Q. Given the limiting molar conductivity as $\bigwedge_{m}^{0}\ (HCl) = 425.9\mathrm{\Omega}^{- 1}\ {cm}^{2\ }{mol}^{- 1}$ $\bigwedge_{m}^{0}\ (NaCl) = 126.4\mathrm{\Omega}^{- 1}\ {cm}^{2\ }{mol}^{- 1}$ $\bigwedge_{m}^{0}\ \left( CH_{3}COONa \right) = 91\mathrm{\Omega}^{- 1}\ {cm}^{2\ }{mol}^{- 1}$ The molar conductivity , at infinite dilution, of acetic acid ($in\ \mathrm{\Omega}^{- 1}{cm}^{2\ }{mol}^{- 1}$) will be
A) 481.5
B) 390.5
C) 299.5
D) 516.9
Explanation
(b)
Sum of molar conductivity of reactants = sum of molar conductivity of products
Therefore, for the reaction
${CH}_{3}COOH + NaOH\ \longrightarrow \ {CH}_{3}COONa + HCl$
$\bigwedge_{m}^{0}\ {CH}_{3}COOH = \bigwedge_{m}^{0}\ {CH}_{3}COONa + \ \bigwedge_{m}^{0}\ HCl - \ \bigwedge_{m}^{0}\ NaCl$
= 91 + 425.9 – 126.4
$= \ 390.5\mathrm{\Omega}^{- 1}{cm}^{2\ }{mol}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.