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 find the freezing point of a solution, you need to use the formula for the freezing point depression:
ΔT = K_f × m
Where:
ΔT is the freezing point depression
K_f is the cryoscopic constant (a constant value depending on the solvent)
m is the molality of the solute in the solution
First, let's calculate the molality (m) of the CaCl2 solution.
Molality (m) is defined as the number of moles of solute (CaCl2) per kilogram of solvent (water). We have 2.20 mol of CaCl2 dissolved in 1.00 L of water. However, we need to convert this volume to mass in kilograms, as molality requires mass.
To convert the volume to mass, we need to know the density of water.
The density of water is approximately 1.00 g/mL. Since 1 L is equivalent to 1000 mL, the mass of 1.00 L of water is 1000 g.
Therefore, the molality (m) of the CaCl2 solution is:
m = (moles of solute) / (mass of solvent in kg)
= (2.20 mol) / (1.00 kg)
= 2.20 m
Now, we need to determine the cryoscopic constant, K_f, for water.
The cryoscopic constant for water is 1.86 °C/m.
Finally, we can substitute the values into the freezing point depression formula:
ΔT = K_f × m
= 1.86 °C/m × 2.20 m
= 4.092 °C
Therefore, the freezing point of this CaCl2 solution at 1 atm of pressure is lowered by 4.092 °C.