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The conductivity of 0.001 mol L-1 solution of CH3COOH is 3.905× 10-5 S cm-1. Calculate molar conductivity and degree of dissociation - Chemistry

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The conductivity of 0.001 mol L-1 solution of CH3COOH is 3.905× 10-5 S cm-1. Calculate its molar conductivity and degree of dissociation (α) Given λ°(H+)= 349.6 S cm2 mol-1 and λ°(CH3COO)= 40.9S cm2mol-1.

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Solution

Given:

Conductivity (k) = 3.905 x 10-5 S cm-1

Concentration ofelectrolyte (c) = 0.001 mol L-1

 `^^_m=k/cxx1000=39.05S cm^2mol^(-1)`

`^^_m^@=lambda_(CH_3COO^-)^@+lambda_(H+)^@`

 = 40.9 + 349.6

= 390.5 S cm2 mol-1

Degree of dissociation = `39.05/390.5=0.1`

Concept: Conductance of Electrolytic Solutions - Variation of Conductivity and Molar Conductivity with Concentration
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