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

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(C) Given: $M = 0.5 \, mol \, L^{-1} = 500 \, mol \, m^{-3}$,$R = 50 \, \Omega$,$l = 2.2 \, cm = 0.022 \, m$,$A = 4.4 \, cm^2 = 4.4 	imes 10^{-4} \,

Vedclass · Accepted Answer

$0.002$

Vedclass · Answer

(C) Given: $M = 0.5 \, mol \, L^{-1} = 500 \, mol \, m^{-3}$,$R = 50 \, \Omega$,$l = 2.2 \, cm = 0.022 \, m$,$A = 4.4 \, cm^2 = 4.4 	imes 10^{-4} \, m^2$. Cell constant $G^* = \frac{l}{A} = \frac{0.022 \, m}{4.4 	imes 10^{-4} \, m^2} = 50 \, m^{-1}$. Conductivity $\kappa = \frac{1}{R} 	imes G^* = \frac{1}{50 \, \Omega} 	imes 50 \, m^{-1} = 1 \, S \, m^{-1}$. Molar conductivity $\Lambda_m = \frac{\kappa}{M} = \frac{1 \, S \, m^{-1}}{500 \, mol \, m^{-3}} = 0.002 \, S \, m^2 \, mol^{-1}$.

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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