A 2.2-g sample of a small protein having a molecular weight of 42,000 g/mol is dissolved in 31.2 mL of water at 28°C.What is the osmotic pressure of the solution? (R = 0.0821 L · atm/(K · mol))
A)32 mmHg
B)1300 mmHg
C)2.9 mmHg
D)18,000 mmHg
E)0.041 mmHg

Question

Quizplus · Accepted Answer

The osmotic pressure of the solution can be calculated using the formula:π = MRTwhere:π is the osmotic pressure in atmospheresM is the molarity of the solution in moles per literR is the ideal gas constant in L · atm/(K · mol)T is the temperature in KelvinFirst, we need to calculate the molarity of the solution:M = moles of solute / liters of solutionmoles of solute = mass of solute / molecular weight of solutemass of solute = 2.2 gmolecular weight of solute = 42,000 g/molmoles of solute = 2.2 g / 42,000 g/mol = 5.24 × 10^-5 molliters of solution = 31.2 mL = 0.0312 LM = 5.24 × 10^-5 mol / 0.0312 L = 0.00168 MNow we can calculate the osmotic pressure:π = MRTπ = (0.00168 M) × (0.0821 L · atm/(K · mol)) × (28°C + 273.15 K)π = 0.032 atm = 32 mmHgTherefore, the osmotic pressure of the solution is 32 mmHg.

A 22-G Sample of a Small Protein Having a Molecular Weight