The resistance of a conductivity cell filled | vedclass

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Question

(A) Step $1$: Calculate the cell constant $(G^*)$.
Conductivity $(\kappa) = G^* 	imes 	ext{Conductance} = G^* / R$.
For $0.1 \ M$ $KCl$: $\kappa =

Vedclass · Accepted Answer

$124$

Vedclass · Answer

(A) Step $1$: Calculate the cell constant $(G^*)$.
Conductivity $(\kappa) = G^* 	imes 	ext{Conductance} = G^* / R$.
For $0.1 \ M$ $KCl$: $\kappa = 1.29 \ S \ m^{-1} = 1.29 	imes 10^{-2} \ S \ cm^{-1}$.
$G^* = \kappa 	imes R = (1.29 	imes 10^{-2} \ S \ cm^{-1}) 	imes (100 \ \Omega) = 1.29 \ cm^{-1}$.
Step $2$: Calculate the conductivity of $0.02 \ M$ $KCl$ solution.
$\kappa = G^* / R = 1.29 \ cm^{-1} / 520 \ \Omega \approx 0.00248 \ S \ cm^{-1}$.
Step $3$: Calculate molar conductivity $(\Lambda_m)$.
$\Lambda_m = (\kappa 	imes 1000) / M = (0.00248 	imes 1000) / 0.02 = 2.48 / 0.02 = 124 \ S \ cm^2 \ mol^{-1}$.

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