JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
The conductivity of a weak acid \(HA\) of concentration \(0.001\, \mathrm{~mol}\, \mathrm{~L}^{-1}\) is \(2.0 \times 10^{-5}\, \mathrm{~S} \,\mathrm{~cm}^{-1} .\) If \(\Lambda_{\mathrm{m}}^{\circ}\) \((\mathrm{HA})=190 \,\mathrm{~S} \,\mathrm{~cm}^{2} \mathrm{~mol}^{-1}\), the ionization constant \(\left(\mathrm{K}_{\mathrm{a}}\right)\) of \(\mathrm{HA}\) is equal to \(....\,\times 10^{-6} .\) (Round off to the Nearest Integer)
- A \(18\)
- B \(12\)
- C \(61\)
- D \(14\)
Answer & Solution
Correct Answer
(B) \(12\)
Step-by-step Solution
Detailed explanation
\(\wedge_{\mathrm{m}}=1000 \times \frac{\kappa}{\mathrm{M}}\) \(=1000 \times \frac{2 \times 10^{-5}}{0.001}=20 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)…
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