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Chemistry · General chemistry II · Worked example

Molar mass of a protein from osmotic pressure

A solution of 0.500 g of a protein in enough water to make 100.0 mL has an osmotic pressure of 1.86 torr at 25.0 °C. Find the protein’s molar mass.

Π=MRT⇒n=Π⁢VR⁢T

Convert to the units of R

With R = 0.08206 L·atm/(mol·K), use atmospheres, litres and kelvin. The protein is a nonelectrolyte, so i = 1.

Π=1.86760 atm=2.447×10−3 atm,T=298.15 K,V=0.1000 L

Moles of protein

Solve Π = (n/V)RT for n.

n=(2.447×10−3)⁢(0.1000)(0.08206)⁢(298.15)=1.000×10−5 mol

Molar mass

Divide the mass of the sample by the moles.

M=0.500 g⁡1.000×10−5 mol=5.00×104 g/mol

Result

The protein’s molar mass is about 5.00 × 10⁴ g/mol.

Your turn

Why would a freezing-point measurement be a poor way to find this protein’s molar mass?

Show the answer and explanation

The freezing-point change would be only about 0.0002 °C.

The molality is about 1.00 × 10⁻⁵ mol ÷ 0.100 kg = 1.0 × 10⁻⁴ m, so ΔTf ≈ 1.86 × 1.0 × 10⁻⁴ ≈ 1.9 × 10⁻⁴ °C: far too small to measure. The same solution gives an easily measured 1.86 torr of osmotic pressure.

ΔTf⁡≈(1.86)⁢(1.0×10−4)≈1.9×10−4 ∘C

Keep exploring

In the Colligative studio, enter 1.00 × 10⁻⁵ mol in 100.0 mL at 25 °C and confirm the osmotic pressure is 1.86 torr.

Return to the concept →
Sources and scope

Authored study material. Tool results depend on the stated inputs and model assumptions.

  • Tro, Chemistry: A Molecular Approach, 4th ed., §13.5 Expressing Solution Concentration, pp. 585–592 (molality, p. 588)
  • Tro, Chemistry: A Molecular Approach, 4th ed., §13.6 Colligative Properties: Vapor Pressure Lowering, Freezing Point Depression, Boiling Point Elevation, and Osmotic Pressure, pp. 593–605 (volatile solutes, p. 597; osmotic pressure, pp. 603–604)
  • Tro, Chemistry: A Molecular Approach, 4th ed., §13.7 Colligative Properties of Strong Electrolyte Solutions, pp. 605–607
  • OpenStax Chemistry 2e — Colligative properties