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NEET · Chemistry · STD 12 - 2. Electrochemistry
Following limiting molar conductivities are given as \(\lambda_{\mathrm{m}\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)}^{0}=\mathrm{x} \;\mathrm{S}\; \mathrm{cm}^{2} \mathrm{mol}^{-1}\) \(\lambda_{\mathrm{m}\left(\mathrm{K}_{2} \mathrm{SO}_{4}\right)}^{0}=\mathrm{y} \;\mathrm{S\;cm}^{2} \mathrm{mol}^{-1}\) \(\lambda_{\mathrm{m}(\mathrm{CH_3} \mathrm{COOK})}^{0}=\mathrm{z}\; \mathrm{S\;cm}^{2} \mathrm{mol}^{-1}\) \(\lambda_{\mathrm{m}}^{0}\left(\mathrm{in}\; \mathrm{S} \;\mathrm{cm}^{2} \mathrm{mol}^{-1}\right)\) for \(\mathrm{CH}_{3} \mathrm{COOH}\) will be
- A \(x + y + 2 z\)
- B \(x + y - z\)
- C \(x + y + z\)
- D \(\frac{x-y}{2}+z\)
Answer & Solution
Correct Answer
(D) \(\frac{x-y}{2}+z\)
Step-by-step Solution
Detailed explanation
\(\mathrm{CH}_{3} \mathrm{COOH} \rightarrow \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}^{+}\dots (1)\) \(\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{H}^{+}+\mathrm{SO}_{4}^{-2}\dots (2)\)
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