Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1. Calculate its molar conductivity - Chemistry

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Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1. Calculate its molar conductivity and if `Lambda_m^0` for acetic acid is 390.5 S cm2 mol−1, what is its dissociation constant?

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Solution

Given, κ = 7.896 × 10−5 S m−1

c = 0.00241 mol L−1

Then, molar conductivity, `Lambda_m = k/c`

= `(7.896xx10^(-5) S cm^(-1))/(0.00241 " mol L"^(-1))xx1000 cm^(3)/L`

= 32.76S cm2 mol−1

Again, `Lambda_m^0` = 390.5 S cm2 mol−1

Now, alpha = `Lambda_m/Lambda_m^0` = `(32.76 " S ""cm"^2""mol^(-1))/(390.5 " S cm"^2"" mol^(-1))`

= 0.084

∴Dissociation constant ,` K_alpha = (calpha^2)/(1-alpha)`

=`((0.00241 "mol L"^(-1))(0.084)^2)/((1-0.084))`

= 1.86 × 10−5 mol L−1

Concept: Conductance of Electrolytic Solutions - Variation of Conductivity and Molar Conductivity with Concentration
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NCERT Class 12 Chemistry
Chapter 3 Electrochemistry
Q 11 | Page 92
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