Question

By how many degrees is the freezing point decreased for a sugar solution lowered by dissolving 375.00 g of sucrose (342.3 g/mol) in 1500.0 g of water (Kf of water = 1.86 °C • kg solvent/mol solute)?

Answer 1

To calculate the decrease in freezing point caused by dissolving 375.00 g of sucrose in 1500.0 g of water, we can use the equation ΔTf = Kf * m * i.

1. Calculate the moles of sucrose: 1.095 mol.

2. Calculate the molality: 0.730 mol/kg.

3. The van't Hoff factor for sucrose is 1.

4. Use the equation ΔTf = 1.86 °C • kg solvent/mol solute * 0.730 mol/kg * 1.

5. ΔTf = 1.360 °C.

Therefore, the freezing point is decreased by 1.360 °C when 375.00 g of sucrose is dissolved in 1500.0 g of water.