(C) Given: $M = 0.5 \, mol \, L^{-1} = 500 \, mol \, m^{-3}$,$R = 50 \, \Omega$,$l = 2.2 \, cm = 0.022 \, m$,$A = 4.4 \, cm^2 = 4.4 \times 10^{-4} \,

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

(C) Given: $M = 0.5 \, mol \, L^{-1} = 500 \, mol \, m^{-3}$,$R = 50 \, \Omega$,$l = 2.2 \, cm = 0.022 \, m$,$A = 4.4 \, cm^2 = 4.4 \times 10^{-4} \, m^2$. Cell constant $G^* = \frac{l}{A} = \frac{0.022 \, m}{4.4 \times 10^{-4} \, m^2} = 50 \, m^{-1}$. Conductivity $\kappa = \frac{1}{R} \times G^* = \frac{1}{50 \, \Omega} \times 50 \, m^{-1} = 1 \, S \, m^{-1}$. Molar conductivity $\Lambda_m = \frac{\kappa}{M} = \frac{1 \, S \, m^{-1}}{500 \, mol \, m^{-3}} = 0.002 \, S \, m^2 \, mol^{-1}$.

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
$0.001$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Electrochemistry

Subtopic

Conductor and Conductance and Cell constant

Calculate molar conductivity at infinite dilution for $NaBr$ if molar conductivity at infinite dilution for $NaCl$,$KBr$ and $KCl$ are $126$,$152$ and $150 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively.

DifficultMHT CET 2022

View Solution

What is the molar conductivity of $0.1 \ M$ $NaCl$ if its conductivity is $1.06 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$?

EasyMHT CET 2020

View Solution

If $\Lambda^{0}_{NaOAc} = 91 \, S \, cm^{2} \, mol^{-1}$ and $\Lambda^{0}_{HCl} = 496.2 \, S \, cm^{2} \, mol^{-1}$,what additional value is required to calculate $\Lambda^{0}_{HOAc}$?

Medium

View Solution

$\Lambda _m^\circ$ for $NaCl$,$HCl$ and $NaAc$ are $126.4$,$425.9$ and $91.0 \ S \ cm^2 \ mol^{-1}$ respectively. Calculate $\Lambda _m^\circ$ for $HAc$.

Medium

View Solution

The specific conductance of a $0.01 \, M \, KCl$ solution at $25 \, ^{\circ}C$ is $0.0014 \, \Omega^{-1} \, cm^{-1}$. Its equivalent conductance is ..... (in $\Omega^{-1} \, cm^{2} \, eq^{-1}$).

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

Calculate molar conductivity at infinite dilution for $NaBr$ if molar conductivity at infinite dilution for $NaCl$,$KBr$ and $KCl$ are $126$,$152$ and $150 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively.

What is the molar conductivity of $0.1 \ M$ $NaCl$ if its conductivity is $1.06 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$?

If $\Lambda^{0}_{NaOAc} = 91 \, S \, cm^{2} \, mol^{-1}$ and $\Lambda^{0}_{HCl} = 496.2 \, S \, cm^{2} \, mol^{-1}$,what additional value is required to calculate $\Lambda^{0}_{HOAc}$?

$\Lambda _m^\circ$ for $NaCl$,$HCl$ and $NaAc$ are $126.4$,$425.9$ and $91.0 \ S \ cm^2 \ mol^{-1}$ respectively. Calculate $\Lambda _m^\circ$ for $HAc$.

The specific conductance of a $0.01 \, M \, KCl$ solution at $25 \, ^{\circ}C$ is $0.0014 \, \Omega^{-1} \, cm^{-1}$. Its equivalent conductance is ..... (in $\Omega^{-1} \, cm^{2} \, eq^{-1}$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module