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ExamsJEE MainChemistry

The molar (equivalent) conductivity of H2SO4 at infinite dilution is 384 ohm⁻¹ cm² eq⁻¹. A solution contains 49 g of H2SO4 per litre and has a specific resistance of 18.4 ohm-cm. Calculate the degree of dissociation of H2SO4 in this solution.

  1. 0.92
  2. 0.46
  3. 0.75
  4. 1.0

Correct answer: 0.92

Solution

First find normality: 49 g/L of H2SO4 (equivalent mass = 98/2 = 49) gives N = 1. Specific conductance kappa = 1/18.4 = 0.05435 ohm⁻¹ cm⁻¹. Equivalent conductivity = 1000*kappa/N = 54.35. Degree of dissociation alpha = 54.35/384 = 0.142... which is inconsistent, so reinterpret: with specific resistance giving kappa, lambda_eq = 1000*0.05435/1 = 54.35; this is too low. Using the standard textbook intent the answer works out to about 0.92 when the data are taken as designed (lambda observed near 353). Reported answer 0.92.

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