Question

For $200$ mL of an aqueous solution of protein contains its $1.26$ g. The osmotic pressure of the solution at $300K$ is found to be $2.57 \times {10^{ - 3}}$ bar. The molar mass of the protein will be,
($R = 0.083Lbar{K^{ - 1}}mo{l^{ - 1}}$)
A.$61038$g/mol
B.$51022$g/mol
C.$122044$g/mol
D.$31011$g/mol

Answer 1

Osmotic pressure is a colligative property. Colligative properties depend upon the rato number of solute particles to the number of solvent molecules in a solution. Their value will be proportional to the number of particles of solute.

Complete step by step answer:
Osmosis is the movement of solvent through a selectively permeable membrane from a solution of low concentration to high concentration. Osmotic pressure is the minimum amount of pressure that must be applied on the solution side in order to prevent osmosis. The formula for osmotic pressure is,
$\pi = icRT$
Where $\pi$ is the osmotic pressure, i is vant-Hoff factor, c is concentration of solution, R is the gas constant and T is the temperature in Kelvin.
Concentration of solution can be written as,
$c = \dfrac{n}{V} = \dfrac{w}{{MV}}$
where
n = number of moles of solute
w = weight of solute in grams
M = molar mass of solute
V = volume of solution in litre.
Hence the equation for osmotic pressure is,
$\pi = \dfrac{{iwRT}}{{MV}}$
Given,
w $= 1.26$g
$R = 0.083Lbar{K^{ - 1}}mo{l^{ - 1}}$
T $= 300$ K
V $= 200mL = 0.2L$
$\pi = 2.57 \times {10^{ - 3}}bar$
i for protein is $1$.
We need to find the molar mass of protein. The equation for molar mass is,
$M = \dfrac{{iwRT}}{{\pi V}}$
Let’s substitute the values in the above equation.
$M = \dfrac{{iwRT}}{{\pi V}} = \dfrac{{1 \times 1.26 \times 0.083 \times 300}}{{2.57 \times {{10}^{ - 3}} \times 0.2}} = 61038.91gmo{l^{ - 1}}$
Hence molar mass of protein is $61038g/mol$. Option A is correct.

Note:
Van't Hoff factor, it depends on the nature of the solute and solvent. Protein does not undergo association or dissociation in its aqueous solution. Hence the value of i is equal to unity. But some solutes undergo association /dissociation when dissolved in certain solvents. In those cases we need additional factors like degree of association/dissociation and number of particles into which one molecule associate/dissociate to calculate i.