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`.

Answer

`30.96 Pa`

Explanation

`W_(2)=1.0g, Mw_(2)=185000g mol^(-1)`
`T=37^(@)C=37+273=310K`
`R=8.314 K Pa L K ^(-1) mol^(-1)`
`=8.314xx10^(3) Pa L K^(-1)mol^(-1)`
`V=450mL=0.450 L,` osmotic pressure `(pi)=?`
`n_(2)=(W_(2))/(Mw_(2))=(1.0g)/(185,000g mol^(-1))=(1)/(185000)mol`
Applying the formula , we get
`pi=MRT=(n_(2))/(V(i n L))xxRT`
`=(1)/(185000xx0.450L)xx8.31410^(3)Pa L K^(-1)mol^(-1)xx310K` ltbr. `=30.96Pa`