Answer:
The freezing point of the aqueous sucrose solution = -0.166 °C.
Step-by-step explanation:
To determine the freezing point of an aqueous solution, we can use the formula:
ΔTf = Kf · m
where ΔTf is the freezing point depression, Kf is the freezing point depression constant for the solvent (water), and m is the molality of the solution (moles of solute per kilogram of solvent).
The freezing point depression constant for water is 1.86 °C/m.
To calculate the molality, we need to know the molar mass of sucrose, which is 342.3 g/mol.
First, we need to calculate the number of moles of sucrose in the solution:
moles of sucrose = mass of sucrose / molar mass of sucrose
moles of sucrose = 0.24 mol/kg * 342.3 g/mol = 0.082 mol
Next, we need to calculate the mass of water in the solution:
mass of water = total mass of solution - mass of sucrose
mass of water = 1000 g - 0.24 mol/kg * 342.3 g/mol * 1000 g = 917.7 g
Now we can calculate the molality of the solution:
molality = moles of sucrose / kg of water
molality = 0.082 mol / 0.9177 kg = 0.0893 m
Finally, we can calculate the freezing point depression:
ΔTf = Kf · m
ΔTf = 1.86 °C/m · 0.0893 m = 0.166 °C
The freezing point depression is the difference between the freezing point of pure water (0 °C) and the freezing point of the solution. Therefore, the freezing point of the solution is:
freezing point = 0 °C - ΔTf
freezing point = 0 °C - 0.166 °C = -0.166 °C
So the freezing point of the aqueous sucrose solution is -0.166 °C.