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The equivalent conductivity of a monobasic acid at infinite dilution is 348 $o h m^{- 1} c m^{2} e q^{- 1} .$ If the resistivity of the solution containing 15g acid (M.Wt-49) in 1 lit is 18.5 ohm cm, what is the degree of dissociation of acid

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50.7%

60.4%

25.35%

40%

answer is A.

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Detailed Solution

Resistivity $= \frac{1}{K}$

$∴ K = \frac{1}{18.5} o h m^{- 1} c m^{- 1}$

$λ_{c} = \frac{K \times 1000}{N} = \frac{1}{18.5} \times \frac{1000}{15} \times 49 = \frac{λ_{c}}{λ^{0}} = \frac{1 \times 49 \times 1000}{18.5 \times 15 \times 348} = 0.507$

$∴ % of ionization = 50.7%$

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