What is the degree of dissociation of 0.025 M methanoic acid if its molar conductivity is 46.1 S cm² molâ»Â¹, given la
Given: What is the degree of dissociation of 0.025 M methanoic acid if its molar conductivity is 46.1 S cm² molâ»Â¹, given lambdaâ°(H^+) = 349.6 S cm² mol^{-1 and lambdaâ°(HCOO^-) = 54.6 S cm² mol^{-1 ? These values define the system as per NCERT data. Formula: Lambda_mâ° = 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_mâ° = 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 kgâ»Â¹ Kâ»Â¹
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.