MHT CET · Chemistry · Ionic Equilibrium
What is degree of dissociation of \(\mathrm{CH}_3 \mathrm{COOH}\) if \(\wedge^{\circ}\left(\mathrm{CH}_3 \mathrm{COO}^{-}\right)=50 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\), \(\wedge^{\circ}\left(\mathrm{H}^{+}\right)=350 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\) and molar conductivity of \(5 \times 10^{-2} \mathrm{M} \mathrm{CH}_3 \mathrm{COOH}\) is \(20 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\) ?
- A \(1.25 \times 10^{-4}\)
- B \(1.25 \times 10^{-2}\)
- C \(5 \times 10^{-2}\)
- D \(5 \times 10^{-4}\)
Answer & Solution
Correct Answer
(C) \(5 \times 10^{-2}\)
Step-by-step Solution
Detailed explanation
\(\wedge^{\circ}_{\mathrm{m}}(\mathrm{CH}_3 \mathrm{COOH}) = \wedge^{\circ}(\mathrm{CH}_3 \mathrm{COO}^{-}) + \wedge^{\circ}(\mathrm{H}^{+})\) \(\wedge^{\circ}_{\mathrm{m}}(\mathrm{CH}_3 \mathrm{COOH}) = 50 + 350 = 400 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\)
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