If the specific conductance of 1 , M , H_2SO_ | vedclass

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(A) The formula for equivalent conductivity is $\lambda_{eq} = \frac{1000 	imes \kappa}{N}$,where $\kappa$ is the specific conductance and $N$ is the

Vedclass · Accepted Answer

$1.3 	imes 10^2 \, S \, cm^2 \, eq^{-1}$

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(A) The formula for equivalent conductivity is $\lambda_{eq} = \frac{1000 	imes \kappa}{N}$,where $\kappa$ is the specific conductance and $N$ is the normality of the solution.For $H_2SO_4$,the molarity is $1 \, M$,so the normality $N = M 	imes 	ext{n-factor} = 1 	imes 2 = 2 \, N$.Substituting the values: $\lambda_{eq} = \frac{1000 	imes 26 	imes 10^{-2}}{2} = \frac{260}{2} = 130 \, S \, cm^2 \, eq^{-1} = 1.3 	imes 10^2 \, S \, cm^2 \, eq^{-1}$.

If the specific conductance of $1 \, M \, H_2SO_4$ solution is $26 \times 10^{-2} \, S \, cm^{-1}$,then the equivalent conductivity would be

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