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NVAEasyJEE 2025Electrolytic Conductance & Kohlrausch's Law

JEE Chemistry 2025 Question with Solution

The molar conductance of an infinitely dilute solution of ammonium chloride was found to be 185 S cm−1 mol−1185 \, \text{S cm}^{-1} \, \text{mol}^{-1} and the ionic conductance of hydroxyl and chloride ions are 170 S cm−1 mol−1170 \, \text{S cm}^{-1} \, \text{mol}^{-1} and 70 S cm−1 mol−170 \, \text{S cm}^{-1} \, \text{mol}^{-1}, respectively. If molar conductance of 0.02 M0.02 \, \text{M} solution of ammonium hydroxide is 85.5 S cm−1 mol−185.5 \, \text{S cm}^{-1} \, \text{mol}^{-1}, its degree of dissociation is given by x×10−1x \times 10^{-1}. The value of xx is _____ . (Nearest integer)

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given:

  • Λm∞(NH4Cl)=185 S cm−1 mol−1\Lambda_m^\infty(\text{NH}_4\text{Cl}) = 185 \, \text{S cm}^{-1} \, \text{mol}^{-1}
  • λ∞(OH−)=170 S cm−1 mol−1\lambda^\infty(\text{OH}^-) = 170 \, \text{S cm}^{-1} \, \text{mol}^{-1}
  • λ∞(Cl−)=70 S cm−1 mol−1\lambda^\infty(\text{Cl}^-) = 70 \, \text{S cm}^{-1} \, \text{mol}^{-1}
  • Λm(NH4OH)=85.5 S cm−1 mol−1\Lambda_m(\text{NH}_4\text{OH}) = 85.5 \, \text{S cm}^{-1} \, \text{mol}^{-1}

Find: The value of xx if degree of dissociation is written as x×10−1x \times 10^{-1}.

Using Kohlrausch's law:

Λm∞(NH4OH)=Λm∞(NH4Cl)−λ∞(Cl−)+λ∞(OH−)\Lambda_m^\infty(\text{NH}_4\text{OH}) = \Lambda_m^\infty(\text{NH}_4\text{Cl}) - \lambda^\infty(\text{Cl}^-) + \lambda^\infty(\text{OH}^-) =185−70+170= 185 - 70 + 170 =285 S cm−1 mol−1= 285 \, \text{S cm}^{-1} \, \text{mol}^{-1}

Now use the relation:

Λm=αΛm∞\Lambda_m = \alpha \Lambda_m^\infty

So,

α=ΛmΛm∞\alpha = \frac{\Lambda_m}{\Lambda_m^\infty} =85.5285= \frac{85.5}{285} =0.3= 0.3

Given that

α=x×10−1\alpha = x \times 10^{-1}

we get

0.3=x×10−10.3 = x \times 10^{-1}

Hence,

x=3x = 3

Therefore, the value of xx is 33.

The second extracted approach shows an inconsistent substitution involving 0.020.02, but the first solution correctly applies Kohlrausch's law and gives x=3x = 3.

Using Ionic Conductances Explicitly

Given: The limiting molar conductance of NH4_4Cl and ionic conductances of Cl^-$$** and **OH^-$$ are provided.

Find: Degree of dissociation of NH4_4OH in the form x×10−1x \times 10^{-1}.

For ammonium hydroxide,

Λm∞(NH4OH)=λ∞(NH4+)+λ∞(OH−)\Lambda_m^\infty(\text{NH}_4\text{OH}) = \lambda^\infty(\text{NH}_4^+) + \lambda^\infty(\text{OH}^-)

From ammonium chloride,

Λm∞(NH4Cl)=λ∞(NH4+)+λ∞(Cl−)\Lambda_m^\infty(\text{NH}_4\text{Cl}) = \lambda^\infty(\text{NH}_4^+) + \lambda^\infty(\text{Cl}^-)

Subtracting λ∞(Cl−)\lambda^\infty(\text{Cl}^-) and adding λ∞(OH−)\lambda^\infty(\text{OH}^-) gives

Λm∞(NH4OH)=185−70+170=285\Lambda_m^\infty(\text{NH}_4\text{OH}) = 185 - 70 + 170 = 285

Then,

α=85.5285=0.3=3×10−1\alpha = \frac{85.5}{285} = 0.3 = 3 \times 10^{-1}

So, the required integer is 33.

Common mistakes

  • Using the given concentration 0.02 M0.02 \, \text{M} directly in the formula for degree of dissociation is incorrect here because the solution uses Λm=αΛm∞\Lambda_m = \alpha \Lambda_m^\infty. First find Λm∞\Lambda_m^\infty of NH4_4OH, then divide observed molar conductance by it.

  • Adding 170170 and 7070 directly to get NH4_4OH is wrong because 7070 is for Cl^-$$**, not for **NH$$_4^+$$**. Use **NH$$_4$$Cl** as the reference salt and replace **Cl^-** by **OH$^- through Kohlrausch's law.$$

  • Writing α=0.3\alpha = 0.3 and then reporting the answer as 0.30.3 is incomplete because the question asks for xx in x×10−1x \times 10^{-1}. Convert 0.30.3 to 3×10−13 \times 10^{-1}, so x=3x = 3.

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