1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42 × 10⁻³ bar.
The molar mass of the biopolymer is _____________ × 10⁴ g mol⁻¹. (Round off to the Nearest Integer)
[Use : R = 0.083 L bar mol⁻¹ K⁻¹]

Solution

Chemistry ma’am · 4 steps

Osmotic pressure is the clean way to weigh a big polymer: rearrange π V = nRT and solve for M.

1. Write moles as w/M. Use π = CRT with C = n/V and n = w/M. Rearranged, this gives M directly.

   M = wRT/(π V)

2. Match the units. R is in L bar mol⁻¹ K⁻¹, so the volume must be in litres: V = 100 mL = 0.1 L. The pressure stays in bar: π = 2.42 × 10⁻³ bar.
3. Put in numbers. Numerator: 1.46 × 0.083 × 300 = 36.354. Denominator: 2.42 × 10⁻³ × 0.1 = 2.42 × 10⁻⁴. So M = 36.354/(2.42 × 10⁻⁴) ≈ 1.502 × 10⁵ g mol⁻¹.
4. So. M ≈ 15.02 × 10⁴ g mol⁻¹, which rounds to 15.

The idea to keep. For polymers and proteins, use osmotic pressure to find molar mass: M = wRT/(π V). Always match the units of V and π to the units of R before you multiply.

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