Question

Given:
${\lambda }^{\infty }\left[\frac{1}{2}M{g}^{2+}\right]=53.06oh{m}^{-1}c{m}^{2}mo{l}^{-1}$
${\lambda }^{\infty }\left[\frac{1}{2}S{O}_4^{2-}\right]=80oh{m}^{-1}c{m}^{2}mo{l}^{-1}$
${\lambda }^{\infty }\left[\frac{1}{3}A{l}^{3+}\right]=63oh{m}^{-1}c{m}^{2}mo{l}^{-1}$.
Calculate the values of ${\wedge }_m^{\infty }\left[A{l}_2\right(S{O}_4{)}_3]$ and ${\wedge }^{\infty }\left[MgS{O}_4\right]$.

Answer 1

${\lambda }^{\infty }of\frac{1}{2}M{g}^{2+}=53.06oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

$\frac{1}{2}S{O}_4^{2-}=80oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

$\frac{1}{2}A{l}^{3+}=63oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

$\left(i\right){\lambda }^{\infty }\left[A{l}_2\right(S{O}_4{)}_3]=2{\lambda }_{A{l}^{+3}}^{\infty }+3{\lambda }_{S{O}_4^{2-}}^{\infty }$

$2\times 3\times 63+3\times 2\times 80$

$=858oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

$MgS{O}_4\to M{g}^2+S{O}_4^{2-}$

$\left(ii\right){\lambda }^{\infty }\left[Mg\right(S{O}_4{)}_3]={\lambda }_{M{g}^{+2}}^{\infty }+{\lambda }_{S{O}_4^{2-}}^{\infty }$

$2\times 3\times 53.06+2\times 80$

$=266.12oh{m}^{-1}c{m}^{2}mo{l}^{-1}$