The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm2 mol-1. Calculate its degree of dissoiation and dissociation constant. Given ∧°m(H+) = 349.6 S cm2 mol-1 and ∧°m(HCOO–) = 546 S cm2 mol-1?

The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm2�

1.	The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm2 mol-1. Calculate its degree of dissoiation and dissociation constant. Given ∧°m(H+) = 349.6 S cm2 mol-1 and ∧°m(HCOO–) = 546 S cm2 mol-1?
Answer» ∧°_m (HCOOH^–)= ∧°_m(H⁺) + ∧°_m (HCOO^–) = 349.6 + 54.6 = 404.2 S cm²mol^-1

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The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm2 mol-1. Calculate its degree of dissoiation and dissociation constant. Given ∧°m(H+) = 349.6 S cm2 mol-1 and ∧°m(HCOO–) = 546 S cm2 mol-1?

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