Molar conductances of BaCl_2, H_2SO_4 and HCl | vedclass

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(A) According to Kohlrausch's law,the molar conductance at infinite dilution is the sum of the ionic conductances of the constituent ions.$\lambda_m^\

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$(x_1 + x_2 - 2x_3) / 2$

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(A) According to Kohlrausch's law,the molar conductance at infinite dilution is the sum of the ionic conductances of the constituent ions.$\lambda_m^\infty (BaSO_4) = \lambda_{Ba^{2+}}^\infty + \lambda_{SO_4^{2-}}^\infty$We can express this using the given values:$\lambda_m^\infty (BaSO_4) = \lambda_m^\infty (BaCl_2) + \lambda_m^\infty (H_2SO_4) - 2\lambda_m^\infty (HCl)$$\lambda_m^\infty (BaSO_4) = x_1 + x_2 - 2x_3$Equivalent conductance $(\lambda_e^\infty)$ is related to molar conductance $(\lambda_m^\infty)$ by the formula $\lambda_e^\infty = \frac{\lambda_m^\infty}{n}$,where $n$ is the valency factor. For $BaSO_4$,$n = 2$.Therefore,$\lambda_e^\infty (BaSO_4) = \frac{x_1 + x_2 - 2x_3}{2}$.

Molar conductances of $BaCl_2, H_2SO_4$ and $HCl$ at infinite dilutions are $x_1, x_2$ and $x_3$ respectively. Equivalent conductance of $BaSO_4$ at infinite dilution will be

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