Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving $1.0 \ g$ of polymer of molar mass $185,000$ in $450 \ mL$ of water at $37^{\circ} C$.

(N/A) Given:
Mass of polymer,$w = 1.0 \ g$
Molar mass of polymer,$M = 185,000 \ g \ mol^{-1}$
Volume of solution,$V = 450 \ mL = 0.45 \ L$
Temperature,$T = 37 + 273 = 310 \ K$
Gas constant,$R = 8.314 \ Pa \ m^3 \ K^{-1} \ mol^{-1} = 8.314 \ Pa \ L \ K^{-1} \ mol^{-1} \times 10^{-3} \ m^3 \ L^{-1}$ (using $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)
Using the formula for osmotic pressure $\pi = CRT = \frac{n}{V} RT = \frac{w}{M \times V} RT$
$\pi = \frac{1.0 \ g}{185,000 \ g \ mol^{-1} \times 0.45 \ L} \times 8.314 \ Pa \ L \ K^{-1} \ mol^{-1} \times 310 \ K$
$\pi = \frac{2577.34}{83250} \ Pa \approx 0.03096 \ Pa$
Rounding to appropriate significant figures,the osmotic pressure is approximately $0.031 \ Pa$.