The molar conductance of an infinitely dilute solution of ammonium chloride was found to be 185 S and the ionic conductance of hydroxyl and chloride ions are 170 and 70 S , respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is

Question

The molar conductance of an infinitely dilute solution of ammonium chloride was found to be 185 S ${cm}^{2}$ ${mol}^{- 1}$ and the ionic conductance of hydroxyl and chloride ions are 170 and 70 S ${cm}^{2}$ ${mol}^{- 1}$ , respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is 85.5 S ${cm}^{2}$ ${mol}^{- 1}$ , its degree of dissociation is given by $x \times 10^{- 1}$ .

The value of $x$ is _____. (Nearest integer) [2025]

Smart Achievers · Accepted Answer

(3)

By Kohlrausch law:

${mol}^{- 1}$

$= (λ_{{NH}_{4}^{+}}^{\circ} + λ_{{Cl}^{-}}^{\circ}) + λ_{{OH}^{-}}^{\circ} - λ_{{Cl}^{-}}^{\circ}$

$= Λ_{{NH}_{4} Cl}^{\circ} + λ_{{OH}^{-}}^{\circ} - λ_{{Cl}^{-}}^{\circ}$

$= (185 + 170 - 70) {S cm}^{2} {mol}^{- 1}$

$= 285 {S cm}^{2} {mol}^{- 1}$

$α = \frac{Λ_{{NH}_{4} OH}}{Λ_{{NH}_{4} OH}^{\circ}} = \frac{85.5}{285} = 0.3 = 3 \times 10^{- 1}$

Solution

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Solution

Ans. (3) By Kohlrausch law: ΛNH4OH∘=λNH4+∘+λOH-∘ =(λNH4+∘+λCl-∘)+λOH-∘-λCl-∘ =ΛNH4Cl∘+λOH-∘-λCl-∘ =(185+170-70)S cm2 mol-1 =285 S cm2 mol-1 α=ΛNH4OHΛNH4OH∘=85.5285=0.3=3×10-1

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