Which of the following expressions for conduc | vedclass

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(B) Conductivity $(\kappa)$ is defined as the reciprocal of resistivity $(\varrho)$. Thus,$\kappa = \frac{1}{\varrho}$.For a given electrolytic cell,t

Vedclass · Accepted Answer

$\kappa = G \cdot \frac{1}{a}$

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(B) Conductivity $(\kappa)$ is defined as the reciprocal of resistivity $(\varrho)$. Thus,$\kappa = \frac{1}{\varrho}$.For a given electrolytic cell,the relationship between conductivity $(\kappa)$ and conductance $(G)$ is given by $\kappa = G \cdot G^*$,where $G^*$ is the cell constant,defined as $\frac{l}{A}$.Evaluating the options:$A$: $\kappa = \frac{1}{\varrho}$ is the correct definition of conductivity.$B$: $\kappa = G \cdot \frac{1}{a}$ is incorrect because the correct relation is $\kappa = G \cdot a$ (assuming $a$ is the cell constant).$C$: $\kappa = \frac{1}{R} \cdot \frac{l}{A}$ is correct because $G = \frac{1}{R}$ and $\frac{l}{A}$ is the cell constant.$D$: $\kappa = \Lambda_m \cdot c$ is correct based on the definition of molar conductivity,$\Lambda_m = \frac{\kappa}{c}$.Therefore,the incorrect expression is $B$.

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

Which of the following expressions for conductivity of a solution of an electrolyte is $NOT$ correct?

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