Q17 of 26 Page 1

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%


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