Question

1. 200 mL of an aqueous solution of a protein contain
its 1.26 g. The Osmotic pressure of this solution
at 300 K is found to be 2.57 x10-3 bar. The molar
mass of protein will be :-
(R = 0.083 L bar mol-1 K-1)
(1) 61038 g mol-1 (2) 51022 g mol-1
(3) 122044 g mol-1 (4) 31011 g mol-1
(please provide solution)​

Answer 1

Answer: π=CRT

2.57*10^-3={1.26*1000}/M*200 *8.314 *300

NOW CALCULATED (M ) FROM ABOVE EQUATIONS

Explanation: