The equivalent conductance at infinite dilution of \(\mathrm{NaCl}\), \(\mathrm{HCl}\) and \(\mathrm{CH}_{3} \mathrm{COONa}\) at \(25^{\circ} \mathrm{C}\) are \(126.0\), \(426.0\) and \(91.0 \mathrm{ohm}^{-1} \mathrm{~cm}^{2}\) respectively. The equivalent conductance of acetic acid at infinite dilution at \(25^{\circ} \mathrm{C}\) will be (a) \(643.0\) (b) \(517.0\) (c) \(217.0\) (d) \(391.0\)

We know the equivalent conductance of \(\mathrm{NaCl}\), \(\mathrm{HCl}\), and \(\mathrm{CH}_{3} \mathrm{COONa}\), which are respectively represented by the ions Na\(^{+}\), Cl\(^{-}\), H\(^{+}\), and \(\mathrm{CH}_{3} \mathrm{COO^{-}}\). As \(\mathrm{NaCl}\) dissociates into Na\(^{+}\) and Cl\(^{-}\), we can say that the equivalent conductance of Na\(^{+}\) and Cl\(^{-}\) is 126.0. Similarly, \(\mathrm{HCl}\) dissociates into H\(^{+}\) and Cl\(^{-}\), so the equivalent conductance of H\(^{+}\) and Cl\(^{-}\) is 426.0. \(\mathrm{CH}_{3} \mathrm{COONa}\) dissociates into Na\(^{+}\) and \(\mathrm{CH}_{3} \mathrm{COO^{-}}\), so the equivalent conductance of Na\(^{+}\) and \(\mathrm{CH}_{3} \mathrm{COO^{-}}\) is 91.0

From the given data the conductance of Na\(^{+}\) can be obtained from the relation: Conductance of Na\(^{+}\) = Conductance of \(\mathrm{NaCl}\) - Conductance of Cl\(^{-}\), which simplifies to Conductance of Na\(^{+}\) = 126.0 - (426.0 - 126.0), so the Conductance of Na\(^{+}\) = -174. Therefore, the conductance of \(\mathrm{CH}_{3} \mathrm{COO^{-}}\) can be determined from the relation: Conductance of \(\mathrm{CH}_{3} \mathrm{COO^{-}}\) = Conductance of \(\mathrm{CH}_{3} \mathrm{COONa}\) - Conductance of Na\(^{+}\), which simplifies to Conductance of \(\mathrm{CH}_{3} \mathrm{COO^{-}}\) = 91.0 - (-174), so the Conductance of \(\mathrm{CH}_{3} \mathrm{COO^{-}}\) = 265.

Finally, the equivalent conductance of acetic acid (\(\mathrm{CH}_{3} \mathrm{COOH}\)) can be found by adding the conductance of H\(^{+}\) and \(\mathrm{CH}_{3} \mathrm{COO^{-}}\) that have been determined in the previous steps. Therefore, the equivalent conductance of \(\mathrm{CH}_{3} \mathrm{COOH}\) = Conductance of H\(^{+}\) + Conductance of \(\mathrm{CH}_{3} \mathrm{COO^{-}}\) = 426.0 + 265.0 which gives us 691.0