Question

Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185000 in 450 mL of water at 37^o C.

Answer 1

Mass of polymer (W_B) = 1.0 g

Molar mass of polymer (M_B) = 185000 g mol^-1

Volume of solution (V) = 450 mL = 0.450 L

Temperature (T) = 37 + 273 = 310 K

Solution constant (R) = 8.314 x 10³ Pa L K^-1 mol^-1

Osmotic pressure $\left(\pi \right) = \text{CRT}$

$= \frac{\text{W}_{\text{B}} \times \text{R} \times \text{T}}{\text{M}_{\text{B}} \times \text{V}}$

$\pi=\frac{(1.0 \mathrm{~g}) \times\left(8.314 \times 10^3 \mathrm{~Pa} \mathrm{~L} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right) \times(310 \mathrm{~K})}{\left(185000 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times(0.450 \mathrm{~L})}$

= 30.95 Pa