The resistance of $0.01 \ N$ $NaCl$ solution at $25 \ ^oC$ is $200 \ \Omega$. The cell constant of the conductivity cell is $1 \ cm^{-1}$. The equivalent conductance is:

  • A
    $5 \times 10^2 \ \Omega^{-1} cm^2 eq^{-1}$
  • B
    $6 \times 10^3 \ \Omega^{-1} cm^2 eq^{-1}$
  • C
    $7 \times 10^4 \ \Omega^{-1} cm^2 eq^{-1}$
  • D
    $8 \times 10^5 \ \Omega^{-1} cm^2 eq^{-1}$

Explore More

Similar Questions

What is the conductivity of $0.05 \ M \ BaCl_2$ solution if its molar conductivity is $220 \ \Omega^{-1} \ cm^2 \ mol^{-1}$?

The limiting molar conductivities $\wedge ^0$ for $NaCl$,$KBr$,and $KCl$ are $126$,$152$,and $150 \ S \ cm^2 \ mol^{-1}$ respectively. The $\wedge ^0$ for $NaBr$ is ............ $S \ cm^2 \ mol^{-1}$.

The conductivity of $0.04 \ M \ BaCl_2$ solution is $0.0112 \ \Omega^{-1} \ cm^{-1}$ at $25^{\circ} C$. What is its molar conductivity?

Molar conductivities $(\Lambda ^o_m)$ at infinite dilution of $NaCl$,$HCl$ and $CH_3COONa$ are $126.4$,$425.9$ and $91.0 \ S \ cm^2 \ mol^{-1}$ respectively. $(\Lambda ^o_m)$ for $CH_3COOH$ will be .......... $S \ cm^2 \ mol^{-1}$.

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

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