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CUET · CHEMISTRY · PYQ PAPER 2025

The conductivity of a 0.001 \(\mathrm{mol} \mathrm{L}^{-1}\) solution of a weak monobasic acid is \(4.0 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1}\). Calculate its dissociation constant if \(\Lambda_m^{\circ}\) for the weak acid is \(400.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\).

  1. A \(1.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\)
  2. B \(1.1 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\)
  3. C \(1.5 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)
  4. D \(1.1 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(1.1 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)

Step-by-step Solution

Detailed explanation

\(\Lambda_m = \frac{\kappa \times 1000}{C} = \frac{4.0 \times 10^{-5} \mathrm{~S} \mathrm{~cm}^{-1} \times 1000}{0.001 \mathrm{~mol} \mathrm{~L}^{-1}} = 40 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)…
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