Q. Limiting molar conductivity of NH₄OH, i.e. Λ°m(NH₄OH), is equal to
A) Λ°m(NH₄Cl) + Λ°m(NaOH) − Λ°m(NaCl)
B) Λ°m(NH₄Cl) + Λ°m(NaCl) − Λ°m(NaOH)
C) Λ°m(NaOH) + Λ°m(NaCl) − Λ°m(NH₄Cl)
D) Λ°m(NH₄OH) + Λ°m(NH₄Cl) − Λ°m(HCl)
Explanation
[Solution from source] By Kohlrausch's law, Λ°m(NH₄OH) = λ°(NH₄⁺) + λ°(OH⁻) = [λ°(NH₄⁺) + λ°(Cl⁻)] + [λ°(Na⁺) + λ°(OH⁻)] − [λ°(Na⁺) + λ°(Cl⁻)] = Λ°m(NH₄Cl) + Λ°m(NaOH) − Λ°m(NaCl). The combination subtracting NaOH would cancel the OH⁻ contribution instead. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.