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The conductivity of `0.00241 M` acetic acid is `7.896xx10^(-5)Scm^(-1)`. Calculate its molar conductivity. If `wedge_(m)^(@)` for acetic acid is `390.

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The conductivity of 0.00241M acetic acid is 7.896×10-5Scm-1 . Calculate its molar conductivity. If ∧ ∘ m for acetic acid is 390.5Scm2mol-1 , what is its dissociation constant ? Select the correct answer from above options

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Correct Answer - 1.86×10-5 ∧ c m = k×1000 Molarity = (7.896×10-5Scm-1)×1000cm3L-1 0.00241molL-1 =32.76Scm2mol-1 α= ∧ c m ∧ ∘ m = 32.76 390.5 =8.4×10-2 Ka= cα2 1-α = 0.00241×(8.4×10-2)2 1-0.084 =1.86×10-5

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