Osmotic Pressure problem with a Protein?
A protein solution containing 90.0 mg of the protein (dissolved in water) displays an osmotic pressure of 319.88 Pa at 310.15 K. If the volume of the protein solution is 27.0 mL, what is the molar mass of the protein?
A protein solution containing 90.0 mg of the protein (dissolved in water) displays an osmotic pressure of 319.88 Pa at 310.15 K. If the volume of the protein solution is 27.0 mL, what is the molar mass of the protein?
1 Answer
#M ~~ "26900 g/mol"#
Doesn't this look just like the ideal gas law? It's no coincidence...
#Pi = icRT# where:
#Pi# is the osmotic pressure, i.e. the pressure needed to prevent solvent flow across the semi-permeable membrane to the side with more solute...#c# is the concentration in... guess what?#"mol/L"# . Just like the ideal gas law,#n//V# ...#R# is the universal gas constant, which is, guess what?#"0.082057 L"cdot"atm/mol"cdot"K"# .#T# is the temperature in#"K"# .- Take a guess at what
#i# is. It relates to boiling point elevation and freezing point depression...
In this case we don't know what
#Pi = (1) n/VRT#
#=> n = (PiV)/(RT)#
#= (319.88 cancel"Pa" xx (cancel"1 atm")/(101325 cancel"Pa") cdot 27.0 cancel"mL" xx "1 L"/(1000 cancel"mL"))/("0.082057 L"cdot"atm/mol"cdot"K" cdot "310.15 K")#
#= 3.349 xx 10^(-6) "mols protein"#
As a result, the molar mass is
#color(blue)(M) = (90.0 cancel"mg" xx "1 g"/(1000 cancel"mg"))/(3.349 xx 10^(-6) "mols")#
#=# #"26872 g/mol"# #-># #color(blue)("26900 g/mol")#
