The molar conductivity of a 0.02 M solution | vedclass

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Question

(D) Given: Concentration $C = 0.02 \ M = 0.02 \ mol \ L^{-1} = 0.02 	imes 1000 \ mol \ m^{-3} = 20 \ mol \ m^{-3}$.Cell constant $G^* = 129 \ m^{-1}$

Vedclass · Accepted Answer

$520$

Vedclass · Answer

(D) Given: Concentration $C = 0.02 \ M = 0.02 \ mol \ L^{-1} = 0.02 	imes 1000 \ mol \ m^{-3} = 20 \ mol \ m^{-3}$.Cell constant $G^* = 129 \ m^{-1}$.Molar conductivity $\Lambda_m = 124 	imes 10^{-4} \ S \ m^2 \ mol^{-1}$.Formula: $\Lambda_m = \frac{\kappa}{C}$,where $\kappa$ is conductivity.$\kappa = \Lambda_m 	imes C = (124 	imes 10^{-4} \ S \ m^2 \ mol^{-1}) 	imes (20 \ mol \ m^{-3}) = 2480 	imes 10^{-4} \ S \ m^{-1} = 0.248 \ S \ m^{-1}$.Resistance $R = \frac{G^*}{\kappa} = \frac{129 \ m^{-1}}{0.248 \ S \ m^{-1}} \approx 520.16 \ \Omega$.Rounding to the nearest integer,$R = 520 \ \Omega$.

The molar conductivity of a $0.02 \ M$ solution of an electrolyte is $124 \times 10^{-4} \ S \ m^2 \ mol^{-1}$. What is the resistance of the same solution (in ohms),kept in a cell with a cell constant of $129 \ m^{-1}$?

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