Final answer:
The freezing point of the aqueous solution is approximately -0.23°C.
Step-by-step explanation:
To calculate the freezing point of an aqueous solution, we can use the formula:
ΔT = Kf × m
where ΔT is the change in freezing point, Kf is the freezing point depression constant, and m is the molality of the solution.
Given that 20.0 g of Ca(NO3)2 is dissolved in 500.0 g of water, we first need to calculate the molality of the solution.
Molality (m) = moles of solute / mass of solvent (in kg)
To calculate the moles of Ca(NO3)2, we divide the given mass by the molar mass of Ca(NO3)2 (164.09 g/mol). This gives us:
moles of Ca(NO3)2 = 20.0 g / 164.09 g/mol
Next, we convert the mass of water to kg:
mass of water = 500.0 g / 1000 = 0.5 kg
Now we can calculate the molality:
molality (m) = moles of solute / mass of solvent (in kg)
Plugging in the values:
molality (m) = (20.0 g / 164.09 g/mol) / 0.5 kg
Using the freezing point depression constant for water (Kf = 1.86 °C/mol/kg), we can calculate the change in freezing point:
ΔT = 1.86 °C/mol/kg × (20.0 g / 164.09 g/mol) / 0.5 kg
Simplifying the expression:
ΔT = 1.86 °C/mol/kg × 0.1215 mol/kg
ΔT = 0.2279 °C
To find the freezing point of the solution, we subtract the change in freezing point from the freezing point of pure water (0°C):
Freezing point = 0°C - 0.2279 °C
Freezing point = -0.2279 °C