Calculate the degree of dissociation (\alpha) of acetic acid if its molar conductivity (\\Lambda_m) is 39.05 {\S} {\cm} {\mol}^{-1}. Given \lambda^{\circ} ( {\H}^{+}) =349.6 {\S} {\cm}^2 {\mol}^{-1} and \lambda^{\circ} ( {\CH}_3 {\COO}^{-}) =40.9 {\S} {\cm}^2 {\mol}^{-1}.

CBSE Class 12 Chemistry 2017 Delhi Set 1 Solved Paper

Question : 9 of 26

Marks: +1, -0

Calculate the degree of dissociation

(α)

of acetic acid if its molar conductivity

(\Λ_m)

39.05 {\S} {\cm}

{\mol}^{-1}

.
Given

λ^{∘} ( {\H}^{+}) =349.6 {\S} {\cm}^2 {\mol}^{-1}

and

λ^{∘} ( {\CH}_3 {\COO}^{-})

=40.9 {\S} {\cm}^2 {\mol}^{-1}

Solution:

\Λ°_{ {\CH}_3 {\COOH}} =λ°_{ {\CH}_3 {\COO}^{-}}+λ°_{ {\H}^{+} }

\;=40.9+349.6

\;=390.5 {\S} {\cm}^2 \/ {\mol}

N, \;\; α\;=\Λ_{ {\m}} \/ \Λ°_{ {\m}}

\;=39.05 \/ 390.5=0.1

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