Question

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

• A. $97.65,111.0$ and $242.8$ $S.c{m}^2$ $mo{l}^{-1}$
• B. $195.3,182.0$ and $26.2$ $S.c{m}^2$ $mo{l}^{-1}$
• C. $271.6,91.0$ and $126.4$ $S.c{m}^2$ $mo{l}^{-1}$
• D. $119.0,1024.5$ and $9.2$ $S.c{m}^2$ $mo{l}^{-1}$

Answer 1

The correct option is C $271.6,91.0$ and $126.4$ $S.c{m}^2$ $mo{l}^{-1}$

We know,

According to Kohlraush's law at infinite dilution,

${\Lambda }_{m\left(CaC{l}_2\right)}^0={\Lambda }_{C{a}^{2+}}^0+2{\Lambda }_{C{l}^{-}}^0$

${\Lambda }_{m\left(CaC{l}_2\right)}^0=119+2\times 76.3=271.6$ $S$ $c{m}^2$ $mo{l}^{-1}$

Also,

${\Lambda }_{m\left(C{H}_{3}COONa\right)}^0={\Lambda }_{C{H}_{3}CO{O}^{-}}^0+{\Lambda }_{N{a}^{+}}^0$

${\Lambda }_{m\left(C{H}_{3}COONa\right)}^0=40.9+50.1=91$ $S$ $c{m}^2$ $mo{l}^{-1}$

Similarly,

${\Lambda }_{m\left(NaCl\right)}^0={\Lambda }_{N{a}^{+}}^0+{\Lambda }_{C{l}^{-}}^0$

${\Lambda }_{m\left(NaCl\right)}^0=50.1+76.3=126.4$ $S$ $c{m}^2$ $mo{l}^{-1}$