Question

Which of the following expressions correctly represents the equivalent conductance at infinite dilution of \( \text{Al}_2(\text{SO}_4)_3 \)? Given that \( \Lambda^\circ_{\text{Al}^{3+}} \) and \( \Lambda^\circ_{\text{SO}_4^{2-}} \) are the equivalent conductances at infinite dilution of the respective ions?

• A. \( 2\Lambda^\circ_{\text{Al}^{3+}} + 3\Lambda^\circ_{\text{SO}_4^{2-}} \)
• B. \( \Lambda^\circ_{\text{Al}^{3+}} + \Lambda^\circ_{\text{SO}_4^{2-}} \)
• C. \( \left( \Lambda^\circ_{\text{Al}^{3+}} + 3\Lambda^\circ_{\text{SO}_4^{2-}} \right) \times 6 \)
• D. \( \frac{1}{3}\Lambda^\circ_{\text{Al}^{3+}} + \frac{1}{2}\Lambda^\circ_{\text{SO}_4^{2-}} \)

Hint:

For ionic compounds, the equivalent conductance at infinite dilution is the sum of the conductances of the individual ions in the solution.

Answer 1

The Correct Option is B

The equivalent conductance of a compound at infinite dilution is the sum of the individual conductances of its ions. Since \( \text{Al}_2(\text{SO}_4)_3 \) dissociates into two \( \text{Al}^{3+} \) ions and three \( \text{SO}_4^{2-} \) ions, the total conductance is simply the sum of the individual ion conductances: \[ \Lambda^\circ_{\text{Al}_2(\text{SO}_4)_3} = 2\Lambda^\circ_{\text{Al}^{3+}} + 3\Lambda^\circ_{\text{SO}_4^{2-}} \] The correct expression is therefore option (a), which represents the sum of the conductances of each ion.

Step 2: Conclusion.
The equivalent conductance at infinite dilution of \( \text{Al}_2(\text{SO}_4)_3 \) is \( 2\Lambda^\circ_{\text{Al}^{3+}} + 3\Lambda^\circ_{\text{SO}_4^{2-}} \), corresponding to option (a).