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If the molar conductivities at infinite dilution of NH₄CI, NaOH and NaCl respectively are 120.8, 210.4 and 110 S cm² mo1^-1. Then the limiting molar conductivity of NH₄OH is equal to

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199.6 S cm² mo1^-1

228.4 S cm² mo1^-1

389.9 S cm² mo1^-1

221.2 S cm² mo1^-1

answer is B.

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Detailed Solution

$Λ_{m ({NH}_{4} OH)}^{\infty} = Λ_{m ({NH}_{4} Cl)}^{\infty} + Λ_{m (N_{a} OH)}^{\infty} - Λ_{m (NaCl)}^{\infty}$

= 120.8 + 210.4 – 110.0

$Λ_{m ({NH}_{4} OH)}^{\infty} = 221 .2 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}$

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