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ΛXm0 HX2O is equal to:

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प्रश्न

\[\ce{Λ^0_m H2O}\] 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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उत्तर

(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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अध्याय 3: Electrochemistry - Exercises [पृष्ठ ३७]

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एनसीईआरटी एक्झांप्लर Chemistry Exemplar [English] Class 12
अध्याय 3 Electrochemistry
Exercises | Q II. 22. | पृष्ठ ३७

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