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
See you post above.
To determine the freezing point of the solution, we can use the equation for the freezing point depression:
ΔTf = i * Kf * m
where:
ΔTf is the change in freezing point,
i is the van't Hoff factor (the number of particles per formula unit that the solute dissociates into),
Kf is the cryoscopic constant (a constant specific to the solvent),
m is the molality of the solution (moles of solute per kilogram of solvent).
First, let's calculate the molality (m) of the solution:
Given that the density of the solution is 1.50 g/mL, we need to convert it to kg/L:
1 mL = 1 gram, and 1 L = 1000 mL,
So, the density of the solution is 1.50 g/mL * (1 kg / 1000 g) * (1000 mL / 1 L) = 1.50 kg/L
Next, we need to calculate the molality (m) of the CaCl2 solution:
m = moles of solute / mass of solvent (in kg)
We are given the concentration of CaCl2 as 0.050M, which means there are 0.050 moles of CaCl2 in 1 liter (1000 g) of the solution.
So, moles of CaCl2 = 0.050 mol
Now, we calculate the mass of the solvent, which is the solution minus the solute:
Mass of solvent = volume * density = 1.00 L * 1.50 kg/L = 1.50 kg
Now we can calculate the molality:
m = 0.050 mol / 1.50 kg = 0.0333 mol/kg
Next, we need to determine the van't Hoff factor (i). CaCl2 dissociates into three particles in water: Ca2+ and two Cl- ions. Therefore, the van't Hoff factor (i) for CaCl2 is 3.
Finally, we need to know the cryoscopic constant (Kf) for the solvent at 25°C. Since the solvent is not specified, we cannot determine the exact value for Kf. Each solvent has a specific Kf value. For water, the typical value of Kf is approximately 1.86 °C/m.
Now we can substitute the values into the equation:
ΔTf = i * Kf * m
ΔTf = 3 * 1.86 °C/m * 0.0333 mol/kg
ΔTf = 0.186 °C
The freezing point of the solution will be lowered by 0.186 °C. Therefore, the freezing point of the solution will be the freezing point of the pure solvent (water) minus 0.186 °C.