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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.
Solution
Given,
Number of moles of solute dissolved (n) = `(1.0 "g")/(185,000 "g mol"^(-1)) = 1/(185,000) "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, `pi = "n"/"V""RT"`
= `1/(185,000) "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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