Dear Student, The conductivity at infinite dilution of NH4Cl (λ∞NH4Cl) = λNH4+ + λCl- =130 Ω-1cm-2eq-1 (i)The conductivity at infinite dilution of NaOH (λ∞NaOH) = λNa+ + λOH- = 218 Ω-1cm-2eq-1 (ii)The conductivity at infinite dilution of NaCl (λ∞NaCl) = λNa+ + λCl- = 120 Ω-1cm-2eq-1 (iii)The conductivity at infinite dilution of NH4OH (λ∞NH4OH) is= λNH4+ + λCl- =(i) + (ii) - (iii) = (130+218-120) Ω-1cm-2eq-1 = 228 Ω-1cm-2eq-1 The degree of dissociation of NH4OH, α = λNH4OHλ∞NH4OH=10228 = 0 043Hence, correct answer is (2)

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Dear Student,

The conductivity at infinite dilution of {NH}_{4} Cl ({λ^{\infty}}_{{NH}_{4} Cl}) = λ_{{NH}_{4}^{+}} + λ_{{Cl}^{-}} = 130 Ω^{- 1} {cm}^{- 2} {eq}^{- 1} . . . (i) The conductivity at infinite dilution of NaOH ({λ^{\infty}}_{NaOH}) = λ_{{Na}^{+}} + λ_{{OH}^{-}} = 218 Ω^{- 1} {cm}^{- 2} {eq}^{- 1} . . . (ii) The conductivity at infinite dilution of NaCl ({λ^{\infty}}_{NaCl}) = λ_{{Na}^{+}} + λ_{{Cl}^{-}} = 120 Ω^{- 1} {cm}^{- 2} {eq}^{- 1} . . . (iii) The conductivity at infinite dilution of {NH}_{4} OH ({λ^{\infty}}_{{NH}_{4} OH}) is = λ_{{NH}_{4}^{+}} + λ_{{Cl}^{-}} = (i) + (ii) - (iii) = (130 + 218 - 120) Ω^{- 1} {cm}^{- 2} {eq}^{- 1} = 228 Ω^{- 1} {cm}^{- 2} {eq}^{- 1} The degree of dissociation of {NH}_{4} OH, α = \frac{λ_{{NH}_{4} OH}}{{λ^{\infty}}_{{NH}_{4} OH}} = \frac{10}{228} = 0.043 Hence, correct answer is (2)

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