The molar conductivity of `0.25 mol L^(-1)` methanoic acid is `46.1 S cm^(2) mol^(-1)`. Calculate the degree of dissociation constant. Given `: lambda

Question

The molar conductivity of 0.25molL-1 methanoic acid is 46.1Scm2mol-1. Calculate the degree of dissociation constant. Given :λ ∘ (H⊕) =349.6Scm2mol-1 and λ ∘ (CHM3COOc-) =54.6Scm2mol-1 Select the correct answer from above options

Answer 1

Correct Answer - 3.67×10-4 ∧ ∘ m(HCOOH) =∧ ∘ (H⊕) +∧ ∘ (HCOOc-) =349.6+54.6Scm2mol-1 =404.2Scm2mol-1 ∧ ∘ m =46.1Scm2mol-1( Given ) ∴α= ∧ ∘ m ∧ ∘ m = 46.1 404.2 =0.114 HCOOH ⇔ HCOOc- + H⊕ Initial conc c molL-1 - - Conc at Eq c(1-α) calpha calpha Ka= calpha.calpha c(1-α) = cα2 1-α = 0.025×(0.114)2 1-0.114 =3.67×10-4