Question:
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°C.
Solution:
It is given that:
Volume of water, V = 450 mL = 0.45 L
Temperature, T = (37 + 273)K = 310 K
Number of moles of the polymer, $n=\frac{1}{185000} \mathrm{~mol}$
We know that:
Osmotic pressure, $\pi=\frac{n}{V} \mathrm{R} T$
$=\frac{1}{185000} \mathrm{~mol} \times \frac{1}{0.45 \mathrm{~L}} \times 8.314 \times 10^{3} \mathrm{~Pa} \mathrm{~L} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \times 310 \mathrm{~K}$
= 30.98 Pa
= 31 Pa (approximately)