A solution of 7.50 mg of a small protein in 5.00 mL aqueous solution has an osmotic pressure of 5.15 torr at 23.1°C.

Question

A solution of 7.50 mg of a small protein in 5.00 mL aqueous solution has an osmotic pressure of 5.15 torr at 23.1°C.



Answer 1

What is the question? Are you asked to determine the molar mass of the small protein? If so, here is how to go about that.

π = iMRT

π = osmotic pressure = 5.15 torr x 1 atm / 760 torr = 6.78x10^-3 atm

i = van't Hoff factor = 1 for a protein

M = molarity = moles / liter = ?

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = 23.1C + 273.15 = 296.25K

Solving for M, we have...

M = π / iRT = 6.78x10^-3 atm / (1)(0.0821 Latm/Kmol)(296.25K)

M = 2.79x10^-4 mol / L

Since we have 5.00 ml or 0.00500 L (and not 1 L), we can find the moles present:

2.79x10^-4 mol / L x 0.00500 L = 1.39x10^-6 moles of protein

Now, we calculate the molar mass:

molar mass = gram / moles

grams = 7.50 mg x 1 g / 1000 mg = 0.00750 g

moles = 1.39x10^-6 moles

molar mass = 0.00750 moles / 1.39x10^-6 moles

molar mass = 5396 g / mole