300 ~mL of an aqueous solution of a protein contains 2.52 ~g of the protein. If osmotic pressure of such a solution at 300 ~K is 5.04 10⁻³ bar, the molar mass of the protein in g mol ⁻¹ is

Step-by-step solution

p=i M R T aligned & p= osmotic pressure =5.04 10⁻³ bar =5.04 ~atm & i= van't Hoff constant =1 for undissociating & molecules like protein aligned molecules like protein aligned & M= molarity = moles solute liter solution & = weight in g molecular weight 1 liter = 2.52 x 1000 300 = 25.2 3 x aligned aligned T & = Temperature =300 ~K R & = Gas constant =0.08206 ~L ~atm / mol K . aligned aligned 5.04 10⁻³ & =1 25.2 3 x 0.08206 300 3 x & = 25.2 0.08206 300 5.04 10⁻³ x & = 25.2 0.08206 300 5.04 3 10⁻³ = 6.2037 100 1512 1