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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Applying the relation `pi = CRT = n/V RT` Number of moles of solute dissolved `(n) = (10g)/(185,000 g "mol"^(-1)) = (1)/(185,000)` `V = 450 mL = 0.450 L, T = 37^@C = 37 + 273 = 310 K` The osmotic pressure is to be given in pascals. ` R = 8.314 kPa LK^(-1) "mol"^(-1) = 8.314 xx 10^3 Pa LK^(-1) "mol"^(-1)` Substituting the values, we get `pi = (1)/(185000) "mol" xx (1)/(0.45 L) xx 8.314 xx 10^3 Pa LK^(-1) "mol"^(-1) xx 310 K = 30.96 Pa `
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