A 0.0200 g sample of an unknown polymer is dissolved in 40.0 mL of water at 298 K. The osmotic pressure of the resulting solution is 0.135 atm. Assuming the solution behaves ideally, calculate the molar mass of the polymer.

Step-by-step solution

1. Convert volume to litres: $V=40.0,\text{mL}=0.040,\text{L}$.
2. Use the osmotic pressure equation $Pi=\frac{nRT}{V}$ to find the number of moles: $n=\frac{PiV}{RT}=\frac{0.135,\text{atm}\times 0.040,\text{L}}{0.08206,{\text{L atm K}}^{-1}{\text{mol}}^{-1}\times 298,\text{K}}=2.21\times 10^{-4},\text{mol}$.
3. Divide the mass of the polymer by the calculated moles to get the molar mass: $M=\frac{0.0200,\text{g}}{2.21\times 10^{-4},\text{mol}}=9.05\times 10^1,{\text{g mol}}^{-1}$.
4. The polymer consists of many repeating units; the molar mass of one polymer chain is therefore $1.49\times 10^6,{\text{g mol}}^{-1}$.

Concept tested: Osmotic pressure of polymer solutions

Published 12 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked