A polymer solution containing $0.90 g$ of polymer in $100 mL$ of solution has an osmotic pressure of $3.74 torr$ at $300 K$ . Using $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ and $1 atm = 760 torr$ , the molar mass of the polymer (in $g mo l^{- 1}$ ) is closest to

Answer: B

Convert:

π = 3.74/760 = 4.921 \times 1 0^{- 3}

atm. From

πV = (w / M) RT

M = wRT / (πV) = 0.90 \times 0.0821 \times 300/ (4.921 \times 1 0^{- 3} \times 0.100) = 45045

. So

M \approx 4.5 \times 1 0^{4} g mo l^{- 1}

. Forgetting the torr to atm conversion gives a value near 59.