Q8 of 35 Page 1

The conductivity of 10-3 mol /L acetic acid at 25°C is 4.1 × 10-5 S cm-1.

Calculate its degree of dissociation, if for acetic acid at 250C is 390.5 S cm2 mol-1.

Given: Conductivity (K) = 4.1 × 10-5 S cm-1


C = 10-3 mol /L


= 390.5 S cm2 mol-1


First we will calculate molar conductivity () by applying the formula given below:


=


⇒ =


⇒ = 41 S cm2 mol-1


Now, to calculate the degree of dissociation, we apply the formula:



⇒ α =


⇒ α = 0.105


Thus, degree of dissociation is 0.105


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