Question

What will be the molar conductivity of $A{l}^{3+}$ ions at infinite dilution if molar conductivity of $A{l}_2(S{O}_4{)}_3$ is $858$ $S.c{m}^2$ $mo{l}^{-1}$ and ionic conductance of $S{O}_4^{2-}$ is $160$ $S.c{m}^2$ $mo{l}^{-1}$ at infinite dilution?

• A. $189$ $S.c{m}^2$ $mo{l}^{-1}$
• B. $698$ $S.c{m}^2$ $mo{l}^{-1}$
• C. $1018$ $S.c{m}^2$ $mo{l}^{-1}$
• D. $429$ $S.c{m}^2$ $mo{l}^{-1}$

Answer 1

The correct option is A $189$ $S.c{m}^2$ $mo{l}^{-1}$

Since it is infinite dilution we can write,

${\Lambda }_{A{l}_{2}S{O}_{\$}}^0=2{\Lambda }_{A{l}^{3+}}^0+3{\Lambda }_{S{O}_4^{3-}}^0$

$⟹{\Lambda }_{A{l}^{3+}}^0=\frac{{\Lambda }_{A{l}_2(S{O}_4{)}_3}^0-3{\Lambda }_{S{O}_4^{2-}}^0}{2}$

$⟹{\Lambda }_{A{l}^{3+}}^0=\frac{858-(3\times 160)}{2}=189S.c{m}^{2}mo{l}^{-1}$