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If the equivalent conductance of 1 M benzoic acid is 12.8 $o h m^{- 1} c m^{2} e q^{- 1}$ and if the conductance of benzoate ion and H⁺ ions are 42 and 288.42 $o h m^{- 1} c m^{2} e q^{- 1}$ respectively, its degree of dissociation is

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39%

3.9%

0.35%

0.039%

answer is B.

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Detailed Solution

$α = \frac{Λ_{m}}{{Λ^{0}}_{m}} = \frac{12.8}{(42 + 288.42)} = 0.0387$

Percentage dissociation
= 0.0387 × 100 $\approx$ 3.9%

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