Class 12 Chemistry - Chapter Electrochemistry NCERT Solutions | Conductivity of 0.00241 M acetic acid is
Conductivity of 0.00241 M acetic acid is 7.896 × 10 - 5 S cm - 1. Calculate its molar conductivity and if Amº for acetic acid is 390.5 S cm2 mol - 1, what is its dissociation constant?
Given,K = 7.896 × 10 - 5 S m - 1
c = 0.00241 mol L - 1
Then, molar conductivity, Am = K/c
= 32.76S cm2 mol - 1
Again, Amº = 390.5 S cm2 mol - 1
Now, 
= 0.084
Dissociation constant, 
= (0.00241 mol L-1)(0.084)2 / (1-0.084)
= 1.86 × 10 - 5 mol L - 1
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