Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12

# Conductivity of 0.00241 M acetic acid is 7.896 × 10−5 S cm−1. Calculate its molar conductivity - Chemistry

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?

#### 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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#### APPEARS IN

NCERT Class 12 Chemistry
Chapter 3 Electrochemistry
Q 11 | Page 92