The resistance of a 0.5 , M solution of an el | vedclass

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(C) The cell constant $G^*$ is given by $G^* = \frac{l}{a} = \frac{2.2 \, cm}{4.4 \, cm^2} = 0.5 \, cm^{-1} = 50 \, m^{-1}$.Conductivity $\kappa = \fr

Vedclass · Accepted Answer

$0.002$

Vedclass · Answer

(C) The cell constant $G^*$ is given by $G^* = \frac{l}{a} = \frac{2.2 \, cm}{4.4 \, cm^2} = 0.5 \, cm^{-1} = 50 \, m^{-1}$.Conductivity $\kappa = \frac{1}{R} 	imes G^* = \frac{1}{50 \, \Omega} 	imes 50 \, m^{-1} = 1 \, S \, m^{-1}$.Molar conductivity $\Lambda_m = \frac{\kappa}{C}$.Given $C = 0.5 \, M = 0.5 \, mol \, L^{-1} = 0.5 	imes 10^3 \, mol \, m^{-3} = 500 \, mol \, m^{-3}$.$\Lambda_m = \frac{1 \, S \, m^{-1}}{500 \, mol \, m^{-3}} = 0.002 \, S \, m^2 \, mol^{-1}$.Thus,option $C$ is correct.

$c^{1/2} / (mol \, L^{-1})^{1/2}$	$\Lambda_m / S \, cm^2 \, mol^{-1}$
$0.000198$	$148.61$
$0.000309$	$148.29$
$0.000521$	$147.81$
$0.000989$	$147.09$

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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