Question

The osmotic pressure of a solution formed by dissolving 0.231 g of the protein in 150 mL of water at 298 K is found to be 2.054 torr.

Answer 1

Answer:

13924g/mol is the molar mass of the protein

Explanation:

What is the molar mass of the protein.

The osmotic pressure of a solution, π, is defined as:

π = c*R*T

Where π is osmotic pressure = 2.054torr* (1atm/760torr) = 2.70x10⁻³atm

c is molar concentration of the protein,

R is gas constant = 0.082atmL/molK

And T is absolute temperature = 298K

Solving the molar concentration:

c = π / RT

c = 2.70x10⁻³atm / 0.082atmL/molK*298K

c = 1.106x10⁻⁴ moles protein / L

The moles of protein in 150mL = 0.150L:

0150L * (1.106x10⁻⁴ moles protein / L) = 1.659x10⁻⁵ moles protein

The molar mass is the ratio of mass and moles:

0.231g / 1.659x10⁻⁵ moles protein =

13924g/mol is the molar mass of the protein