0.5 N solution of a salt placed between two | vedclass

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(C) The formula for equivalent conductivity is $\Lambda_{eq} = \kappa 	imes \frac{1000}{N}$,where $\kappa = \frac{1}{R} 	imes \frac{l}{A}$.Given: $R

Vedclass · Accepted Answer

$40$

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(C) The formula for equivalent conductivity is $\Lambda_{eq} = \kappa 	imes \frac{1000}{N}$,where $\kappa = \frac{1}{R} 	imes \frac{l}{A}$.Given: $R = 25 \ \Omega$,$l = 2.0 \ cm$,$A = 4.0 \ cm^2$,$N = 0.5 \ N$.First,calculate the cell constant: $\frac{l}{A} = \frac{2.0}{4.0} = 0.5 \ cm^{-1}$.Next,calculate the conductivity $\kappa$: $\kappa = \frac{1}{25} 	imes 0.5 = 0.02 \ \Omega^{-1} \ cm^{-1}$.Finally,calculate equivalent conductivity: $\Lambda_{eq} = 0.02 	imes \frac{1000}{0.5} = 40 \ \Omega^{-1} \ cm^2 \ eq^{-1}$.

$0.5 \ N$ solution of a salt placed between two platinum electrodes $2.0 \ cm$ apart and of area of cross section $4.0 \ cm^2$ has a resistance of $25 \ \Omega$. Calculate the equivalent conductivity of solution ................. $\Omega^{-1} \ cm^2 \ eq^{-1}$

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