Molar conductivity of NH_4OH can be calculated by Kohlrausch's law as: ^ _ NH_4OH = 1 2 ^…

Chemistry · Class 12 · CBSEElectrochemistry

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?

  1. L°NH4OH = Ba(OH)2 + NH4Cl BaCl2
  2. L°NH4OH = BaCl2 + L°NH4ClBa(OH)2
  3. NH4OH =
  4. L°NH4OH =

Answer

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

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