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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.
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Solution
Given: Number of moles of solute dissolved (n) = `(1.0 g)/(185000 g "mol"^(-1)) = 1/(185000)` mol
V = 450 mL = 0.45 L
T = 37°C = (37 + 273) K = 310 K
R = 8.314 k Pa L K−1 mol−1 = 8.314 × 103 Pa L K−1 mol−1
Osmotic pressure (π) = `n/V RT`
= `1/(185000) "mol" xx 1/(0.45 L) xx 8.314 xx 10^3 Pa L K^(-1) "mol"^(-1) xx 310 K`
= 30.96 Pa
= 31 Pa (approximately)
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