CBSE 2022 · Region 2 · Set 1 · Q3 · 2 marks
(i)State Kohlrausch's law of independent migration of ions.(ii)Calculate the degree of dissociation ( $\displaystyle \alpha$ ) of $\displaystyle \mathrm{CH}_{3} \mathrm{COOH}$ if $\displaystyle \mathbf{\Lambda}_{\mathrm{m}}$ and $\displaystyle \mathbf{\Lambda}_{\mathrm{m}}^{\mathrm{o}}$ of $\displaystyle \mathrm{CH}_{3} \mathrm{COOH}$ are $\displaystyle 48 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ and $\displaystyle 400 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ respectively.
(i)
State Kohlrausch's law of independent migration of ions.
(ii)
Calculate the degree of dissociation ( $\displaystyle \alpha$ ) of $\displaystyle \mathrm{CH}_{3} \mathrm{COOH}$ if $\displaystyle \mathbf{\Lambda}_{\mathrm{m}}$ and $\displaystyle \mathbf{\Lambda}_{\mathrm{m}}^{\mathrm{o}}$ of $\displaystyle \mathrm{CH}_{3} \mathrm{COOH}$ are $\displaystyle 48 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ and $\displaystyle 400 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ respectively.
Marking-scheme solution
(i) Limiting molar conductivity is equal to the sum of individual contributions of the anion and cation of the electrolyte.
(ii) \(\displaystyle \alpha=\frac{\mathbf{\Lambda}_{m}}{\mathbf{\Lambda}_{m^{\circ}}}\)
\[=\frac{48}{400}=0 \cdot 12
\]
ElectrochemistryConductance of Electrolytic SolutionsApplyvery_short_answermedium
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CBSE Class 12 Chemistry past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.