Q. The molar conductivities ${\Lambda{^\circ}}_{NaOAc}$ and ${\Lambda{^\circ}}_{HCl}$ at infinite dilution in water at 25℃ are 91.0 and 426.2 S ${cm}^{2}/mol$ respectively. To calculate ${\Lambda{^\circ}}_{HOAc}$, the additional value required is
A) ${\Lambda{^\circ}}_{H_{2}O}$
B) ${\Lambda{^\circ}}_{KCl}$
C) ${\Lambda{^\circ}}_{NaOH}$
D) ${\Lambda{^\circ}}_{NaCl}$
Explanation
(d)
According to Kohlrausch’s law
O
$\parallel$
$\lbrack NaOAc = \ {CH}_{3}\ C - O^{-}Na^{+}\rbrack$
$\Lambda_{{CH}_{3}COOH}^{\circ} = \ \Lambda_{{CH}_{3}COO^{-}}^{\circ} + \Lambda_{H^{+}}^{\circ}$ …(i)
$\Lambda_{HCl}^{\circ} = \ \Lambda_{H^{+}}^{\circ} + \ \Lambda_{{Cl}^{-}}^{\circ}$ ..(ii)
${\ \ \Lambda}_{{CH}_{3}COONa}^{\circ} = \ \Lambda_{{CH}_{3}COO^{-}}^{\circ} + \ \Lambda_{{Na}^{+}}^{\circ}\$ …(iii)
Thus, on adding (ii) and (iii), if $\Lambda_{{Na}^{+}}^{\circ}$ and $\Lambda_{{Cl}^{-}}^{\circ}$ are subtracted we can obtained the value of $\Lambda_{HOAc}^{\circ}.$
Thus, additional value required is $\Lambda_{NaCl}^{\circ}.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.