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Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving `1.0g` of polymer of molar mass `185,000` in `450mL` of water at `37^(@)C`. |
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Answer» Correct Answer - `30.96 Pa` `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` |
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