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

`13980.4`

Explanation

`pi = cRT = (w_(2))/(M_(2)V) RT or M_(2) = (w_(2)RT)/(piV)`
`w_(2) = 100 xx 10^(-3)` g, R = 0.0821 L atm `mol^(-1) K^(-1)` T = 298 K
`pi = (13.3)/(760) atm, V = (10)/(1000) L`
`M_(2) = ((100 xx 10^(-3)g) xx (0.0821 L atm mol^(-1)K^(-1))xx(298K))/(((13.3)/(760) atm) xx ((10)/(1000)L))`