What is the freezing point depression of a sugar solution prepared 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)?

Question

Quizplus · Accepted Answer

The formula for calculating freezing point depression is:

∆Tf = Kf × m

where ∆Tf is the change in freezing point, Kf is the freezing point depression constant for the solvent (in this case, water), and m is the molality of the solution (moles of solute per kg of solvent).

First, we need to calculate the molality of the solution:

moles of sucrose = mass ÷ molar mass = 375.00 g ÷ 342.3 g/mol = 1.094 mol

mass of water = 1500.0 g

molality = moles of solute ÷ kg of solvent = 1.094 mol ÷ 1.5 kg = 0.729 mol/kg

Now, we can plug in the values and solve for ∆Tf:

∆Tf = Kf × m = 1.86 °C∙kg/mol × 0.729 mol/kg = 1.36 °C

However, we need to remember that the freezing point depression is negative, so the correct answer is:
B) -1.36 °C

What Is the Freezing Point Depression of a Sugar Solution