An artificial cell is made by encapsulating 0.2 M glucose solution within a semipermeable membrane. The osmotic pressure developed when the artificial cell is placed within a 0.05 M solution of NaCl at 300 K is ________ × 10⁻¹ bar. (nearest integer).
[Given : R=0.083 L bar mol⁻¹ K⁻¹ ]
Assume complete dissociation of NaCl

Solution

Chemistry ma’am · 4 steps

Osmotic pressure here comes from the difference in particle concentration across the membrane.

1. Count particles inside. Glucose does not dissociate, so i = 1. The particle concentration inside the cell is 1 × 0.2 = 0.2 M.
2. Count particles outside. NaCl → Na⁺ + Cl⁻ dissociates completely, so i = 2. The particle concentration outside is 2 × 0.05 = 0.1 M.

   π = Δ(iC) RT = (0.2 - 0.1) × 0.083 × 300

3. Put in numbers. 0.083 × 300 = 24.9, and 0.1 × 24.9 = 2.49 bar. In the units asked, that is 24.9 × 10⁻¹ bar.
4. So. The osmotic pressure is 24.9 × 10⁻¹ bar, which rounds to 25.

The common slip: 37. Leaving out the van't Hoff factor of NaCl makes the difference 0.15 M and gives 37.35 × 10⁻¹ bar, reported as 37, not 25.

The idea to keep. When a membrane separates two solutions, find iC on each side first and subtract. Water moves because of the difference, so π = Δ(iC)RT.

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