200 `cm^3` of an aqueous solution of a protein contains 1.26g of the protein . The osmotic pressure of such a solution at 300K is found to be `2.7 xx 10^-3` bar. Calculate the molar mass of the protein (R=0.083 L bar `mol^-1 K^-1`)

Answer

`58100 g mol^(-1) `

Explanation

`V = 200 cm^(3) = 0 . 2 L`
`W_(B) = 1 . 26 g, pi = 2 . 7 xx 10^(-3)` bar
T = 300 K
` R = 0 . 0 83" L bar " mol^(-1) K^(-1)` `pi = (n)/(V) RT`
`n = ( pi xx V)/( R xx T)`
`= (( 2 . 7 xx 10^(-3) b a r) xx ( 0 . 2L))/( ( 0 . 0 83 " L bar " mol^(-1)K^(-1)) xx (300 K))`
`= (5 . 4 xx 10^(-4))/( 2 . 49) = 2 . 169 xx 10 ^(-4) mol`
`n = (W_(B))/( M_(B))`
`= (12 . 6)/( 2 . 169 ) xx 10^(5) g mol^(-1)`
`= 5 . 809 xx 10^(5)`
`= 580900 g mol^(-1)`
`= 580 . 9 kg mol ^(-1)`