Here's how to determine the boiling point of a 3.60 mole aqueous sucrose solution:

Understanding the Concept

* Boiling Point Elevation: When a solute (like sucrose) is dissolved in a solvent (like water), the boiling point of the solution increases compared to the pure solvent. This is known as boiling point elevation.

* Molality: The concentration of the solution is expressed in molality (moles of solute per kilogram of solvent). We need to know the molality to calculate the boiling point elevation.

* The Formula: The boiling point elevation (ΔT_b) is calculated using the following formula:

ΔT_b = K_b * m

Where:

* ΔT_b is the boiling point elevation

* K_b is the molal boiling point elevation constant (for water, K_b = 0.512 °C/m)

* m is the molality of the solution

Calculations

1. Assume a Mass of Water: For simplicity, let's assume we have 1 kg (1000 g) of water.

2. Calculate Molality:

* Moles of sucrose = 3.60 moles

* Mass of water = 1 kg

* Molality (m) = (3.60 moles) / (1 kg) = 3.60 mol/kg

3. Calculate Boiling Point Elevation:

* ΔT_b = (0.512 °C/m) * (3.60 mol/kg) = 1.84 °C

4. Determine the Boiling Point:

* The normal boiling point of water is 100 °C.

* The boiling point of the solution is 100 °C + 1.84 °C = 101.84 °C

Therefore, the boiling point of a 3.60 mole aqueous sucrose solution would be approximately 101.84 °C.

Important Notes:

* This calculation assumes ideal behavior of the solution, meaning there are no significant interactions between the solute and solvent molecules that would affect the boiling point.

* In reality, the actual boiling point may differ slightly due to non-ideal behavior.

* It's crucial to work with the molality of the solution, not molarity, when calculating boiling point elevation. Molality is based on the mass of the solvent, which remains constant regardless of temperature changes.