An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to

1. increase in ionic mobility of ions

2. 100% ionization of electrolyte at normal dilution

3. increase in both, i.e, number of ions and ionic mobility of ions

4. increase in the number of ions

The correct expression that  represents the equivalent conductance at infinite dilution of $A{l}_2{\left(S{O}_4\right)}_3$ is:

(Given that ${\wedge }_{A{l}^{3+}}^{°} and {\wedge }_{S{O}_4^{2-}}^{°}$ are the equivalent conductances at infinite dilution of the respective ions)

1. ${\wedge }_{A{l}^{3+}}^{°} + {\wedge }_{S{O}_4^{2-}}^{°}$

2. $\left({\wedge }_{A{l}^{3+}}^{°} + {\wedge }_{S{O}_4^{2-}}^{°}\right)\times 6$

3. $\frac{1}{3}{\wedge }_{A{l}^{3+}}^{°} +\frac{1}{2} {\wedge }_{S{O}_4^{2-}}^{°}$

4. $2{\wedge }_{A{l}^{3+}}^{°} +3 {\wedge }_{S{O}_4^{2-}}^{°}$