Lambda_{m}^0 for NaCl,HCl and NaAc are 126.4 | vedclass

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(D) According to Kohlrausch's law of independent migration of ions,the molar conductivity of an electrolyte at infinite dilution is the sum of the mol

Vedclass · Accepted Answer

$390.5 \ S \ cm^2 \ mol^{-1}$

Vedclass · Answer

(D) According to Kohlrausch's law of independent migration of ions,the molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of its constituent ions. We are given: $\Lambda_{m}^0(NaCl) = 126.4 \ S \ cm^2 \ mol^{-1}$ $\Lambda_{m}^0(HCl) = 425.9 \ S \ cm^2 \ mol^{-1}$ $\Lambda_{m}^0(NaAc) = 91.0 \ S \ cm^2 \ mol^{-1}$ We need to find $\Lambda_{m}^0(HAc)$,which is given by: $\Lambda_{m}^0(HAc) = \Lambda_{m}^0(H^+) + \Lambda_{m}^0(Ac^-)$ Using the given values: $\Lambda_{m}^0(HAc) = \Lambda_{m}^0(HCl) + \Lambda_{m}^0(NaAc) - \Lambda_{m}^0(NaCl)$ $\Lambda_{m}^0(HAc) = 425.9 + 91.0 - 126.4$ $\Lambda_{m}^0(HAc) = 516.9 - 126.4 = 390.5 \ S \ cm^2 \ mol^{-1}$ Therefore,the correct option is $D$.

$\Lambda_{m}^0$ for $NaCl$,$HCl$ and $NaAc$ are $126.4 \ S \ cm^2 \ mol^{-1}$,$425.9 \ S \ cm^2 \ mol^{-1}$ and $91.0 \ S \ cm^2 \ mol^{-1}$ respectively. Calculate $\Lambda^0$ for $HAc$.

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