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NEET 2025 Chemistry Conductivity MCQ Question

Type: MCQ-numerical-Medium-Class 12

If the molar conductivity (Λₘ) of a 0.050 mol L⁻¹ solution of a monobasic weak acid is 90 S cm² mol⁻¹, its extent (degree) of dissociation will be

[Assume Λₘ⁰ = 349.6 S cm² mol⁻¹ and Λ⁰ = 50.4 S cm² mol⁻¹.]

A

0.115

B

0.125

C

0.225

D

0.215

Correct Answer

Option C

Detailed Explanation

Explanation

(A) 🔴 ✗ Incorrect

α=90400=0.225\alpha=\frac{90}{400}=0.225

So, 0.115 is not the correct value.

(B) 🔴 ✗ Incorrect The limiting molar conductivity is 349.6+50.4=400349.6+50.4=400, giving α=0.225\alpha=0.225, not 0.125.

(C) 🟢 ✓ Correct

α=ΛmΛm∘=90349.6+50.4=0.225\alpha=\frac{\Lambda_m}{\Lambda_m^\circ} =\frac{90}{349.6+50.4} =\boxed{0.225}

(D) 🔴 ✗ Incorrect Using the given ionic conductivities, Λm∘=400\Lambda_m^\circ=400. Hence, the degree of dissociation is 0.225, not 0.215.

⚡ Quick Revision

  • Λm∘\Lambda_m^\circ → Sum of ionic conductivities
  • α=ΛmΛm∘\alpha=\frac{\Lambda_m}{\Lambda_m^\circ}
  • Weak electrolyte: Λm<Λm∘\Lambda_m < \Lambda_m^\circ
  • Here: 90/400 = 0.225

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