Limiting molar conductivity of NH_4OH [i.e.,L | vedclass

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(D) According to Kohlrausch's law of independent migration of ions,the limiting molar conductivity of a weak electrolyte can be calculated using the l

Vedclass · Accepted Answer

$\Lambda ^o_{m(NH_4Cl)} + \Lambda ^o_{m(NaOH)} - \Lambda ^o_{m(NaCl)}$

Vedclass · Answer

(D) According to Kohlrausch's law of independent migration of ions,the limiting molar conductivity of a weak electrolyte can be calculated using the limiting molar conductivities of strong electrolytes.For $NH_4OH$,we can express it as:$\Lambda ^o_{m(NH_4OH)} = \lambda ^o_{NH_4^+} + \lambda ^o_{OH^-}$Using strong electrolytes $NH_4Cl$,$NaOH$,and $NaCl$:$\Lambda ^o_{m(NH_4Cl)} = \lambda ^o_{NH_4^+} + \lambda ^o_{Cl^-}$$\Lambda ^o_{m(NaOH)} = \lambda ^o_{Na^+} + \lambda ^o_{OH^-}$$\Lambda ^o_{m(NaCl)} = \lambda ^o_{Na^+} + \lambda ^o_{Cl^-}$By performing the operation $\Lambda ^o_{m(NH_4Cl)} + \Lambda ^o_{m(NaOH)} - \Lambda ^o_{m(NaCl)}$,we get:$(\lambda ^o_{NH_4^+} + \lambda ^o_{Cl^-}) + (\lambda ^o_{Na^+} + \lambda ^o_{OH^-}) - (\lambda ^o_{Na^+} + \lambda ^o_{Cl^-}) = \lambda ^o_{NH_4^+} + \lambda ^o_{OH^-} = \Lambda ^o_{m(NH_4OH)}$Therefore,the correct expression is $\Lambda ^o_{m(NH_4Cl)} + \Lambda ^o_{m(NaOH)} - \Lambda ^o_{m(NaCl)}$.

Limiting molar conductivity of $NH_4OH$ [i.e.,$\Lambda ^o_{m(NH_4OH)}$] is equal to:

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