The resistance of 0.05, M solution of oxalic | vedclass

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Question

(A) The formula for equivalent conductance is $\Lambda_{eq} = \frac{\kappa 	imes 1000}{N}$.First,calculate the specific conductance $(\kappa)$:$\kapp

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$100$

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(A) The formula for equivalent conductance is $\Lambda_{eq} = \frac{\kappa 	imes 1000}{N}$.First,calculate the specific conductance $(\kappa)$:$\kappa = \frac{1}{R} 	imes 	ext{cell constant} = \frac{1}{200\, \Omega} 	imes 2.0\, cm^{-1} = 0.01\, S\, cm^{-1}$.Next,determine the normality $(N)$ of the oxalic acid solution.For oxalic acid $(H_2C_2O_4)$,the n-factor is $2$.$N = Molarity 	imes n	ext{-factor} = 0.05\, M 	imes 2 = 0.1\, N$.Finally,calculate the equivalent conductance:$\Lambda_{eq} = \frac{0.01\, S\, cm^{-1} 	imes 1000}{0.1\, eq\, L^{-1}} = 100\, S\, cm^2\, eq^{-1}$.

The resistance of $0.05\, M$ solution of oxalic acid is $200\, \Omega$ and the cell constant is $2.0\, cm^{-1}$. The equivalent conductance in $S\, cm^2\, eq^{-1}$ will be:

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