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 \(400.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\).
- A \(1.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\)
- B \(1.1 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\)
- C \(1.5 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)
- D \(1.1 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}\)
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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