HA is a weak electrolyte. At 25⁰C, the degree of dissociation of 0.5mol $L^{- 1} HA is 0 .1$ What is its molar conductivity $(in S {cm}^{2} {mol}^{- 1}) ?$ (The
limiting molar conductivity of HA is 390 S cm² mol^–1)

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390

answer is A.

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Detailed Solution

$α = \frac{\land_{m}^{c}}{\land_{m}^{\infty}} 0.1 = \frac{\land_{m}^{c}}{390} M o l a r c o n d u c t i v i t y (\land_{m}^{c}) = 39 S c m^{2} m o l^{- 1}$

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