Question

Which of the following expressions correctly represents the equivalent conductance at infinite dilution of $A{l}_2(S{O}_4{)}_3$? Given that ${\wedge }_{A{l}^{3+}}^{\circ }$ and ${\wedge }_{S{O}_4^{2-}}^{\circ }$ are the equivalent conductances at infinite dilution of the respective ions?

• A. $2{\wedge }_{A{l}^{3+}}^{\circ }+3{\wedge }_{S{O}_4^{2-}}^{\circ }$
• B. ${\wedge }_{A{l}^{3+}}^{\circ }+{\wedge }_{S{O}_4^{2-}}^{\circ }$
• C. $({\wedge }_{A{l}^{3+}}^{\circ }+3{\wedge }_{S{O}_4^{2-}}^{\circ })\times 6$
• D. $\frac{1}{3}{\wedge }_{A{l}^{3+}}^{\circ }+\frac{1}{2}{\wedge }_{S{O}_4^{2-}}^{\circ }$

Answer 1

The correct option is B ${\wedge }_{A{l}^{3+}}^{\circ }+{\wedge }_{S{O}_4^{2-}}^{\circ }$

$A{l}_2(S{O}_4{)}_3\rightleftharpoons 2A{l}^{3+}+3S{O}_4^{2-}$

We can calculate the equivalent conductance only for ions, so the equivalent conductance at infinite dilution,

${\wedge }_{eq}^{\infty }={\wedge }_{A{l}^{3+}}^{\circ }+{\wedge }_{S{O}_4^{2-}}^{\circ }$

So, the correct option is $B$