An aqueous solution of volume 300 cm³ contains 0.63 g of protein. The osmotic pressure of the solution at 300 K is 1.29 mbar. The molar mass of the protein is ___________ g mol⁻¹
Given : R = 0.083 L bar K⁻¹ mol⁻¹

Solution

Chemistry ma’am · 4 steps

Osmotic pressure suits proteins: rearrange π V = w/MRT for M, with units matched to R.

1. Set up the formula. Osmotic pressure is π = CRT, and the molarity is C = w/MV. Rearranging gives the molar mass directly.

   M = wRT/(π V)

2. Match units to R. R is in L bar, so convert both quantities. The pressure is π = 1.29 mbar = 1.29 × 10⁻³ bar, and the volume is V = 300 cm³ = 0.3 L.
3. Put in numbers. The numerator is 0.63 × 0.083 × 300 = 15.687. The denominator is 1.29 × 10⁻³ × 0.3 = 3.87 × 10⁻⁴. So M = 15.687/(3.87 × 10⁻⁴) ≈ 40535 g mol⁻¹.
4. So. The molar mass of the protein is 40535 g mol⁻¹.

The common slip: 40.53. Leaving the pressure in mbar, or the volume in cm³, makes the answer 1000 times too small and gives 40.53.

The idea to keep. For osmotic pressure, put pressure and volume in the same units as R before you substitute. A large molar mass like this one is normal for a protein.

Answer from the NTA final answer key. Solution by PrepWiser.