A 2.3-g sample of a small protein having a molecular weight of 62,000 g/mol is dissolved in 38.8 mL of water at 28°C.What is the osmotic pressure of the solution? (R = 0.0821 L · atm/K·mol; 1 atm = 760 mmHg)
A) 18 mmHg
B) 1100 mmHg
C) 1.7 mmHg
D) 32,000 mmHg
E) 0.024 mmHg

Question

Quizplus · Accepted Answer

The osmotic pressure of the 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 KThe molarity of the solution is:$$M = \frac{moles\ of\ solute}{liters\ of\ solution}$$The number of moles of solute is:$$moles\ of\ solute = \frac{mass\ of\ solute}{molecular\ weight\ of\ solute}$$Substituting the given values into the equations, we get:$$moles\ of\ solute = \frac{2.3\ g}{62,000\ g/mol} = 3.71 	imes 10^{-5}\ mol$$$$M = \frac{3.71 	imes 10^{-5}\ mol}{38.8 	imes 10^{-3} L} = 9.56 	imes 10^{-4}\ M$$$$T = 28°C + 273.15 = 301.15\ K$$Substituting the values of M, R, and T into the equation for osmotic pressure, we get:$$\pi = (9.56 	imes 10^{-4}\ M)(0.0821\ L · atm/K·mol)(301.15\ K) = 0.018\ atm$$Converting atm to mmHg, we get:$$0.018\ atm 	imes 760\ mmHg/atm = 18\ mmHg$$Therefore, the osmotic pressure of the solution is 18 mmHg.

A 23-G Sample of a Small Protein Having a Molecular Weight