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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]

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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%.

This problem has been solved

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