Conductivity of 2.5 × 10-4 M methanoic acid is 5.25 × 10-5 S cm-1.
Calculate its molar conductivity and degree of dissociation.
Given:
λ(H+) = 349.5S cm2mole-1 and λ(HCOO-) = 50.5 Scm2mol-1
We know molar conductivity is calculated by
λ m = ![]()
λm =
= 210 cm2mol-1.
λ 0HCOOH = λ0H+ + λ(HCOO-) = (349.5 + 50.5) = 400 Scm2mol-1
=
= 0.525
Or, α = 52.5%
Couldn't generate an explanation.
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