The resistance of a $1 \, N$ solution of a salt is $50 \, \Omega$. The two platinum electrodes in the solution are $2.1 \, cm$ apart and each has an area of $4.2 \, cm^2$. Calculate the equivalent conductivity of the solution.

  • A
    $10$
  • B
    $11$
  • C
    $12$
  • D
    $13$

Explore More

Similar Questions

The specific conductance of a saturated solution of silver bromide is $\kappa \ S \ cm^{-1}$. The limiting ionic conductances for $Ag^{+}$ and $Br^{-}$ are $x$ and $y$ $S \ cm^2 \ mol^{-1}$ respectively. Then the solubility of silver bromide (in $g/L$) will be $(Ag = 108, Br = 80)$.

Difficult
View Solution

Calculate the concentration of silver nitrate solution if molar conductivity and conductivity of silver nitrate at $25^{\circ} \text{C}$ are respectively $120 \text{ } \Omega^{-1} \text{cm}^2 \text{mol}^{-1}$ and $0.0024 \text{ } \Omega^{-1} \text{cm}^{-1}$. (in $\text{ M}$)

The values of conductivity of some materials at $298.15 \ K$ in $S \ m^{-1}$ are $2.1 \times 10^3$,$1.0 \times 10^{-16}$,$1.2 \times 10$,$3.91$,$1.5 \times 10^{-2}$,$1 \times 10^{-1}$,$1.0 \times 10^3$. The number of conductors among the materials is............

Let $C_{NaCl}$ and $C_{BaSO_4}$ be the conductances (in $S$) measured for saturated aqueous solutions of $NaCl$ and $BaSO_4,$ respectively,at a temperature $T.$ Which of the following is false?

The molar conductivity of a $0.02 \ M$ solution of an electrolyte is $124 \times 10^{-4} \ S \ m^2 \ mol^{-1}$. What is the resistance of the same solution (in ohms),kept in a cell with a cell constant of $129 \ m^{-1}$?

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