The molar conductivities $\wedge_{\mathrm{NaOAc}}^{\circ}$ and $\wedge_{\mathrm{HCl}}^{\circ}$ at infinite dilution in water at $25^{\circ} \mathrm{C}$ are 91.0 and $426.2 \mathrm{~S} \mathrm{~cm}^{2} / \mathrm{mol}$ respectively. To calculate $\wedge_{\mathrm{HOAc}}^{\circ}$, the additional value required is

(D) $\wedge_{\mathrm{NaCl}}^{\circ}$

(A) $\wedge_{\mathrm{H}_{2} \mathrm{O}}^{\circ}$

(B) $\wedge_{\mathrm{KCl}}^{\circ}$

(C) $\wedge_{\mathrm{NaOH}}^{\circ}$

JEE Main · 2006 · 30 Apr 2006 · Offline · Redox Reactions and Electrochemistry

The molar conductivities

\land_{NaOAc}^{\circ}

and

\land_{HCl}^{\circ}

at infinite dilution in water at

2 5^{\circ} C

are 91.0 and

426.2 S cm^{2} / mol

respectively. To calculate

\land_{HOAc}^{\circ}

, the additional value required is

Answer: (D)

Kohlrausch's law lets the limiting molar conductivity of acetic acid be assembled from strong-electrolyte values, since

Λ_{HOAc}^{\circ} = Λ_{NaOAc}^{\circ} + Λ_{HCl}^{\circ} - Λ_{NaCl}^{\circ}

. Adding the acetate-plus-sodium and hydrogen-plus-chloride contributions and subtracting the sodium-plus-chloride pair leaves exactly

Λ_{H^{+}}^{\circ} + Λ_{OA c^{-}}^{\circ}

, the quantity wanted. Beyond the two values already supplied, the additional datum needed is therefore

Λ_{NaCl}^{\circ}

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option