Question

Limiting molar conductivity for some ions is given below (in $S \,cm^2\, mol^{-1}$) :
$Na^{+}-50.1$, $Cl^{-}-76.3$, $H^{+}-349.6$, $CH_{3}COO^{-}-40.9$, $Ca^{2+}-119.0$.
What will be the limiting molar conductivities $\left(\Lambda^{\circ}_{m}\right)$ of $CaCl_{2}$, $CH_{3}COONa$ and $NaCl$ respectively ?

• A. $97.65$, $111.0$ and $242.8\, S\, cm^2 \,mol^{-1}$
• B. $197.3$, $182.0$ and $26.2\, S\, cm^2 \,mol^{-1}$
• C. $271.6$, $91.0$ and $126.4\, S\, cm^2 \,mol^{-1}$
• D. $119.0$, $1024.5$ and $9.2\, S\, cm^2 \,mol^{-1}$

Answer 1

Answer: (C)

$\Lambda^{\circ}_{m\,CaCl_2}=\lambda^{\circ}_{Ca^{2+}}+2\lambda^{\circ}_{Cl^{-}}$
$=119.0+2\times76.3$
$=271.6\,S\,cm^{2}\,mol^{-1}$
$\Lambda^{\circ}_{m\,CH_3COONa}=\lambda^{\circ}_{CH_3COO^{-}}+\lambda^{\circ}_{Na^{+}}$
$=40.9+50.1$
$=91\,S\,cm^{2}\,mol^{-1}$
$\Lambda^{\circ}_{m\,NaCl}=\lambda^{\circ}_{Na^{+}}+\lambda^{\circ}_{Cl^{-}}$
$=50.1+76.3$
$=126.4\,S\,cm^{2}\,mol^{-1}$