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.

Answer 1

Answer :

Given that,

Volume of water = V = 450 mL = 0.45

Temperature, T = (37 + 273)K = 310K

R = 8.314 kpaL $K^{-1}$ $mol^{-1}$

Now,

πV = nRT

⇒ π × 0.45 × $10^{-3} m^{3}$

= $\frac{1}{185000}$ × 8.314 Pa × 310 K

⇒ π = $\frac{8.314 × 310}{185000 × 0.45 × 10^{-3} m^{3}}$

= $\frac{8.314 × 310 × 100}{185 × 45}$

= 30.96 Pa

Answer 2

Given, weight of polymers , w = 1g
molar mass of polymers , M = 185,000 g/mol
volume of water , V = 450 mL =0.45L
temperature , T = 37° C = 37 + 273 = 310K

we know osmotic pressure, $\bf{\pi=CRT}$
where π is the osmotic pressure,
C is the concentration of solute
and T is the temperature .
we also know, C = no of mole/volume
= w/MV
= 1/(185,000 × 450)

now, π = 1/(185,000 × 450) × 8.314 × 10³ × 310
[ R = 8.314 × 10³ Pascal.L/mol/K]
= 30.98 Pa ≈ 31 Pascal

hence, osmotic pressure = 31 Pa