1.

100 mg of a protein was disoved in just enough water to make 10 mL of the solution. If the solution has an osmotic pressure of 13.3 mm Hg at `25^(@)C`, what is the mass of prtein `(R=0.0821 L atm mol^(-1)K^(-1))`

Answer» Osmotic pressure `(pi)=(W_(B)xxRxxT)/(M_(B)xxV)` or `M_(B)=(W_(B)xxRxxT)/(pixxV)`
`pi=13.3 , Hg=13.3mm Hg=(13.3)/(760) atm, V=10 mL=10/1000=0.01L`
`W_(B)=100mg=100/1000=0.1g, T=25^(@)C=25+273=298K`
`R=0.0821 L atm mol^(-1)K^(-1)`
`M_(B)=((0.1g)xx0.0821(atm K^(-1)Mol^(-1))xx298K)/(((13.3)/760atm)xx(0.01 L))`
=`13980.45 g mol^(-1)=1.45xx10^(4)g mol^(-1)`


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