A \(0.10-m\) NaCl aqueous solution has a higher boiling point than a $0.10-m \mathrm{MgSO}_{4}$ aqueous solution. Explain why.

Boiling point elevation is a colligative property, which means it depends on the number of solute particles in a solution, and not their specific nature. The boiling point elevation can be calculated using the following formula: \[ΔT_b = i K_b molality \]Where ΔTb is the elevation in boiling point, i is the van't Hoff factor (number of solute particles formed per formula unit of solute), Kb is the boiling point elevation constant, and molality is the number of moles of solute per kilogram of solvent.

The van't Hoff factor (i) represents the number of particles formed when a solute dissociates in a solvent. For NaCl, when it dissolves in water, it dissociates into Na+ and Cl- ions. Therefore, the van't Hoff factor for NaCl is 2. For MgSO4, when it dissolves in water, it dissociates into Mg2+ and SO42- ions. Therefore, the van't Hoff factor for MgSO4 is also 2.

Both NaCl and MgSO4 solutions have the same molality (0.10 m) and van't Hoff factor (2). However, NaCl has a higher boiling point elevation than MgSO4. This can be explained by the fact that dissociation of MgSO4 is not complete, i.e., some MgSO4 molecules don't dissociate into ions in the solution, leading to a lower effective concentration of particles contributing to the boiling point elevation. In the case of NaCl, there is almost complete dissociation, so the effective concentration of particles is higher, leading to a higher boiling point elevation. Thus, a 0.10 m NaCl aqueous solution has a higher boiling point than a 0.10 m MgSO4 aqueous solution because NaCl dissociates more completely into ions than MgSO4, leading to a greater effective concentration of solute particles and a higher boiling point elevation.