The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrode is 120 cm with an area of cross section of $1 {cm}^{2}$ . The conductance of this solution was found to be $5 \times 10^{- 7} S$ . The pH of the solution is 4. The value of limiting molar conductivity $(Λ_{m}^{0})$ of this weak monobasic acid in aqueous solution is $Z \times 10^{2} S {cm}^{- 1} {mol}^{- 1}$ . The value of Z is

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Detailed Solution

$\begin{array}{l} K = conductance \times cell constant k = 5 \times 10^{- 7} S \times (\frac{120}{1}) \frac{cm}{{cm}^{2}} = 600 \times 10^{- 7} {Scm}^{- 1} \\ Λ_{m} = \frac{600 \times 10^{- 7} \times 1000}{C} = \frac{600 \times 10^{- 7} \times 1000}{0 .0015 M} = 40 \\ pH = 4 \\ (H^{+}) = Cα = 10^{- 4} \\ \frac{Λ_{M}}{Λ_{m}^{0}} = α \\ Λ_{m}^{0} = \frac{Λ_{m}}{α} = \frac{40}{1 / 15} = 600 \\ \frac{600 S - {cm}^{2} {mol}^{- 1}}{2 \times 10^{2}} = 6 \times 10^{2} \end{array}$