(A) The formula for equivalent conductivity is $\lambda_{eq} = \frac{1000 \times \kappa}{N}$,where $\kappa$ is the specific conductance and $N$ is the

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

$1.3 \times 10^2 \, S \, cm^2 \, eq^{-1}$

(A) The formula for equivalent conductivity is $\lambda_{eq} = \frac{1000 \times \kappa}{N}$,where $\kappa$ is the specific conductance and $N$ is the normality of the solution.For $H_2SO_4$,the molarity is $1 \, M$,so the normality $N = M \times \text{n-factor} = 1 \times 2 = 2 \, N$.Substituting the values: $\lambda_{eq} = \frac{1000 \times 26 \times 10^{-2}}{2} = \frac{260}{2} = 130 \, S \, cm^2 \, eq^{-1} = 1.3 \times 10^2 \, S \, cm^2 \, eq^{-1}$.

Class 12 Chemistry Electrochemistry Conductor and Conductance and Cell constant

Medium

If the specific conductance of $1 \, M \, H_2SO_4$ solution is $26 \times 10^{-2} \, S \, cm^{-1}$,then the equivalent conductivity would be

A
$1.3 \times 10^2 \, S \, cm^2 \, eq^{-1}$
B
$1.6 \times 10^2 \, S \, cm^2 \, eq^{-1}$
C
$13 \, S \, cm^2 \, mol^{-1}$
D
$1.3 \times 10^3 \, S \, cm^2 \, mol^{-1}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Electrochemistry

Subtopic

Conductor and Conductance and Cell constant

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

$A$ conductivity cell has been calibrated with $0.01 \ M$ $1:1$ electrolyte solution (specific conductance,$k = 1.25 \times 10^{-3} \ S \ cm^{-1}$) in the cell and the measured resistance was $800 \ \Omega$ at $25^{\circ}C$. The cell constant will be (in $cm^{-1}$)

MediumWBJEE 2013

View Solution

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:

Medium

View Solution

Which of the following graphs represents the variation of molar conductance $(\Lambda_m)$ of an electrolyte with $\sqrt{C}$ for weak and strong electrolytes,respectively?

Medium

View Solution

$\wedge _m^o$ for $NaCl$,$HCl$ and $NaA$ are $126.4$,$425.9$ and $100.5 \ S \ cm^2 \ mol^{-1}$,respectively. If the conductivity of $0.001 \ M \ HA$ is $5 \times 10^{-5} \ S \ cm^{-1}$,degree of dissociation of $HA$ is

DifficultJEE MAIN 2019

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

If the specific conductance of $1 \, M \, H_2SO_4$ solution is $26 \times 10^{-2} \, S \, cm^{-1}$,then the equivalent conductivity would be

Similar Questions

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 $,cm^{-1}$)

$A$ conductivity cell has been calibrated with $0.01 \ M$ $1:1$ electrolyte solution (specific conductance,$k = 1.25 \times 10^{-3} \ S \ cm^{-1}$) in the cell and the measured resistance was $800 \ \Omega$ at $25^{\circ}C$. The cell constant will be (in $cm^{-1}$)

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:

Which of the following graphs represents the variation of molar conductance $(\Lambda_m)$ of an electrolyte with $\sqrt{C}$ for weak and strong electrolytes,respectively?

$\wedge _m^o$ for $NaCl$,$HCl$ and $NaA$ are $126.4$,$425.9$ and $100.5 \ S \ cm^2 \ mol^{-1}$,respectively. If the conductivity of $0.001 \ M \ HA$ is $5 \times 10^{-5} \ S \ cm^{-1}$,degree of dissociation of $HA$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module