Question

Molar conductivities $\left(\Lambda_{ m }^{\circ}\right)$ at infinite dilution of $NaCl , HCl$ and $CH _{3} COONa$ are $126.4,425.9$ and $91.0\, S\, cm ^{2} mol ^{-1}$ respectively. $\Lambda_{ m }^{\circ}$ for $CH _{3} COOH$ will be

• A. $ 180.5 \, S \, cm^2 \, mol^{-1} $
• B. $290.8 \, S \, cm^2 \, mol^{-1} $
• C. $ 390.5 \, S \, cm^2 \, mol^{-1} $
• D. $ 425.5 \, S \, cm^2 \, mol^{-1} $

Answer 1

Answer: (C)

$CH _{3} COONa + HCl \longrightarrow NaCl + CH _{3} COOH$
$\Lambda_{ m \left( CH _{3} COOH \right)}^{\circ}=\Lambda_{ m \left( CH _{3} COONa \right)}^{\circ}+\Lambda_{ m ( HCl )}^{\circ}-\Lambda_{ m ( NaCl )}^{\circ}$
$\Lambda_{ m \left( CH _{3} COOH \right)}^{\circ} =(91.0+425.9-126.4) S\, cm ^{2} mol ^{-1}$
$\Lambda_{ m \left( CH _{3} COOH \right)}^{\circ} =390.5\, S\, cm ^{2} mol ^{-1}$