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Ionic Equilibrium question

2025 · 4 Apr · Shift 2 · Q24
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  5. /2025 · 4 Apr · Shift 2 · Q24

Ionic Equilibrium question

2025 · 4 Apr · Shift 2 · Q24

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
The molar conductance of an infinitely dilute solution of ammonium chloride was found to be 185 S cm2 mol−1185 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}185 S cm2 mol−1 and the ionic conductance of hydroxyl and chloride ions are 170 and 70 S cm2 mol−170 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}70 S cm2 mol−1, respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is 85.5 S cm2 mol−185.5 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}85.5 S cm2 mol−1, its degree of dissociation is given by x×10−1x \times 10^{-1}x×10−1. The value of xxx is ‾\underline{\hspace{2cm}}​ . (Nearest integer)
Numerical answer
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Correct answer: 3

  1. Use Kohlrausch’s law to find Λm∞\Lambda_m^\inftyΛm∞​ for NH4OH\mathrm{NH_4OH}NH4​OH

Given:

  • Λm∞(NH4Cl)=185 S cm2 mol−1\Lambda_m^\infty(\mathrm{NH_4Cl}) = 185\ \mathrm{S\,cm^2\,mol^{-1}}Λm∞​(NH4​Cl)=185 Scm2mol−1
  • λ∞(OH−)=170 S cm2 mol−1\lambda^\infty(\mathrm{OH^-}) = 170\ \mathrm{S\,cm^2\,mol^{-1}}λ∞(OH−)=170 Scm2mol−1
  • λ∞(Cl−)=70 S cm2 mol−1\lambda^\infty(\mathrm{Cl^-}) = 70\ \mathrm{S\,cm^2\,mol^{-1}}λ∞(Cl−)=70 Scm2mol−1

First, find ionic conductance of NH4+\mathrm{NH_4^+}NH4+​:

λ∞(NH4+)=Λm∞(NH4Cl)−λ∞(Cl−)\lambda^\infty(\mathrm{NH_4^+}) = \Lambda_m^\infty(\mathrm{NH_4Cl}) - \lambda^\infty(\mathrm{Cl^-})λ∞(NH4+​)=Λm∞​(NH4​Cl)−λ∞(Cl−) =185−70=115 S cm2 mol−1= 185 - 70 = 115\ \mathrm{S\,cm^2\,mol^{-1}}=185−70=115 Scm2mol−1

Now,

Λm∞(NH4OH)=λ∞(NH4+)+λ∞(OH−)\Lambda_m^\infty(\mathrm{NH_4OH}) = \lambda^\infty(\mathrm{NH_4^+}) + \lambda^\infty(\mathrm{OH^-})Λm∞​(NH4​OH)=λ∞(NH4+​)+λ∞(OH−) =115+170=285 S cm2 mol−1= 115 + 170 = 285\ \mathrm{S\,cm^2\,mol^{-1}}=115+170=285 Scm2mol−1
  1. Use the relation for degree of dissociation of a weak electrolyte

For weak electrolyte,

α=ΛmΛm∞\alpha = \frac{\Lambda_m}{\Lambda_m^\infty}α=Λm∞​Λm​​

Given:

Λm=85.5 S cm2 mol−1\Lambda_m = 85.5\ \mathrm{S\,cm^2\,mol^{-1}}Λm​=85.5 Scm2mol−1

So,

α=85.5285=0.3\alpha = \frac{85.5}{285} = 0.3α=28585.5​=0.3
  1. Match with the form x×10−1x \times 10^{-1}x×10−1
0.3=3×10−10.3 = 3 \times 10^{-1}0.3=3×10−1

Hence,

x=3x = 3x=3
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