At 27° C, a solution containing 2.5 g of solute in 250.0 mL of solution exerts an osmotic pressure of 400 Pa. The molar mass of the solute is ___________ g mol⁻¹ (Nearest integer)
(Given : R=0.083 L bar K⁻¹ mol⁻¹)

Solution

Chemistry ma’am · 4 steps

With R in L bar, the pressure must be in bar and the volume in litres before you solve for M.

1. Match the units. R is in L bar K⁻¹ mol⁻¹. Since 1 bar = 10⁵ Pa, π = 400 Pa = 4 × 10⁻³ bar. Also V = 250.0 mL = 0.25 L and T = 27 + 273 = 300 K.
2. Write van 't Hoff. Osmotic pressure is π = CRT, and the molarity is C = w/MV. Rearrange for the molar mass.

   M = wRT/(π V) = (2.5 × 0.083 × 300)/((4 × 10⁻³) × 0.25)

3. Put in numbers. The numerator is 2.5 × 0.083 × 300 = 62.25. The denominator is 4 × 10⁻³ × 0.25 = 1 × 10⁻³.
4. So. M = 62.25/10⁻³ = 62250 g mol⁻¹, so the answer is 62250.

The common slip: 62. Putting the volume in as 250 mL instead of 0.25 L makes the denominator 1000 times too large and gives 62, not 62250.

The idea to keep. In π V = w/MRT, let the units of R decide everything: L bar means pressure in bar and volume in litres. A very small osmotic pressure points to a very large molar mass, such as a polymer or protein.

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