Question

can some one help to solve the question 4 bellow. the answer is also bellow, show all work thanks

4. When a 0.51 g sample of an isolated protein was dissolved in a 100.0 mL solution at 25 °C, the osmotic pressure against the buffer solution without protein was 4.5 mmHg. Determine the molecular weight of this protein. What was the concentration of protein (as a molarity) in the solution?

solution: Obtain molar mass21100g/mol

Answer 1

Osmotic pressure, π = MRT

where M is molarity, R is the universal gas constant = 0.0821 L atm/mol K, T = 25 0C = 25+273 K = 298 K

π = 4.5 mm Hg

1 atm = 760 mmHg ⇒ 1 mmHg = 1/760 atm

So, 4.5 mmHg = 4.5/760 = 0.00592 atm

π = 0.00592 atm = M* 0.0821*298

Molarity , M = 0.00592/(0.0821*298) = 0.000242 M

Molarity = number of moles / volume of the solution in liter

0.000242 = number of moles / 0.1

So, number of moles = 0.000242 *0.1 = 0.0000242

Number of moles = mass / molar mass

molar mass = mass / number of moles = 0.51/0.0000242 = 21,077 g / mol