Here's how to calculate the change in freezing point of water when 35.5 g of sucrose is dissolved in 55.0 g of water:

1. Understand the Concepts

* Freezing Point Depression: When a solute (like sucrose) is dissolved in a solvent (like water), the freezing point of the solution is lower than the freezing point of the pure solvent. This is a colligative property, meaning it depends on the amount of solute, not its identity.

* Freezing Point Depression Equation: ΔT_f = K_f * m * i

* ΔT_f = change in freezing point (°C)

* K_f = molal freezing point depression constant (for water, K_f = 1.86 °C/m)

* m = molality of the solution (moles of solute per kg of solvent)

* i = van't Hoff factor (for sucrose, i = 1, as it doesn't dissociate in solution)

2. Calculate Molality (m)

* Molar mass of sucrose (C₁₂H₂₂O₁₁): 12(12.01 g/mol) + 22(1.01 g/mol) + 11(16.00 g/mol) = 342.34 g/mol

* Moles of sucrose: 35.5 g / 342.34 g/mol = 0.103 moles

* Mass of water in kg: 55.0 g / 1000 g/kg = 0.055 kg

* Molality: 0.103 moles / 0.055 kg = 1.87 mol/kg

3. Calculate the Change in Freezing Point (ΔT_f)

* ΔT_f = K_f * m * i

* ΔT_f = 1.86 °C/m * 1.87 mol/kg * 1

* ΔT_f = 3.48 °C

Therefore, the freezing point of the solution will be lowered by 3.48 °C compared to the pure water.