Lambda_{m(HAc)}^0 is equal to . . . . . . . | vedclass

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(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity of a weak electrolyte at infinite dilution can be calculated

Vedclass · Accepted Answer

$\Lambda_{m(HCl)}^0 + \Lambda_{m(NaAc)}^0 - \Lambda_{m(NaCl)}^0$

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(B) According to Kohlrausch's law of independent migration of ions,the molar conductivity of a weak electrolyte at infinite dilution can be calculated using the molar conductivities of strong electrolytes.For acetic acid $(HAc)$,the expression is:$\Lambda_{m(HAc)}^0 = \lambda_{H^+} + \lambda_{Ac^-}$Using strong electrolytes $HCl$,$NaAc$,and $NaCl$:$\Lambda_{m(HCl)}^0 = \lambda_{H^+} + \lambda_{Cl^-}$$\Lambda_{m(NaAc)}^0 = \lambda_{Na^+} + \lambda_{Ac^-}$$\Lambda_{m(NaCl)}^0 = \lambda_{Na^+} + \lambda_{Cl^-}$Therefore,$\Lambda_{m(HCl)}^0 + \Lambda_{m(NaAc)}^0 - \Lambda_{m(NaCl)}^0 = (\lambda_{H^+} + \lambda_{Cl^-}) + (\lambda_{Na^+} + \lambda_{Ac^-}) - (\lambda_{Na^+} + \lambda_{Cl^-}) = \lambda_{H^+} + \lambda_{Ac^-} = \Lambda_{m(HAc)}^0$.Thus,the correct option is $B$.

$\Lambda_{m(HAc)}^0$ is equal to . . . . . . .

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