(B) The formula for specific conductivity $(\kappa)$ is given by: $\kappa = \frac{1}{R} \times \left(\frac{\ell}{a}\right)$,where $\frac{\ell}{a}$ is

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(B) The formula for specific conductivity $(\kappa)$ is given by: $\kappa = \frac{1}{R} \times \left(\frac{\ell}{a}\right)$,where $\frac{\ell}{a}$ is the cell constant.Given: $\kappa = 0.012 \, \Omega^{-1} \, cm^{-1}$ and $R = 56 \, \Omega$.Rearranging the formula to find the cell constant: $\text{Cell constant} = \kappa \times R$.$\text{Cell constant} = 0.012 \, \Omega^{-1} \, cm^{-1} \times 56 \, \Omega = 0.672 \, cm^{-1}$.Therefore,the correct option is $B$.

Class 12 Chemistry Electrochemistry Conductor and Conductance and Cell constant

Easy

The specific conductivity of $N/10$ $KCl$ solution at $20 \, ^oC$ is $0.012 \, \Omega^{-1} \, cm^{-1}$ and the resistance of the solution in the cell at $20 \, ^oC$ is $56 \, \Omega$. The cell constant is ........... $cm^{-1}$.

A
$4.616$
B
$0.672$
C
$2.173$
D
$3.324$

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Class 12

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Electrochemistry

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Conductor and Conductance and Cell constant

The limiting molar conductivities $\lambda^{0}$ for $NaCl$,$KBr$,and $KCl$ are $126$,$152$,and $150 \ S \ cm^{2} \ mol^{-1}$ respectively. What is the $\lambda^{0}$ for $NaBr$ in $S \ cm^{2} \ mol^{-1}$?

Easy

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What benefit do we get by forming a thin layer of $Pt$ particles on the electrodes of a conductivity cell?

Medium

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What is the value of $\Lambda _m^o$ $(NaCl)$ at $298 \ K$?

Medium

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At $25^o C$,the molar conductivity of a $0.1 \, N \, CH_3COOH$ solution is $80 \, S \, cm^2 \, mol^{-1}$ and the molar conductivity at infinite dilution is $400 \, S \, cm^2 \, mol^{-1}$. What is the degree of dissociation $(\alpha)$ of $CH_3COOH$?

Easy

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The conductivity of a solution of concentration $0.1 \ mol \ L^{-1}$ of a weak monobasic acid $(HA)$ (in $S \ cm^{-1}$) is (Given $\Lambda_{HA}^{\circ}=400 \ S \ cm^2 \ mol^{-1}$ and degree of dissociation $(\alpha)$ of $HA=0.02$)

MediumTS EAMCET 2023

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The specific conductivity of $N/10$ $KCl$ solution at $20 \, ^oC$ is $0.012 \, \Omega^{-1} \, cm^{-1}$ and the resistance of the solution in the cell at $20 \, ^oC$ is $56 \, \Omega$. The cell constant is ........... $cm^{-1}$.

Similar Questions

The limiting molar conductivities $\lambda^{0}$ for $NaCl$,$KBr$,and $KCl$ are $126$,$152$,and $150 \ S \ cm^{2} \ mol^{-1}$ respectively. What is the $\lambda^{0}$ for $NaBr$ in $S \ cm^{2} \ mol^{-1}$?

What benefit do we get by forming a thin layer of $Pt$ particles on the electrodes of a conductivity cell?

What is the value of $\Lambda _m^o$ $(NaCl)$ at $298 \ K$?

At $25^o C$,the molar conductivity of a $0.1 \, N \, CH_3COOH$ solution is $80 \, S \, cm^2 \, mol^{-1}$ and the molar conductivity at infinite dilution is $400 \, S \, cm^2 \, mol^{-1}$. What is the degree of dissociation $(\alpha)$ of $CH_3COOH$?

The conductivity of a solution of concentration $0.1 \ mol \ L^{-1}$ of a weak monobasic acid $(HA)$ (in $S \ cm^{-1}$) is (Given $\Lambda_{HA}^{\circ}=400 \ S \ cm^2 \ mol^{-1}$ and degree of dissociation $(\alpha)$ of $HA=0.02$)

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