Question

Limiting molar conductivity for some ions is given below (in $Sc{m}^{2}mo{l}^{-1}$) :
$N{a}^{+}-50.1$, $C{l}^{-}-76.3$, $H^{+}-349.6$, $C{H}_{3}CO{O}^{-}-40.9$, $C{a}^{2+}-119.0$.
What will be the limiting molar conductivities $\left({\Lambda }_m^{\circ }\right)$ of $CaC{l}_2$, $C{H}_{3}COONa$ and $NaCl$ respectively ?

• A. $97.65$, $111.0$ and $242.8Sc{m}^{2}mo{l}^{-1}$
• B. $197.3$, $182.0$ and $26.2Sc{m}^{2}mo{l}^{-1}$
• C. $271.6$, $91.0$ and $126.4Sc{m}^{2}mo{l}^{-1}$
• D. $119.0$, $1024.5$ and $9.2Sc{m}^{2}mo{l}^{-1}$

Answer 1

Answer: (C)

${\Lambda }_{mCaC{l}_2}^{\circ }={\lambda }_{C{a}^{2+}}^{\circ }+2{\lambda }_{C{l}^{-}}^{\circ }$
$=119.0+2\times 76.3$
$=271.6Sc{m}^{2}mo{l}^{-1}$
${\Lambda }_{mC{H}_{3}COONa}^{\circ }={\lambda }_{C{H}_{3}CO{O}^{-}}^{\circ }+{\lambda }_{N{a}^{+}}^{\circ }$
$=40.9+50.1$
$=91Sc{m}^{2}mo{l}^{-1}$
${\Lambda }_{mNaCl}^{\circ }={\lambda }_{N{a}^{+}}^{\circ }+{\lambda }_{C{l}^{-}}^{\circ }$
$=50.1+76.3$
$=126.4Sc{m}^{2}mo{l}^{-1}$