Question

Conductivity of 0.0025 M acetic acid is 8.0 × 10∧-5 S cm .Calculate its molar conductivity if ∧° for acetic acid is 390.5 S cm² mol . what will be the degree of dissociation of acetic acid?

Answer 1

Molar conductivity =
$(8.0 \times 1 {0}^{ - 5} \times 1000) \div 0.0025 = 32$
Degree of dissociation =
$32 \div 390 =0.082$