Question

1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42 x 10^–3 bar. The molar mass of the biopolymer is ...... x 104 g mol^–1 . (Round off to the Nearest Integer) [Use : R = 0.083 L bar mol^–1 K ^–1]

Answer 1

Answer is : 15

π = CRT ; π = osmotic pressure

C = molarity 

T = Temperature of solution let the molar mass be M gm / mol 2.42 x 10^–3 bar =

M = 15.02 x 10⁴ g/mol