(D) The conductivity $\kappa$ is related to resistance $R$ and cell constant $G^{*}$ by the formula: $\kappa = \frac{1}{R} \cdot G^{*}$For the same co

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(D) The conductivity $\kappa$ is related to resistance $R$ and cell constant $G^{*}$ by the formula: $\kappa = \frac{1}{R} \cdot G^{*}$For the same conductivity cell,$G^{*}$ is constant,therefore $\kappa \cdot R = G^{*} = \text{constant}$.For the $KCl$ solution: $\kappa_{KCl} \cdot R_{KCl} = 0.14 \times 4.19 = 0.5866 \, S$.For the $HCl$ solution: $\kappa_{HCl} \cdot R_{HCl} = 0.5866 \, S$.$\kappa_{HCl} = \frac{0.5866}{1.03} \approx 0.5695 \, S m^{-1}$.Converting to the required units: $0.5695 \, S m^{-1} = 56.95 \times 10^{-2} \, S m^{-1}$.Rounding to the nearest integer,we get $57 \times 10^{-2} \, S m^{-1}$.

Class 12 Chemistry Electrochemistry Conductor and Conductance and Cell constant

Difficult

$A$ $KCl$ solution of conductivity $0.14 \, S m^{-1}$ shows a resistance of $4.19 \, \Omega$ in a conductivity cell. If the same cell is filled with an $HCl$ solution,the resistance drops to $1.03 \, \Omega$. The conductivity of the $HCl$ solution is $....... \, \times 10^{-2} \, S m^{-1}$. (Round off to the Nearest Integer).

JEE MAIN 2021 All JEE MAIN

A
$28$
B
$107$
C
$67$
D
$57$

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Electrochemistry

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Conductor and Conductance and Cell constant

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At $298 \text{ K}$, the specific conductance $(\kappa)$ of a $0.0020 \text{ M } NaCl$ solution is $2.50 \times 10^{-4} \text{ S cm}^{-1}$. Calculate the molar conductivity $(\Lambda_m)$ of the solution in $\text{S cm}^2 \text{mol}^{-1}$.

DifficultMHT CET 2026

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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

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The equivalent conductance at infinite dilution $\wedge^0$ for electrolytes $BA$ and $CA$ are $140$ and $120 \ S \ cm^2 \ eq^{-1}$,respectively. The equivalent conductance at infinite dilution for $BX$ is $198 \ S \ cm^2 \ eq^{-1}$. The $\wedge^0$ (in $S \ cm^2 \ eq^{-1}$) of $CX$ is:

Medium

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What will be the molar conductance at infinite dilution for $NH_4OH$ if at infinite dilution the molar conductances of $Ba(OH)_2$,$BaCl_2$,and $NH_4Cl$ are $523.28$,$280.0$,and $129.8 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively?

Medium

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At infinite dilution, the equivalent conductances of $Ba^{2+}$ and $Cl^{-}$ ions are $127 \ \Omega^{-1} \ cm^2 \ eqv^{-1}$ and $76 \ \Omega^{-1} \ cm^2 \ eqv^{-1}$ respectively. The equivalent conductivity of $BaCl_2$ at infinite dilution is:

Medium

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At $298 \text{ K}$, the specific conductance $(\kappa)$ of a $0.0020 \text{ M } NaCl$ solution is $2.50 \times 10^{-4} \text{ S cm}^{-1}$. Calculate the molar conductivity $(\Lambda_m)$ of the solution in $\text{S cm}^2 \text{mol}^{-1}$.

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 equivalent conductance at infinite dilution $\wedge^0$ for electrolytes $BA$ and $CA$ are $140$ and $120 \ S \ cm^2 \ eq^{-1}$,respectively. The equivalent conductance at infinite dilution for $BX$ is $198 \ S \ cm^2 \ eq^{-1}$. The $\wedge^0$ (in $S \ cm^2 \ eq^{-1}$) of $CX$ is:

What will be the molar conductance at infinite dilution for $NH_4OH$ if at infinite dilution the molar conductances of $Ba(OH)_2$,$BaCl_2$,and $NH_4Cl$ are $523.28$,$280.0$,and $129.8 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively?

At infinite dilution, the equivalent conductances of $Ba^{2+}$ and $Cl^{-}$ ions are $127 \ \Omega^{-1} \ cm^2 \ eqv^{-1}$ and $76 \ \Omega^{-1} \ cm^2 \ eqv^{-1}$ respectively. The equivalent conductivity of $BaCl_2$ at infinite dilution is:

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