A solution is prepared by dissolving $0.0100 g$ of an unknown polymer in $20.0 mL$ of water at $298 K$ . The measured osmotic pressure of the solution is $0.085 atm$ . Assuming the polymer behaves as an ideal dilute solution, calculate its molar mass.

hard 3 marks Class 12 · Chemistry · Solutions

Answer

M_{polymer} \approx 1.44 \times 1 0^{2} g mol^{- 1}

The correct answer is $M_{polymer} \approx 1.44 \times 1 0^{2} g mol^{- 1}$ — Express the final answer with appropriate significant figures.

Step-by-step solution

Write the osmotic pressure equation for an ideal solution: $π = i M R T$ , where $i = 1$ for a non‑electrolyte.
Rearrange to find molarity: $M = \frac{π}{R T}$ .
Insert the values: $R = 0.0821 L atm mol^{- 1} K^{- 1}$ , $T = 298 K$ , $π = 0.085 atm$ .
Calculate $M = \frac{0.085}{0.0821 \times 298} = \frac{0.085}{24.4538} = 0.00348 mol L^{- 1}$ (rounded).
Convert the solution volume to litres: $20.0 mL = 0.0200 L$ .
Find moles of polymer: $n = M \times V = 0.00348 mol L^{- 1} \times 0.0200 L = 6.96 \times 1 0^{- 5} mol$ .
Use the mass of polymer to obtain molar mass: $M_{polymer} = \frac{0.0100 g}{6.96 \times 1 0 ^{- 5} mol} \approx 144 g mol^{- 1}$ .
Express the final answer with appropriate significant figures.

Common mistake: Forgetting to convert the solution volume from mL to L before using

π = M R T

Concept tested: Osmotic pressure and molar mass determination