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Question
\[\ce{Λ^0_m H2O}\] is equal to:
(i) \[\ce{Λ^0_m_{(HCl)} + \ce{Λ^0_m_{(NaOH)} - \ce{Λ^0_m_{(NaCl)}}}}\]
(ii) \[\ce{Λ^0_m_{(HNO_3)} + \ce{Λ^0_m_{(NaNO_3)} - \ce{Λ^0_m_{(NaOH)}}}}\]
(iii) \[\ce{Λ^0_{(HNO_3)} + \ce{Λ^0_m_{(NaOH)} - \ce{Λ^0_m_{(NaNO_3)}}}}\]
(iv) \[\ce{Λ^0_m_{(NH_4OH)} + \ce{Λ^0_m_{(HCl)} - \ce{Λ^0_m_{(NH_4Cl)}}}}\]
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Solution
(i) \[\ce{Λ^0_m_{(HCl)} + \ce{Λ^0_m_{(NaOH)} - \ce{Λ^0_m_{(NaCl)}}}}\]
(iv) \[\ce{Λ^0_m_{(NH_4OH)} + \ce{Λ^0_m_{(HCl)} - \ce{Λ^0_m_{(NH_4Cl)}}}}\]
Explanation:
\[\ce{Λ^0_m_{(H_2O)} = \ce{Λ^0_m_{(HCl)} + \ce{Λ^0_m_{(NaOH)} - \ce{Λ^0_m_{(NaCl)}}}}}\]
\[\ce{Λ^0_{m(H^+)} + \ce{Λ^0_{m(OH^-)} = \ce{Λ^0_{m(H^+)} + \ce{Λ^0_{m(Cl^-)} + \ce{Λ^0_{m(Na^+)} + \ce{Λ^0_{m(OH^-)} - \ce{Λ^0_{m(Na^+)} - \ce{Λ^0_{m(Cl^-)}}}}}}}}}\]
\[\ce{Λ^0_m_{(HNO_3)} + \ce{Λ^0_m_{(NaOH)} - \ce{Λ^0_m_{(NaNO_3)} = \ce{Λ^0_m_{(H_2O)}}}}}\]
\[\ce{Λ^0_{m(H^+)} + \ce{Λ^0_{m(NO_3^-)} + \ce{Λ^0_{m(Na^+)} - \ce{Λ^0_{m(OH^-)} + \ce{Λ^0_{m(Na^+)} + \ce{Λ^0_{m(NO_3^-)} = \ce{Λ^0_{m(H^+)} - \ce{Λ^0_{m(OH^-)}}}}}}}}}\]
\[\ce{Λ^0_m_{(NH_4OH)} + \ce{Λ^0_m_{(HCl)} - \ce{Λ^0_m_{(NH_4Cl)} = \ce{Λ^0_m_{(H_2O)}}}}}\]
However, the sum of molar conductivities of constituent ions gives the molar conductivity of water but here \[\ce{NH4OH}\] is a weak electrolyte of which complete decomposition is not possible.
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