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The specific conductivity of a 0.05 M solution of propionic acid is 3.2 ×10^-2 Sm^-1at 298K. The molar conductivity of the acid at infinite dilution is 386.0 × 10^-4 Sm²mol^-1. Incorrect statements are

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$The molar conductivity of the acid at 0.05 M conc. is 6.4 \times 10^{- 2} (in {Sm}^{2} {mol}^{- 1})$

The degree of dissociation of the acid is 0.332

dissociation of constant of the acid is 1.38 × 10^-5

The ionic mobility of propionate ion $(in m^{2} {volt}^{- 1} \sec^{- 1}),$ given the molar ionic conductance of proton is $350 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}, is 4.2 \times 10^{- 6}$

answer is A, B, D.

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Detailed Solution

(1) $Λ = molar conductance = \frac{k}{C} = \frac{3 .2 \times 10^{- 2} {Sm}^{- 1}}{0 .05 mol {litre}^{- 1} \times 10^{3} lit m^{- 3}}$

$= \frac{3.2 \times 10^{- 2}}{5 \times 10} = 6.4 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$

(2) $α = \frac{Λ}{Λ_{0}} = \frac{6.4 \times 10^{- 4}}{386 \times 10^{- 4}} = 1.66 \times 10^{- 2}$

(3) $K \approx α^{2} c = {(1.66 \times 10^{- 2})}^{2} \times 0.05 = 1.38 \times 10^{- 5}$

(4) $λ_{0} (molar conductance of propionate ion)$

$= 386 \times 10^{- 4} - 350 \times 10^{- 4} = 36 \times 10^{- 4} {Sm}^{2} {mol}^{- 1}$

$(ionic mobility) μ_{anion} = \frac{36 \times 10^{- 4}}{96, 500} = 3 .73 \times 10^{- 8} m^{2} {volt}^{- 1} \sec^{- 1}$

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