WBJEE · Chemistry · Electrochemistry
The specific conductance \((\mathrm{k})\) of \(0.02(\mathrm{M})\) aqueous acetic acid solution at 298 K is \(1.65 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}\). The degree of dissociation of acetic acid is \(\left[\lambda_{\mathrm{H}^{+}}=349.1 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\right.\) and \(\left.\lambda_{\mathrm{CH}_3 \mathrm{COO}^{-}}=40.9 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\right]\)
- A 0.021
- B 0.21
- C 0.012
- D 0.12
Answer & Solution
Correct Answer
(A) 0.021
Step-by-step Solution
Detailed explanation
Hint: Firstly we find \(\lambda_{\mathrm{m}}\left(\mathrm{aq} \mathrm{CH}_3 \mathrm{COOH}\right)\) \(\lambda_{\mathrm{m}}=\frac{\mathrm{K} \times 1000}{\mathrm{C}(\mathrm{M})}=\frac{1.65 \times 10^{-4} \times 1000}{0.02}=8.25 \mathrm{Scm}^2 \mathrm{~mol}^{-1}\)…
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