Dissociation constant nd molar conductance of an acetic acid solution are $1.78 \times 10^{- 5} mol L^{- 1}$ and $48.15 S {cm}^{- 2} {mol}^{- 1}$ respectively. the conductivity of the solution is (considering molar conductance at infinite dilution is $309.5 S {cm}^{- 2} {mol}^{- 1}$ )

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$4.9 \times 10^{- 2} S {cm}^{- 1}$

$4.9 \times 10^{2} S {cm}^{- 1}$

$4.9 \times 10^{- 5} S {cm}^{- 1}$

$4.9 \times 10^{5} S {cm}^{- 1}$

answer is C.

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Detailed Solution

$K_{a} = 1.78 \times 10^{- 5} mol L^{- 1} \land_{m} = 48.15 {scm}^{- 1} {mol}^{- 1}_{\land_{m}^{°} = 390.5 {Scm}^{2} {mol}^{- 1}} α = \frac{\land_{m}}{\land_{m}^{°}} = \frac{48.15}{390.5} = 0.123 K_{a} = \frac{{Cα}^{2}}{(1 - α)} = \frac{C \times {(0.123)}^{3}}{(1 - 0.123)} = \frac{C \times 0.01513}{0.877} C = \frac{1.78 \times 10^{- 5} \times 0.877}{0.01513} mol L^{- 1} = 130.17 \times 10^{- 5} \land_{m} = \frac{1000 \times k}{C} k = \frac{48.45 \times 103.17 \times 10^{- 5}}{1000} = 4.9 \times 10^{- 5}$