Find ∧ m ∞ (in Ω − 1 cm 2 mol − 1 ) for strong electrolyte AB 3 in water at 2 5 ∘ from the following data.

Kohlrausch's Law: Λ m ∞ ( AB 3 ) = Λ m ∞ ( A 3 + ) + 3 Λ m ∞ ( B − ) = 170 Ω − 1 cm 2 mol − 1 .

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Find $\land_{m}^{\infty}$ (in $Ω^{- 1} cm^{2} mol^{- 1}$ ) for strong electrolyte $AB_{3}$ in water at $2 5^{\circ}$ from the following data.

Answer:

Solution:

Kohlrausch's Law: $Λ_{m}^{\infty} (AB_{3}) = Λ_{m}^{\infty} (A^{3 +}) + 3 Λ_{m}^{\infty} (B^{-}) = 170 Ω^{- 1} cm^{2} mol^{- 1}$ .

Stream:JEESubject:ChemistryTopic:ElectrochemistrySubtopic:Kohlrausch's Law

⏱ 2mℹ️ Source: QB

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