The molar conductivity of 0.01 M acetic acid is 50.2 S cm2 mol–1. The degree of dissociation of acetic acid will be[= 349.6 S cm2 mol–1 and = 40.9 S cm2 mol–1]
Question
The molar conductivity of 0.01 M acetic acid is 50.2 S cm2 mol–1. The degree of dissociation of acetic acid will be[= 349.6 S cm2 mol–1 and = 40.9 S cm2 mol–1]
Solution
The degree of dissociation (α) of acetic acid can be calculated using the formula:
α = Λm / Λm°
where: Λm is the molar conductivity of the solution, and Λm° is the molar conductivity at infinite dilution.
Given that Λm = 50.2 S cm2 mol–1 and Λm° = 349.6 S cm2 mol–1, we can substitute these values into the formula:
α = 50.2 / 349.6 = 0.143
Therefore, the degree of dissociation of acetic acid is 0.143 or 14.3%.
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