A quantity of 2.20 mol of CaCl2 are dissolved in 1.00 L of water. What is the freezing point of this solution at 1 atm of pressure?
To determine the freezing point of a solution, you need to use the equation for freezing point depression, which is:
ΔTf = Kf * m
Where:
- ΔTf is the change in freezing point
- Kf is the molal freezing point constant (a constant for a specific solvent)
- m is the molality of the solution (moles of solute per kilogram of solvent)
First, let's calculate the molality of the solution.
Molality (m) = moles of solute / mass of solvent (in kg)
Given:
- Quantity of CaCl2 = 2.20 mol
- Volume of water = 1.00 L
- The molar mass of CaCl2 = 110.98 g/mol
To find the mass of the solvent, we need to convert the volume of water to kilograms:
Mass of water = volume * density of water
Density of water = 1 g/mL (or 1 kg/L)
Mass of water = 1.00 L * 1.00 kg/L = 1.00 kg
Next, calculate the molality:
Molality (m) = 2.20 mol / 1.00 kg = 2.20 mol/kg
Now, you need to know the molal freezing point constant (Kf) for water to find the change in freezing point (ΔTf). For water, Kf is 1.86 °C/m.
Finally, plug in the values into the equation:
ΔTf = (1.86 °C/m) * (2.20 mol/kg)
ΔTf = 4.092 °C
The freezing point of the solution is the freezing point of pure water (0 °C) minus the change in freezing point (ΔTf). Therefore:
Freezing point = 0 °C - 4.092 °C = -4.092 °C
So, the freezing point of the solution is approximately -4.092 °C at 1 atm of pressure.