Limiting molar conductivities for H_2SO_4, K_2SO_4, and CH_3COOK are x, y, z respectively. From the data given above, it can be concluded that ^0_m fo - CHEMISTRY MCQ NEET 2019 2026 | ExamTest.live

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Limiting molar conductivities for H_{2} SO_{4}, K_{2} SO_{4}, and CH_{3} COOK are x, y, z respectively. From the data given above, it can be concluded that Λ_{m}^{0} for CH_{3} COOH will be:

Solution & Step-by-step Explanation

We need Λ_{m}^{0} (CH_{3} COOH) = λ^{0} (CH_{3} COO^{-}) + λ^{0} (H^{+}) .Given: Λ_{m}^{0} (H_{2} SO_{4}) = 2 λ^{0} (H^{+}) + λ^{0} (SO_{4}^{2 -}) = x Λ_{m}^{0} (K_{2} SO_{4}) = 2 λ^{0} (K^{+}) + λ^{0} (SO_{4}^{2 -}) = y Λ_{m}^{0} (CH_{3} COOK) = λ^{0} (CH_{3} COO^{-}) + λ^{0} (K^{+}) = z Calculate: \frac{1}{2} (1) - \frac{1}{2} (2) + (3) = [λ^{0} (H^{+}) + \frac{1}{2} λ^{0} (SO_{4}^{2 -})] - [λ^{0} (K^{+}) + \frac{1}{2} λ^{0} (SO_{4}^{2 -})] + [λ^{0} (CH_{3} COO^{-}) + λ^{0} (K^{+})] = λ^{0} (H^{+}) + λ^{0} (CH_{3} COO^{-}) .The math is: \frac{x}{2} - \frac{y}{2} + z = \frac{x - y}{2} + z .

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Limiting molar conductivities for H_{2} SO_{4}, K_{2} SO_{4}, and CH_{3} COOK are x, y, z respectively. From the data given above, it can be concluded that Λ_{m}^{0} for CH_{3} COOH will be:

x - y + 2 z

x + y + z

\frac{x - y}{2} + z

x - y + z

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