10.0 mg of a certain protein were dissolved in water to make a solution of 3.50 mL at 28.0o C. If the osmotic pressure is 1.30 torr, what is the molar mass of the protein?

Answer 1

pi = MRT
pi = 1.30 torr x (1 atm/760 torr) = ??
M = solve for this
R = 0.08206
T = 273 + 28? (You didn't write units. I asume you omitted the C for 28 C.).
Solve for molarity = M

Then moles = M x L = M from above x 3.50/1000 = xx

Then moles = grams/molar mass or
molar mass = grams/moles.
Plug in grams from the problem (0.010 g) and moles from above. Solve for molar mass.