A solution of $4.0 g$ of a polymer in $500 mL$ of solution has an osmotic pressure of $0.0985 atm$ at $300 K$ ( $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ ). The molar mass of the polymer (in $g mo l^{- 1}$ ) is closest to

Answer: B

πV = \frac{w}{M} RT

, so

M = \frac{wRT}{πV} = \frac{4.0 \times 0.0821 \times 300}{0.0985 \times 0.500} = 2000

, i.e. about

2000 g mo l^{- 1}