JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
The specific conductance of \(0.0025\,M\) acetic acid is \(5 \times 10^{-5}\,S\,cm ^{-1}\) at a certain temperature. The dissociation constant of acetic acid is \(......\,10^{-7}\). (Nearest integer) Consider limiting molar conductivity of \(CH _3 COOH\) as \(400\,S\,cm ^2\,mol ^{-1}\)
- A \(65\)
- B \(64\)
- C \(66\)
- D \(63\)
Answer & Solution
Correct Answer
(C) \(66\)
Step-by-step Solution
Detailed explanation
\(\wedge_{ m }=\frac{ k }{ C } \times 1000\) Given \(k =5 \times 10^{-5} S cm ^{-1}\) \(C =0.0025\,M\) \(\wedge_{ m } =\frac{5 \times 10^{-5} \times 10^3}{0.0025}=\frac{5 \times 10^{-2}}{2.5 \times 10^{-3}}\) \(=20\,S\,cm ^2\,mol ^{-1}\) \(\alpha =\frac{20}{400}=\frac{1}{20}\)…
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