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

Answer: 30.95

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