The osmotic pressure exerted by a solution prepared by dissolving 2.0 g of protein of molar mass 60 kg mol⁻¹ in 200 mL of water at 27°C is ______________ Pa. [integer value]
(use R = 0.083 L bar mol⁻¹ K⁻¹)

Solution

Chemistry ma’am · 4 steps

Convert the protein's molar mass to grams first, then use π = CRT and change bar to pascal.

1. Moles of protein. The molar mass is 60 kg mol⁻¹ = 60 000 g mol⁻¹. So n = 2.0/(60 000) = 3.33 × 10⁻⁵ mol.
2. Molarity. The volume is 200 mL = 0.2 L, so C = (3.33 × 10⁻⁵)/0.2 = 1.67 × 10⁻⁴ mol L⁻¹. The temperature is 27 + 273 = 300 K.

   π = CRT = 1/6000 × 0.083 × 300 = 4.15 × 10⁻³ bar

3. Bar to pascal. With R in L bar mol⁻¹ K⁻¹, π comes out in bar, and 1 bar = 10⁵ Pa. So π = 4.15 × 10⁻³ × 10⁵ Pa.
4. So. The osmotic pressure is 415 Pa.

The common slip: 415000. Taking the molar mass as 60 g mol⁻¹ instead of 60 000 g mol⁻¹ makes the moles 1000 times too large and gives 415 000 Pa.

The idea to keep. Osmotic pressure is the colligative property used for proteins and polymers, because even a tiny concentration gives a pressure you can measure. Match the units of R, then convert to the unit the question asks for.

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