(A) For the $0.2 \, M$ solution:$R_1 = 50 \, \Omega$,$\kappa_1 = 1.4 \, S \, m^{-1}$.Cell constant $G^* = R_1 \times \kappa_1 = 50 \times 1.4 = 70 \,

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(A) For the $0.2 \, M$ solution:$R_1 = 50 \, \Omega$,$\kappa_1 = 1.4 \, S \, m^{-1}$.Cell constant $G^* = R_1 \times \kappa_1 = 50 \times 1.4 = 70 \, m^{-1}$.For the $0.5 \, M$ solution:$R_2 = 280 \, \Omega$.Specific conductance $\kappa_2 = \frac{G^*}{R_2} = \frac{70}{280} = 0.25 \, S \, m^{-1}$.Molar conductivity $\Lambda_m = \frac{\kappa_2}{C} = \frac{0.25 \, S \, m^{-1}}{0.5 \, mol \, L^{-1}}$.Convert concentration to $mol \, m^{-3}$: $0.5 \, mol \, L^{-1} = 0.5 \times 1000 \, mol \, m^{-3} = 500 \, mol \, m^{-3}$.$\Lambda_m = \frac{0.25}{500} = 0.0005 = 5 \times 10^{-4} \, S \, m^2 \, mol^{-1}$.

Class 12 Chemistry Electrochemistry Conductor and Conductance and Cell constant

Resistance of $0.2 \, M$ solution of an electrolyte is $50 \, \Omega$. The specific conductance of the solution is $1.4 \, S \, m^{-1}$. The resistance of $0.5 \, M$ solution of the same electrolyte is $280 \, \Omega$. The molar conductivity of $0.5 \, M$ solution of the electrolyte in $S \, m^2 \, mol^{-1}$ is:

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A
$5 \times 10^{-4}$
B
$5 \times 10^{-3}$
C
$5 \times 10^{3}$
D
$5 \times 10^{2}$

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Conductor and Conductance and Cell constant

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The resistance of a $0.5 \, M$ solution of an electrolyte in a cell was found to be $50 \, \Omega$. If the electrodes in the cell are $2.2 \, cm$ apart and have an area of $4.4 \, cm^2$,then the molar conductivity (in $S \, m^2 \, mol^{-1}$) of the solution is:

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If the conductivity of mercury at $0^{\circ} \, C$ is $1.07 \times 10^{6} \, S \, m^{-1}$ and the resistance of a cell containing mercury is $0.243 \, \Omega$,then the cell constant of the cell is $x \times 10^{4} \, m^{-1}$. The value of $x$ is ...... (Nearest integer).

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$A$ $5.0 \, mmol \, dm^{-3}$ aqueous solution of $KCl$ has a conductance of $0.55 \, mS$ when measured in a cell with a cell constant of $1.3 \, cm^{-1}$. The molar conductivity of this solution is ....... $mS \, m^{2} \, mol^{-1}$. (Round off to the Nearest Integer)

DifficultJEE MAIN 2021

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The resistance of a $0.5 \, M$ solution of an electrolyte in a cell was found to be $50 \, \Omega$. If the electrodes in the cell are $2.2 \, cm$ apart and have an area of $4.4 \, cm^2$,then the molar conductivity (in $S \, m^2 \, mol^{-1}$) of the solution is:

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