If a sample of red blood cells at an osmotic pressure of 7.70 atm is placed in a solution that contains 0.050M calcium chloride (CaCl2) at 25°C,
what is the freezing point of this solution if the density of the solution is 1.50 g/mL
delta T = i*Kf*molality.
You know i = 3 from CaCl2. You know Kf (or you can look it up). I THINK it's about 1.9 or so. All you need is the molality. You have the M. Convert this way.
Assume you have a liter of solution so the mass of solution is 1000 mL x 1.5 g/mL = 1500 grams. How much of that is CaCl2. grams CaCl2 = mols CaCl2 x molar mass CaCl2 = approx 6 grams but you need to verify that and all of the other estimates I've done.
So grams H2O = 1500-g CaCl2 = grams H2O. Divide by 1000 to convert to kg. Then m = 0.05/kg solvent. Finally, substitute this into the dT = i*Kf*m to find delta T and subtract from the normal freezing point of H2O which is zero. By the way I doubt that the density of that CaCl2 solution is anywhere near 1.5 g/mL.
To find the freezing point of the solution, we need to use the formula for freezing point depression:
ΔT = Kf * i * m
Where:
ΔT is the change in temperature (freezing point depression)
Kf is the cryoscopic constant (a property specific to a particular solvent)
i is the van't Hoff factor, which represents the number of particles formed when the solute dissolves
m is the molality of the solution (moles of solute per kilogram of solvent)
First, let's calculate the molality of the calcium chloride solution:
Molarity (M) = moles of solute / volume of solution (L)
Given:
Molarity (M) = 0.050 M
Density of solution = 1.50 g/mL
We can use the density of the solution to convert the volume into mass:
Volume of solution = Mass of solution / Density of solution
Since the volume of the solution is not given, we will assume a convenient volume, such as 100 mL (0.100 L):
0.050 mol/L = moles of solute / 0.100 L
Moles of solute = 0.050 mol/L * 0.100 L = 0.005 mol
Now, let's calculate the molality:
Molality (m) = moles of solute / mass of solvent (kg)
The solvent for calcium chloride is water, and the mass of 0.100 L of water can be calculated using its density:
Mass of solvent = Volume of solvent * Density of water
Since the density of water is about 1 g/mL, the mass of 0.100 L is:
Mass of solvent = 0.100 L * 1.00 g/mL = 0.100 kg
Molality (m) = 0.005 mol / 0.100 kg = 0.050 mol/kg
Now, we need to determine the van't Hoff factor (i). For calcium chloride (CaCl2), i is 3 because it dissociates into three ions: Ca2+ and 2 Cl- ions.
Now, we need the cryoscopic constant (Kf) for water, which is 1.86 °C/m (degrees Celsius per molal solute).
Finally, we substitute the known values into the formula:
ΔT = Kf * i * m
ΔT = 1.86 °C/m * 3 * 0.050 mol/kg
ΔT = 0.278 °C
The freezing point depression (change in temperature) is 0.278 °C below the normal freezing point of pure water. To find the freezing point of the solution, subtract this value from the freezing point of pure water at 25 °C, which is 0 °C.
Freezing point of solution = 0 °C - 0.278 °C = -0.278 °C
Therefore, the freezing point of the calcium chloride solution at 25 °C is approximately -0.278 °C.