Molar conductivity of NH_4OH can be calculated by Kohlrausch's law as: ^ _ NH_4OH = 1 2 ^…
Chemistry · Class 12 · CBSE — Electrochemistry
Molar conductivity of \(NH_4OH\) can be calculated by Kohlrausch's law as:
\[\Lambda^\circ_{NH_4OH} = \frac{1}{2}\Lambda^\circ_{Ba(OH)_2} + \Lambda^\circ_{NH_4Cl} - \frac{1}{2}\Lambda^\circ_{BaCl_2}\]
Which of the following expressions correctly represents this?
- L°NH4OH = L°Ba(OH)2 + L°NH4Cl – L°BaCl2
- L°NH4OH = L°BaCl2 + L°NH4Cl – L°Ba(OH)2
- L°NH4OH =

- L°NH4OH =

Answer
(A) L ° NH 4 OH = L° Ba(OH) 2 + L° NH 4 Cl – L° BaCl 2
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