Question

${\text{200 mL}}$ of an aqueous solution of a protein contains ${\text{1}}{\text{.26 g}}$ of protein. The osmotic pressure of this solution at ${\text{300 K}}$ is found to be ${\text{2}}{\text{.57}} \times {\text{1}}{{\text{0}}^{{\text{ - 3}}}}{\text{ bar}}$ . The molar mass of protein will be
${\text{[R = 0}}{\text{.083 L bar mo}}{{\text{l}}^{{\text{ - 1}}}}{{\text{K}}^{{\text{ - 1}}}}{\text{]}}$
(A) ${\text{51022 g mo}}{{\text{l}}^{{\text{ - 1}}}}$
(B) ${\text{122044 g mo}}{{\text{l}}^{{\text{ - 1}}}}$
(C) ${\text{31011 g mo}}{{\text{l}}^{{\text{ - 1}}}}$
(D) ${\text{61038 g mo}}{{\text{l}}^{{\text{ - 1}}}}$

Answer 1

Use the following expression to calculate the molar mass of the protein.
${\text{M}} = \dfrac{{\text{W}}}{{\pi {\text{V}}}}{\text{RT}}$
In the above expression, substitute the values of the mass, volume, temperature and ideal gas constant and calculate the molar mass of protein.

Complete step by step answer:
Write down the expression for the osmotic pressure of the solution and then rearrange this expression to obtain the expression for the molar mass of protein.

\pi {\text{V = }}\dfrac{{\text{W}}}{{\text{M}}}{\text{RT}} \\\ {\text{M}} = \dfrac{{\text{W}}}{{\pi {\text{V}}}}{\text{RT}} \\\\$$ $${\text{M}} = \dfrac{{\text{W}}}{{\pi {\text{V}}}}{\text{RT}} \\\ {\text{M}} = \dfrac{{1.26}}{{{\text{2}}{\text{.57}} \times {\text{1}}{{\text{0}}^{{\text{ - 3}}}} \times \dfrac{{200}}{{1000}}{\text{ }}}} \times {\text{0}}{\text{.083}} \times {\text{300}} \\\ {\text{M}} = {\text{ 61038 g mo}}{{\text{l}}^{{\text{ - 1}}}} \\\\$$ _**Hence, the correct option is the (D) $${\text{61038 g mo}}{{\text{l}}^{{\text{ - 1}}}}$$.**_ **Note:** The method of the determination of the molar mass of protein by using osmotic pressure is preferred over other methods such as elevation in the boiling point and depression in the freezing point. This is because a very small amount of protein gives a significant amount of osmotic pressure whereas a similar amount of protein will give a very small value of elevation in the boiling point and depression in the freezing point.