Colligative Properties -- Osmotic Pressure and Freezing Point Depression?

Question

Calculate the osmotic pressure at 25°C of an aqueous solution of 1.00 g/L of a protein having molar mass=4.15×104 g/mol.




Calculate the freezing point depression of the above solution, if the density of the solution is 1.00 g/cm3. (units can be either "K" or "degC")

Answer 1

convert the concentration from g/L to mole/L by dividing with the MW: C=1/(4.15*10^4) mole/L

π= CRT =(1/(4.15*10^4))*0.082*298 =5.89*10^-3 atm

If the density of the solution is 1.00 g/mL, then 1L solution is equal to 1 kg solution. In that you have 1 g solute, thus you have 999 g solvent. In this amount of solution you also have 1/(4.15*10^4) mole solute

molality is x mole solute per 1000 g solvent
you have 1/(4.15*10^4) mole in 999 g solvent
thus x =(1000/999)*1/(4.15*10^4) =2.41*10^-5
ΔT = Kf*m =1.86*2.41*10^-5 =4.48*10^-5 degC

Answer 2

find molarity by taking 1.00g/L and dividing it by 4.15X144 g/mol
then plug it into

osmotic pressure = (concentration in molarity)(gass constant .08206)(temp in K)