Given l/a = 0.5 cm^{-1},R = 50 Omega,N = 1. | vedclass

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(A) The cell constant is given by $G^* = l/a = 0.5 \ cm^{-1}$.Resistance $R = 50 \ \Omega$.The conductivity $\kappa$ is given by $\kappa = \frac{1}{R}

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$10$

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(A) The cell constant is given by $G^* = l/a = 0.5 \ cm^{-1}$.Resistance $R = 50 \ \Omega$.The conductivity $\kappa$ is given by $\kappa = \frac{1}{R} 	imes \frac{l}{a} = \frac{1}{50} 	imes 0.5 = 0.01 \ \Omega^{-1} \ cm^{-1}$.The equivalent conductance $\Lambda_{eq}$ is calculated using the formula $\Lambda_{eq} = \frac{\kappa 	imes 1000}{N}$.Substituting the values: $\Lambda_{eq} = \frac{0.01 	imes 1000}{1.0} = 10 \ \Omega^{-1} \ cm^2 \ gm \ eq^{-1}$.

List-$I$ (Symbol of electrical property)	List-$II$ (Units)
$A.$ $\wedge_m$	$I.$ $S\,cm^2\,mol^{-1}$
$B.$ $G$	$II.$ $S$
$C.$ $\kappa$	$III.$ $S\,cm^{-1}$
$D.$ $G^*$	$IV.$ $cm^{-1}$

Given $l/a = 0.5 \ cm^{-1}$,$R = 50 \ \Omega$,$N = 1.0$. The equivalent conductance of the electrolytic cell is ................ $\Omega^{-1} \ cm^2 \ gm \ eq^{-1}$.

Similar Questions

Match the following:
List-$I$ (Symbol of electrical property) List-$II$ (Units)
$A.$ $\wedge_m$ $I.$ $S\,cm^2\,mol^{-1}$
$B.$ $G$ $II.$ $S$
$C.$ $\kappa$ $III.$ $S\,cm^{-1}$
$D.$ $G^*$ $IV.$ $cm^{-1}$

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