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 be2.52xx 10^(-3)bar.The molar mass of protein will be (R=0.083 Lbar mol ^(-1) K^(-1))

200 ml of an aqueous solution of a protein contains 1.26 g of protein.

1.	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 be2.52xx 10^(-3)bar.The molar mass of protein will be (R=0.083 Lbar mol ^(-1) K^(-1))
Answer» `62.22` KG `mol^(-1)` `12444 g mol ^(-1)` `300 g mol ^(-1)` none of these Solution :`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)`