The osmotic pressure of a solution containing 22.7 mg of an unknown protein in 50.0 mL of solution is 2.88 mmHg at 25 C. Determine the molar mass of the protein.

Molar mass for the protein is 2929.03 g/mol

Explanation:

To determine the osmotic pressure (one of the four colligative properties) we apply this formula:

π = M . R . T

π → Pressure

It must be in atm, so let's convert the mmHg to atm

2.88 mmHg . 1 atm / 760 mmHg = 0.00378 atm

T → Absolute T° (K) = T°C + 273 → 25°C + 273 = 298K

R = 0.082 L.atm/mol.K

Let's replace data: 0.00378 atm = M . 0.082 L.atm/mol.K . 298K

0.00378 atm / (0.082 L.atm/mol.K . 298K) = M

1.55×10⁻⁴ mol/L = M

These are the moles of protein that are contained in 1L of solution, but our volume is 50mL. Let's convert it to L

50mL . 1L / 1000 mL = 0.05 L

Now we can determine the moles of protein we used

1.55×10⁻⁴ mol/L . 0.05L = 7.75×10⁻⁶ moles

This moles corresponds to 22.7 mg of mass. Let's convert the mg to g to find out the molar mas mass

22.7 mg . 1g / 1000 mg = 0.0227 g

Molar mass → g/mol → 0.0227 g / 7.75×10⁻⁶ moles = 2929.03 g/mol

the ratio of the conjugate base to the acid required to attain the desired ph.