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Calculated the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0g of polymer of molar mass 185000 in 450 mL of water at 37^@C.

Answer»

Solution :`pi= (W_2RT)/(M_2V)`
`W_2 = 1.0g1, T=273+37=310K`
`M_2` = 185000 V = 450mL = 0.450 L
R = 0.083 L bar `mol^(-1) K^(-1)`
`THEREFORE pi = (1xx0.083xx310)/(185000xx0.45) = 0.000309`bar
1 bar = `10^5` pascal
OSMOTIC pressure, `pi = 0.000309xx10^5 = 30.9 Pa`


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