Question

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

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

Answer 1

Answer: (B)

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

At infinite dilution, when dissociation is complete, each ion makes a definite contribution towards the molar conductance of the electrolyte irrespective of the nature of the other ion with which it is associated.

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

Hence,
${\Lambda }_{A{l}_2(S{O}_4{)}_3}^o={\Lambda }_{A{l}^{3+}}^o$ + ${\Lambda }_{S{O}_4^{2-}}^o$.