200 ml of an aqueous solution of a protein contains 1.26 g of protein. At 300 K, the osmotic pressure of this solution is found to be ` 2.52xx 10^(-3) ` bar.The molar mass of protein will be
` (R=0.083 L bar mol ^(-1) K^(-1)) `

Options

A. `62.22` kg `mol^(-1)`
B. `12444 g mol ^(-1)`
C. `300 g mol ^(-1)`
D. none of these

Answer

A

Explanation

`pi = CRT`
`pi = (W)/(M xxV) xx RT`
` therefore M = (WRT)/(piV) = (1.26 xx 0.083 xx 300)/(2.52 xx 10 ^(-3) xx 0.2)= 62.22 kg mol ^(-1)`