Question

Which of the following expressions correctly represents the equivalent conductance at infinite dilution of ${Al2(SO4)3}$ ? Given that ${\wedge^{\circ}_{Al^{3+}}}$ and ${- \wedge^{\circ}_{SO^{2-}_4}}$ are the equivalent conductances at infinite dilution of the respective ions

• A. ${ \wedge^{\circ}_{Al^{3+}} + \wedge^{\circ}_{SO_4^{2-}}}$
• B. $ \left({\wedge^{\circ}_{Al^{3+}} + \wedge^{\circ}_{SO_4^{2-}}} \right) \times 6$
• C. $ \frac{1}{3}{\wedge^{\circ}_{Al^{3+}} + \frac{1}{2} \wedge^{\circ}_{SO_4^{2-}}} $
• D. ${2 \wedge^{\circ}_{Al^{3+}} + 3 \wedge^{\circ}_{SO_4^{2-}}}$

Answer 1

The Correct Option is A

$A{{l}_{2}}{{(S{{O}_{4}})}_{3}}2A{{l}^{3+}}+3SO_{4}^{2-}$.
Since equivalent conductance's are given only for ions, the equivalent conductance at infinite dilution,
$\Lambda _{eq}^{\infty }=\Lambda _{Al\,3+}^{o}+\Lambda _{so_{4}^{2-}}^{o}$

Answer 2

Ans: The equivalent conduction at the infinite dilution (Λ°), for an electrolyte is calculated by the use of Kohlrausch’s law. It states that, 

At infinite dilution, each of the ion makes a constant contribution towards the equivalent conductivity of an electrolyte, this is irrespective of the nature of the other ion that is present in the solution. 

the expression for the equivalent conductance at infinite dilution of Al2(SO4)3 is:

Λ°(Al2(SO4)3) = 2Λ°(Al^3+) + 3Λ°(SO4^2-) 

Since equivalent conductance's are given only for ions, the equivalent conductance at infinite dilution,

Λ°(Al^3+) + Λ°(SO4^2-)