Question

A solution of protein (extracted from crabs) was prepared by dissolving 0.75 in 125 cm3 of an aqueous solution. At 4°C an osmotic pressure rise of 2.6 mm of the solution was observed. Then molecular weight of protein is ( Assume density of solution is 1.00 g/cm3)

Answer 1

The osmotic pressure of a solution is calculated by:

Pressure = M RT where M is the molarity of the solution, R is the gas constant, and T is the Kelvin temperature.

So, the concentration of this solution can be calculated, but first, convert the pressure to atm and the temperature to K

Now, the pressure was measured in terms of mm H2O, which you must first convert to mm Hg and then to atm:
Because the density of Hg is 13.6 times that of water,

2.6 mm H2O / 13.6 = 0.191 mm Hg
Then, 0.191 mm Hg / 760 = 2.5X10^-4 atm

T = 277 K
R = 0.0821 Latm/molK

Then, 2.5X10^-4 atm = M (0.0821)(277)
M = 1.1 X 10^-5 mol/L

Now, the concentration of the protein solution in terms of g/L is:

0.750 g / 0.125 L = 6 g/L

So, the molar mass of the protein is:

6.00 g/L / 1.1X10^-5 mol/L = 5.4 X 10^5 g/mol