Ncert Exercises & Solutions

2.12 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.

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, $n = \frac{1}{185000} mol$

We know that:

Osmotic pressure, $π = \frac{n}{V} RT$

$= \frac{1}{185000} mol \times \frac{1}{0.45 L} \times 8.314 \times 10^{3} Pa L K^{- 1} {mol}^{- 1} \times 310 K$

= 30.98 Pa

= 31 Pa (approximately)

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