Step 1: Understanding the Concept
Elevation in boiling point is \(\Delta T_b = i K_b m\).
For equimolar solutions in the same solvent, \(K_b\) and \(m\) are equal, so \(\Delta T_b \propto i\).
Step 2: Find i for each salt
Assuming complete dissociation, \(i\) equals the number of ions per formula unit.
\(\text{KCl}\rightarrow \text{K}^{+}+\text{Cl}^{-}\), so \(i=2\).
\(\text{Li}_2\text{SO}_4\rightarrow 2\text{Li}^{+}+\text{SO}_4^{2-}\), so \(i=3\).
\(\text{Al}_2(\text{SO}_4)_3\rightarrow 2\text{Al}^{3+}+3\text{SO}_4^{2-}\), so \(i=5\).
\((\text{NH}_4)_3\text{PO}_4\rightarrow 3\text{NH}_4^{+}+\text{PO}_4^{3-}\), so \(i=4\).
Step 3: Compare the options
The largest value is \(i=5\) for aluminium sulphate, so it shows the maximum boiling point elevation.
Final Answer:
Aluminium sulphate gives the most ions (5), so it has the maximum boiling point elevation.
\[ \boxed{\text{(C) }\text{Aluminium sulphate}} \]