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 `5.04xx10^(-3)` bar, the molar mass of the protein in g `mol^(-1)` is

Options

A. `83.0xx10^(3)`
B. `20.8xx10^(3)`
C. `41.5xx10^(3)`
D. `41.5xx10^(4)`

Answer

C

Explanation

`p=ixxMxxRxxT`
p = osmotic pressure = `5.04xx10^(-3)` bar = 5.04 atm
I = van.t Hoff constant = 1 for undissociating molecules like protein
M = molarity = `("moles solute")/("liter solution")`
= `("weight in g")/("molecular weight")xx1/("liter")=2.52/x xx1000/300=25.2/(3x)`
T = Temperature = 300 K
R = Gas constant = 0.08206 L atm/mol K.
`5.04xx10^(-3)=1xx(25.2)/(3x)xx0.08206xx300`
`3x=(25.2xx0.08206xx300)/(5.04xx10^(-3))`
`x=(25.2xx0.08206xx300)/(5.04xx3xx10^(-3))=(6.2037xx100)/(15.12xx10^(-3))`
= `0.410xx100xx10^(3)=41.0xx10^(3)`