Λ0m(NH4OH) is equal to ______________.
(i) Λ0m(NH4OH) + Λ0m(NH4Cl) - Λ0(HCl)
(ii) Λ0m(NH4Cl)+ Λ0m(NaOH) – Λ0
(NaCl)
(iii) Λ0m(NH4Cl)+ Λ0m (NaCl) – Λ0(NaOH)
(iv) Λ0m(NaOH) + Λ0m (NaCl) –Λ0(NH4Cl)
Λ0m(NH4OH) is equal to ______________.
(i) Λ0m(NH4OH) + Λ0m(NH4Cl) - Λ0(HCl)
(ii) Λ0m(NH4Cl)+ Λ0m(NaOH) – Λ0 (NaCl)
(iii) Λ0m(NH4Cl)+ Λ0m (NaCl) – Λ0(NaOH)
(iv) Λ0m(NaOH) + Λ0m (NaCl) –Λ0(NH4Cl)
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1 Answer
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This is a Multiple Choice Questions as classified in NCERT Exemplar
Ans: Correct Option: (ii)
Given: Salt ammonium hydroxide
Apply Kohlrausch Law
Kohlrausch law of states that limiting molar conductivity of any salt species is equal to the sum of the limiting molar conductivity of cations and anions of the electrolyte
Λ0m (NaOH) = Λ0m (NH4+) + Λ0m (OH-) + Λ0m (Na+) + Λ0m (OH-) - Λ0m (Na+)- Λ0m (Cl-)
Λ0m (NH4OH) = Λ0m (NH4OH) + Λ0m (NaOH) – Λ0 (NaCl)
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