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Practice question

Question

What is the degree of dissociation of 0.025 M methanoic acid if its molar conductivity is 46.1 S cm² mol⁻¹, given lambda⁰(H^+) = 349.6 S cm² mol^{-1 and lambda⁰(HCOO^-) = 54.6 S cm² mol^{-1 ?

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Explanation

Given: What is the degree of dissociation of 0.025 M methanoic acid if its molar conductivity is 46.1 S cm² mol⁻¹, given lambda⁰(H^+) = 349.6 S cm² mol^{-1 and lambda⁰(HCOO^-) = 54.6 S cm² mol^{-1 ? These values define the system as per NCERT data. Formula: Lambda_m⁰ = 349.6 + 54.6 = 404.2 S cm² mol^{-1. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: α = Lambda_m/Lambda_m⁰ = 46.1/404.2 approx 0.114 . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

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