Λ_m^∘ for NaCl, HCl and CH₃COONa are 126.4,425.9 and 91.05S cm²mol^-1 respectively. If conductivity of 0.001028 mol L^-1 acetic acid solution is 4.95x 10^-5S cm^-1, find the degree of dissociation of the acetic acid solution

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0.01233

0.1233

1.233

answer is B.

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Detailed Solution

$λ_{{CH}_{3} COOH}^{\circ} = λ_{{CH}_{3} COONa}^{\circ} + λ_{HCl}^{\circ} - λ_{NaCl}^{\circ}$
=425.9+91.05-126.4=390.55
We know, $Λ_{m} = \frac{k \times 1000}{M}$
On putting values,
$\begin{array}{l} λ_{m} = \frac{(4.95 \times 10^{- 5}) \times 1000}{0.001028} \\ = 48.15 s {cm}^{2} {mol}^{- 1} \\ α = \frac{λ_{m}}{λ_{m}^{\circ}} = \frac{48.15}{390.55} = 0.1233 \end{array}$