Question

Molar conductivities $(\Lambda^{\circ}_m)$ at infinite dilution of ${ NaCl, HCl}$ and ${CH3COONa }$ are $126.4, 425.9$ and ${91.0 \, S\, cm^2 \, mol^{-1}}$ respectively. $(\Lambda^{\circ}_m)$ for ${CH3COOH}$ will be:

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

Answer 1

Answer: (C)

$\ce{ 390.5 \, S \, cm^2 \, mol^{-1} }$

Answer 2

${CH_3COONa + HCl -> NaCl + CH_3COOH }$ 91 + 425.9 = 126.4 + x ${\therefore \; 390.5 \, ohm^{-1} \; cm^2 \; mol^{-1}}$