(C) The cell constant $(G^*)$ is given by the formula: $G^* = \frac{\ell}{A} = \frac{2 \ cm}{4 \ cm^2} = 0.5 \ cm^{-1}$.The conductivity $(\kappa)$ is

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(C) The cell constant $(G^*)$ is given by the formula: $G^* = \frac{\ell}{A} = \frac{2 \ cm}{4 \ cm^2} = 0.5 \ cm^{-1}$.The conductivity $(\kappa)$ is related to resistance $(R)$ by the formula: $\kappa = \frac{1}{R} \times G^*$.Rearranging for resistance $(R)$: $R = \frac{G^*}{\kappa} = \frac{0.5 \ cm^{-1}}{8 \times 10^{-7} \ S \ cm^{-1}}$.$R = \frac{0.5}{8} \times 10^7 \ \Omega = 0.0625 \times 10^7 \ \Omega = 6.25 \times 10^5 \ \Omega$.

Class 12 Chemistry Electrochemistry Conductor and Conductance and Cell constant

Medium

In a conductivity cell,the electrodes are separated by a distance of $2 \ cm$ and have a cross-sectional area of $4 \ cm^2$. If the conductivity of pure water is $8 \times 10^{-7} \ S \ cm^{-1}$,then the resistance of the water is:

A
$8.13 \times 10^{-7} \ \Omega$
B
$6.78 \times 10^{-5} \ \Omega$
C
$6.25 \times 10^{5} \ \Omega$
D
$7.18 \times 10^{7} \ \Omega$

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Conductor and Conductance and Cell constant

Calculate the molar conductivity of $NH_4OH$ at infinite dilution if the molar conductivities of $Ba(OH)_2$,$BaCl_2$,and $NH_4Cl$ at infinite dilution are $520$,$280$,and $129 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively.

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MediumJEE MAIN 2021

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Resistance of a conductivity cell at $298 \ K$ for $0.0100 \ M \ KCl$ is $161.8 \ \Omega$. Resistance becomes $190 \ \Omega$ when $0.005 \ M \ NaOH$ solution is filled in this cell. Calculate: $(i)$ cell constant,$(ii)$ specific conductivity for $NaOH$ solution,and $(iii)$ molar conductivity.

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The values of conductivity of some materials at $298.15 \ K$ in $S \ m^{-1}$ are $2.1 \times 10^3$,$1.0 \times 10^{-16}$,$1.2 \times 10$,$3.91$,$1.5 \times 10^{-2}$,$1 \times 10^{-1}$,$1.0 \times 10^3$. The number of conductors among the materials is............

DifficultJEE MAIN 2024

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Specific conductance of $0.1 \ M \ HNO_3$ is $6.3 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$. The molar conductance of the solution is

MediumKCET 2022

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In a conductivity cell,the electrodes are separated by a distance of $2 \ cm$ and have a cross-sectional area of $4 \ cm^2$. If the conductivity of pure water is $8 \times 10^{-7} \ S \ cm^{-1}$,then the resistance of the water is:

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Calculate the molar conductivity of $NH_4OH$ at infinite dilution if the molar conductivities of $Ba(OH)_2$,$BaCl_2$,and $NH_4Cl$ at infinite dilution are $520$,$280$,and $129 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively.

Resistance of a conductivity cell at $298 \ K$ for $0.0100 \ M \ KCl$ is $161.8 \ \Omega$. Resistance becomes $190 \ \Omega$ when $0.005 \ M \ NaOH$ solution is filled in this cell. Calculate: $(i)$ cell constant,$(ii)$ specific conductivity for $NaOH$ solution,and $(iii)$ molar conductivity.

The values of conductivity of some materials at $298.15 \ K$ in $S \ m^{-1}$ are $2.1 \times 10^3$,$1.0 \times 10^{-16}$,$1.2 \times 10$,$3.91$,$1.5 \times 10^{-2}$,$1 \times 10^{-1}$,$1.0 \times 10^3$. The number of conductors among the materials is............

Specific conductance of $0.1 \ M \ HNO_3$ is $6.3 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$. The molar conductance of the solution is

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