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

Question

Alkyl halide and aryl halide react in the presence of sodium metal and dry ether. The name of the reaction is?

Answer 1

To find the equivalent conductance at infinite dilution of $ \text{Al}_2(\text{SO}_4)_3 $, we can use the concept of equivalent conductance for ionic compounds. The equivalent conductance at infinite dilution, denoted by $ \Lambda^\circ $, is the conductance of all ions produced by one gram equivalent of an electrolyte when dissolved in a solution. According to this concept:

The expression for the equivalent conductance at infinite dilution is given by:

$\Lambda^\circ = \nu_+ \lambda^\circ_{+} + \nu_- \lambda^\circ_{-}$

Where:

$\nu_+$ and $\nu_-$ are the stoichiometric coefficients of the cation and anion respectively.
$\lambda^\circ_{+}$ and $\lambda^\circ_{-}$ are the molar conductances at infinite dilution for the cation and anion respectively.

In the case of $ \text{Al}_2(\text{SO}_4)_3 $:

$\nu_+$ (for $ \text{Al}^{3+} $) = 2
$\nu_-$ (for $ \text{SO}_4^{2-} $) = 3

Therefore, the expression becomes:

$ \Lambda^\circ = 2 \lambda^\circ_{Al^{3+}} + 3 \lambda^\circ_{SO_4^{2-}} $

However, the expression for the equivalent conductance at infinite dilution in terms of equivalent conductance of ions $\wedge^\circ$ is:

$\wedge^\circ = \wedge^\circ_{cation} + \wedge^\circ_{anion}$

Rewriting each ionic conductance in terms of equivalent conductance at infinite dilution:

For $ \text{Al}^{3+} $, the contribution is $ \wedge^\circ_{Al^{3+}} $
For $ \text{SO}_4^{2-} $, the contribution is $ \wedge^\circ_{SO_4^{2-}} $

Thus, the equivalent conductance at infinite dilution for $ \text{Al}_2(\text{SO}_4)_3 $ should be:

$\wedge^\circ = \wedge^\circ_{Al^{3+}} + \wedge^\circ_{SO_4^{2-}}$

Hence, the correct expression is the first option: $\wedge^\circ_{Al^{3+}} + \wedge^\circ_{SO_4^{2-}}$.

Answer 2