(C) The cell constant $G^*$ is given by $G^* = \frac{l}{a} = \frac{2.2 \, cm}{4.4 \, cm^2} = 0.5 \, cm^{-1} = 50 \, m^{-1}$.Conductivity $\kappa = \fr

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) The cell constant $G^*$ is given by $G^* = \frac{l}{a} = \frac{2.2 \, cm}{4.4 \, cm^2} = 0.5 \, cm^{-1} = 50 \, m^{-1}$.Conductivity $\kappa = \frac{1}{R} \times G^* = \frac{1}{50 \, \Omega} \times 50 \, m^{-1} = 1 \, S \, m^{-1}$.Molar conductivity $\Lambda_m = \frac{\kappa}{C}$.Given $C = 0.5 \, M = 0.5 \, mol \, L^{-1} = 0.5 \times 10^3 \, mol \, m^{-3} = 500 \, mol \, m^{-3}$.$\Lambda_m = \frac{1 \, S \, m^{-1}}{500 \, mol \, m^{-3}} = 0.002 \, S \, m^2 \, mol^{-1}$.Thus,option $C$ is correct.

Class 12 Chemistry Electrochemistry Conductor and Conductance and Cell constant

Medium

The resistance of a $0.5 \, M$ solution of an electrolyte in a cell was found to be $50 \, \Omega$. If the electrodes in the cell are $2.2 \, cm$ apart and have an area of $4.4 \, cm^2$,then the molar conductivity (in $S \, m^2 \, mol^{-1}$) of the solution is:

A
$0.2$
B
$0.02$
C
$0.002$
D
none of these

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Electrochemistry

Subtopic

Conductor and Conductance and Cell constant

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:

MediumAP EAMCET 2011

View Solution

Which of the following statements is incorrect with respect to metallic or electronic conductivity?

EasyGUJCET 2014

View Solution

The conductivity of $0.02 \text{ M}$ $KCl$ solution at $298 \text{ K}$ is $0.0123 \text{ } \Omega^{-1} \text{ cm}^{-1}$. If the resistance of the cell containing this solution is $120 \text{ } \Omega$, what is the value of the cell constant (in $\text{ cm}^{-1}$)?

DifficultMHT CET 2026

View Solution

The molar conductivities $\Lambda_{NaOH}^o$ and $\Lambda_{HCl}^o$ at infinite dilution in water at $25 ^\circ C$ are $91.0$ and $426.2 \ S \ cm^2/mol$ respectively. To calculate $\Lambda_{CH_3COOH}^o$,the additional value required is:

DifficultAIEEE 2006

View Solution

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.

Easy

View Solution

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

The resistance of a $0.5 \, M$ solution of an electrolyte in a cell was found to be $50 \, \Omega$. If the electrodes in the cell are $2.2 \, cm$ apart and have an area of $4.4 \, cm^2$,then the molar conductivity (in $S \, m^2 \, mol^{-1}$) of the solution is:

Similar Questions

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:

Which of the following statements is incorrect with respect to metallic or electronic conductivity?

The conductivity of $0.02 \text{ M}$ $KCl$ solution at $298 \text{ K}$ is $0.0123 \text{ } \Omega^{-1} \text{ cm}^{-1}$. If the resistance of the cell containing this solution is $120 \text{ } \Omega$, what is the value of the cell constant (in $\text{ cm}^{-1}$)?

The molar conductivities $\Lambda_{NaOH}^o$ and $\Lambda_{HCl}^o$ at infinite dilution in water at $25 ^\circ C$ are $91.0$ and $426.2 \ S \ cm^2/mol$ respectively. To calculate $\Lambda_{CH_3COOH}^o$,the additional value required is:

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.

Vedclass Test Series

Exam Paper Generator

Online Exam Module