Problem 89P

A solution is made by dissolving 23.88 mg of a protein in water and diluting to a total volume of 20.0 mL. The osmotic pressure of the solution is 3.66 torr at 25°C. What is the molar mass of the protein? (The osmotic pressure of a solution is given by the following equation: Osomotic Pressure = M × R × T, where M is the molarity of the solution, R is the gas constant, and T is temperature in K.)

Step 1

Given:

Mass of protein = 23.88 mg = 0.02388 g

Volume = 20.0 mL = 0.020L

Temperature = 25 oC + 273 = 298K

Osmotic pressure = 3.66 torr

Step-2

Osmotic pressure = M RT

Where M = molarity, R = gas constant and T = temperature in K

0.0048 atm = M 0,0821 L atm mol-1 K-1 298K

Moles (n) = 3.92 10-6 mol

3.92 10-6 mol = 0.02388g / molar mass

Molar mass = 0.02388g / 3.92 10-6 mol

= 6.09103 g/ mol

Thus the molar mass of protein is 6.09103 g/ mol