a normal solution of $KCl$ is $0.002765\,S\,cm^{-1}$. The resistance of the cell is $400\,\Omega$. The cell constant is: (in $,cm^{-1}$)

Difficult
View Solution

Assertion : On increasing dilution,the specific conductance keeps on increasing.
Reason : On increasing dilution,the degree of ionization of a weak electrolyte increases and the mobility of ions also increases.

If the molar conductivities (in $S \ cm^2 \ mol^{-1}$) of $NaCl$,$KCl$,and $NaOH$ at infinite dilution are $126$,$150$,and $250$,respectively,the molar conductivity of $KOH$ (in $S \ cm^2 \ mol^{-1}$) 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