calculate the freezing point and osmotic pressure at 25 degrees celcious of an aqueous solution containing 1.0 g/L of a protein (mm=9.0*10^4g/mol) if the density of the solution is 1.0 g/cm^3
Note the correct spelling of celsius.
delta T = kf*molality
Calculate delta T (then freezing point) knowing Kf and molality.
molality = moles/kg solvent. You have kg solvent (but you will need to use the density) and moles = g/molar mass.
For osmotic pressure, pi = MRT.
don't forget to change T to Kelvin.
To calculate the freezing point and osmotic pressure of the aqueous protein solution, we need to use colligative properties. The two colligative properties we will be considering are freezing point depression and osmotic pressure.
First, let's calculate the freezing point depression using the equation:
∆Tf = Kf * i * m
where ∆Tf is the change in freezing point, Kf is the cryoscopic constant for water (1.86 °C/m), i is the van't Hoff factor, and m is the molality of the solute.
Molality (m) is defined as moles of solute per kilogram (kg) of solvent. In this case, the solvent is water.
Given:
Mass of protein = 1.0 g
Molar mass of protein (mm) = 9.0 * 10^4 g/mol
Density of solution = 1.0 g/cm^3
Temperature (T) = 25 °C = 25 + 273.15 K = 298.15 K
First, we need to calculate the moles of protein using the given mass and molar mass.
Moles of protein = Mass of protein / Molar mass of protein
= (1.0 g) / (9.0 * 10^4 g/mol)
= 1.11 * 10^-5 moles
Next, we need to calculate the mass of water in the solution. Since the density of the solution is 1.0 g/cm^3, the mass of water will be equal to the mass of the solution.
Mass of water = Mass of solution
= 1.0 g
Now, we can calculate the molality:
Molality (m) = Moles of protein / Mass of water (in kg)
= (1.11 * 10^-5 moles) / (1.0 kg)
= 1.11 * 10^-5 mol/kg
Now we can calculate the freezing point depression:
∆Tf = Kf * i * m
Since it's a protein solution, the van't Hoff factor (i) is usually close to 1, assuming the protein does not dissociate into ions. Therefore, we can take i as 1 for simplicity.
∆Tf = (1.86 °C/m) * (1) * (1.11 * 10^-5 mol/kg)
= 1.92 * 10^-5 °C
To calculate the freezing point of the solution, we subtract the freezing point depression (∆Tf) from the freezing point of the pure solvent, which is 0 °C.
Freezing point = 0 °C - 1.92 * 10^-5 °C
= -1.92 * 10^-5 °C
Now, let's calculate the osmotic pressure using the formula:
π = i * M * R * T
where π is the osmotic pressure, i is the van't Hoff factor, M is the molarity of the solution, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.
To calculate molarity, we need to determine the number of moles of the protein and the volume of the solution.
Number of moles of protein = Moles of protein = 1.11 * 10^-5 moles
Volume of solution = Mass of solution / Density of solution
= 1.0 g / 1.0 g/cm^3
= 1.0 cm^3
Molarity (M) = Number of moles of solute / Volume of solution (in L)
= (1.11 * 10^-5 moles) / (1.0 * 10^-3 L)
= 1.11 * 10^-2 M
Now we can calculate the osmotic pressure:
π = (1) * (1.11 * 10^-2 M) * (0.0821 L·atm/(mol·K)) * (298.15 K)
π = 2.82 * 10^-3 atm
Therefore, at 25°C, the freezing point of the aqueous protein solution is approximately -1.92 * 10^-5 °C, and the osmotic pressure is approximately 2.82 * 10^-3 atm.