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JEE Advanced · Chemistry · 17. Electrochemistry

At \(298 \mathrm{~K}\), the limiting molar conductivity of a weak monobasic acid is \(4 \times 10^{2} \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\). At \(298 \mathrm{~K}\), for an aqueous solution of the acid the degree of dissociation is \(\boldsymbol{\alpha}\) and the molar conductivity is \(\mathbf{y} \times 10^{2} \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\). At \(298 \mathrm{~K}\), upon 20 times dilution with water, the molar conductivity of the solution becomes \(3 \mathbf{y} \times 10^{2} \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
The value of y is _________ .

  1. A 0.86
  2. B 0.87
  3. C 0.88
  4. D 0.89
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Answer & Solution

Correct Answer

(A) 0.86

Step-by-step Solution

Detailed explanation

α=λmλm α1=y×10+24×102=y4     ...1
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