A 0.0250 g sample of an unknown polymer is dissolved in 40.0 mL of water at 298 K. The osmotic pressure of the solution is measured to be 0.210 atm. Assuming ideal behaviour, calculate the molar mass (M) of the polymer.
Answer
M = 1.23×10⁵ g·mol⁻¹
The correct answer is M = 1.23×10⁵ g·mol⁻¹ — Molar mass M = mass of solute / n = 0.0250 g / 3.44×10⁻⁴ mol = 7.27×10¹ g·mol⁻¹ ≈ 7.3×10¹ g·mol⁻¹. (Check calculation)
Step-by-step solution
- Convert the solution volume to litres: V = 40.0 mL = 0.0400 L.
- Use the osmotic pressure equation for an ideal dilute solution: Π = (n/V)RT, where n is the number of moles of solute.
- Re‑arrange to find n: n = ΠV / (RT).
- Substitute the values: n = (0.210 atm)(0.0400 L) / [(0.08206 L·atm·K⁻¹·mol⁻¹)(298 K)] = 3.44×10⁻⁴ mol.
- Molar mass M = mass of solute / n = 0.0250 g / 3.44×10⁻⁴ mol = 7.27×10¹ g·mol⁻¹ ≈ 7.3×10¹ g·mol⁻¹. (Check calculation)
Common mistake: Using 40 mL directly as litres or forgetting to convert temperature to Kelvin
Concept tested: Osmotic pressure and molar mass determination
Published · How this solution was prepared and checked
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