(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

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?

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The molar conductivities at infinite dilution of barium chloride,sulphuric acid and hydrochloric acid are $280$,$860$ and $426 \, S \, cm^{2} \, mol^{-1}$ respectively. The molar conductivity at infinite dilution of barium sulphate is $...... \, S \, cm^{2} \, mol^{-1}$ ($Round$ off to the Nearest Integer).

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$\Lambda_{m}^0$ for $NaCl$,$HCl$ and $NaAc$ are $126.4 \ S \ cm^2 \ mol^{-1}$,$425.9 \ S \ cm^2 \ mol^{-1}$ and $91.0 \ S \ cm^2 \ mol^{-1}$ respectively. Calculate $\Lambda^0$ for $HAc$.

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The resistance of $1 \, M$ acetic acid is $250 \, \Omega$. The cell constant of the conductivity cell is $1.15 \, \text{cm}^{-1}$. What is the equivalent conductivity (in $\Omega^{-1} \, \text{cm}^2 \, \text{equiv}^{-1}$) of $1 \, N$ acetic acid?

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The equivalent conductances of $CH_3COONa$,$HCl$,and $CH_3COOH$ at infinite dilution are $91$,$426$,and $391 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ respectively. What is the equivalent conductance of $NaCl$ at infinite dilution?

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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:

Similar Questions

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?

$\Lambda_{m}^0$ for $NaCl$,$HCl$ and $NaAc$ are $126.4 \ S \ cm^2 \ mol^{-1}$,$425.9 \ S \ cm^2 \ mol^{-1}$ and $91.0 \ S \ cm^2 \ mol^{-1}$ respectively. Calculate $\Lambda^0$ for $HAc$.

The resistance of $1 \, M$ acetic acid is $250 \, \Omega$. The cell constant of the conductivity cell is $1.15 \, \text{cm}^{-1}$. What is the equivalent conductivity (in $\Omega^{-1} \, \text{cm}^2 \, \text{equiv}^{-1}$) of $1 \, N$ acetic acid?

The equivalent conductances of $CH_3COONa$,$HCl$,and $CH_3COOH$ at infinite dilution are $91$,$426$,and $391 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ respectively. What is the equivalent conductance of $NaCl$ at infinite dilution?

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