Question

Calculate ${\Lambda }_m^0$ for $CaC{l}_2$ and $MgS{O}_4$ from the data
${\lambda }_{C{a}^{2+}}^0$ = 119.0 S $c{m}^{2}mo{l}^{-1}$
${\lambda }_{C{l}^{\ominus }}^0$ = 76.3 S $c{m}^{2}mo{l}^{-1}$
${\lambda }_{M{g}^{2+}}^0$ = 106.0 S $c{m}^{2}mo{l}^{-1}$
${\lambda }_{S{O}_4^{2-}}^0$ = 160.0 S $c{m}^{2}mo{l}^{-1}$

Answer 1

Kohlraush law states that molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of it's ions at infinite dilution.

Thus, ${\lambda }_0$=${\lambda }_{+}^0$ + ${\lambda }_{-}^0$

For $CaC{l}_2$,

${\lambda }_0$=${\lambda }_{C{a}^{+2}}^0$ + 2${\lambda }_{C{l}^{-}}^0$

=$271.6Sc{m}^{2}mo{l}^{-1}$

For $MgS{O}_4$

${\lambda }_0$=${\lambda }_{M{g}^{+2}}^0$ + ${\lambda }_{S{O}_4^{-2}}^0$

=$266Sc{m}^{2}mo{l}^{-1}$

The correct answer is $266Sc{m}^{2}mo{l}^{-1}$