A 3.0 g sample of a small protein having a molecular weight of 42,000 g/mol is dissolved in 52.8 mL of water at 21°C.What is the osmotic pressure of the solution? (R = 0.0821 L · atm/K·mol,1 atm = 760 mmHg)
A) 25 mmHg
B) 1900 mmHg
C) 1.8 mmHg
D) 23,000 mmHg
E) 0.033 mmHg

Question

Quizplus · Accepted Answer

The osmotic pressure of a solution is given by the equation:$$\pi = MRT$$where:* $\pi$ is the osmotic pressure in atm* M is the molarity of the solution in mol/L* R is the ideal gas constant in L · atm/K·mol* T is the temperature in KWe can calculate the molarity of the solution by dividing the mass of the protein by its molecular weight and then dividing by the volume of the solution in liters:$$M = \frac{3.0 g}{42,000 g/mol} 	imes \frac{1}{0.0528 L} = 1.40 	imes 10^{-5} M$$We can then calculate the osmotic pressure by plugging in the values for M, R, and T:$$\pi = (1.40 	imes 10^{-5} M)(0.0821 L · atm/K·mol)(294 K) = 25 mmHg$$Therefore, the osmotic pressure of the solution is 25 mmHg.

A 30 G Sample of a Small Protein Having a Molecular