(B) Given: Resistance $R = 31.6\,\Omega$,Cell constant $G^* = 0.367\,cm^{-1}$,Molarity $M = 0.5\,M$.Conductance $C = \frac{1}{R} = \frac{1}{31.6} \app

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

(B) Given: Resistance $R = 31.6\,\Omega$,Cell constant $G^* = 0.367\,cm^{-1}$,Molarity $M = 0.5\,M$.Conductance $C = \frac{1}{R} = \frac{1}{31.6} \approx 0.0316\,S$.Specific conductance $\kappa = C \times G^* = 0.0316\,S \times 0.367\,cm^{-1} \approx 0.0116\,S\,cm^{-1}$.Molar conductance $\Lambda_m = \frac{\kappa \times 1000}{M} = \frac{0.0116 \times 1000}{0.5} = \frac{11.6}{0.5} = 23.2\,S\,cm^2\,mol^{-1}$.

Class 12 Chemistry Electrochemistry Conductor and Conductance and Cell constant

Difficult

$A$ $0.5\,M\,NaOH$ solution offers a resistance of $31.6\,\Omega$ in a conductivity cell at room temperature. What shall be the approximate molar conductance of this $NaOH$ solution if the cell constant of the cell is $0.367\,cm^{-1}$? (in $S\,cm^2\,mol^{-1}$)

AIIMS 2010 All AIIMS

A
$234$
B
$23.2$
C
$4645$
D
$5464$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Electrochemistry

Subtopic

Conductor and Conductance and Cell constant

Previous Year

AIIMS 2010 Paper All AIIMS years →

What is the $SI$ unit of conductivity?

EasyMHT CET 2017

View Solution

$A$ conductivity cell dipped in $0.05 \ M$ $KCl$ has resistance $600 \ \Omega$. If conductivity is $0.0012 \ \Omega^{-1} \ cm^{-1}$,what is the value of cell constant (in $cm^{-1}$)?

EasyMHT CET 2024

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 $\text{ M}$)

DifficultMHT CET 2026

View Solution

Given $l/a = 0.5 \ cm^{-1}$,$R = 50 \ \Omega$,$N = 1.0$. The equivalent conductance of the electrolytic cell is ................ $\Omega^{-1} \ cm^2 \ gm \ eq^{-1}$.

Medium

View Solution

The molar conductivity of $0.4 \ M \ KCl$ solution is $2.5 \times 10^2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$. What is the resistivity of the solution?

MediumMHT CET 2021

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

$A$ $0.5\,M\,NaOH$ solution offers a resistance of $31.6\,\Omega$ in a conductivity cell at room temperature. What shall be the approximate molar conductance of this $NaOH$ solution if the cell constant of the cell is $0.367\,cm^{-1}$? (in $S\,cm^2\,mol^{-1}$)

Similar Questions

What is the $SI$ unit of conductivity?

$A$ conductivity cell dipped in $0.05 \ M$ $KCl$ has resistance $600 \ \Omega$. If conductivity is $0.0012 \ \Omega^{-1} \ cm^{-1}$,what is the value of cell constant (in $cm^{-1}$)?

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 $\text{ M}$)

Given $l/a = 0.5 \ cm^{-1}$,$R = 50 \ \Omega$,$N = 1.0$. The equivalent conductance of the electrolytic cell is ................ $\Omega^{-1} \ cm^2 \ gm \ eq^{-1}$.

The molar conductivity of $0.4 \ M \ KCl$ solution is $2.5 \times 10^2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$. What is the resistivity of the solution?

Vedclass Test Series

Exam Paper Generator

Online Exam Module