600 mL of aqueous solution contining 2.5 g of a protein shows an osmotic pressure of 25mm Hg at `27^(@)C` .Determine the molecular mass of protein.

Answer

`3119.8 g mol^(-1)`

Explanation

`pi = cRT = (w_(2))/(M_(2)V) RT or M_(2) = (w_(2)RT)/(piV)`
`w_(2) = 2.5 g , R = 0.0821 L atm mol^(-1) K^(-1)`
` T = 273 + 27 = 300 K`
` pi = (25)/(760) atm ,V = (600)/(1000) L`
`M_(2)= ((2.5 g xx 0.0821 L atm mol^(-1)K^(-1)) xx (300K))/(((25)/(760) atm) xx((600)/(1000)L))`
` = 3119.8 g mol^(-1)`