Question

$200mL$ of an aqueous solution of a protein contains $1.26g$ of protein. The osmotic pressure of this solution at $300K$ is found to be $2.57\times {10}^{-3}bar$ . The molar mass of protein will be:

[R = 0.083 L bar mol${}^{-1}$K${}^{-1}$]

• A. 51022 g mol${}^{-1}$
• B. 122044 g mol${}^{-1}$
• C. 31011 g mol${}^{-1}$
• D. 61038 g mol${}^{-1}$

Answer 1

The correct option is D 61038 g mol${}^{-1}$

The expression for the osmotic pressure of the solution is given below.

$\pi =CRT$

Substitute values in the above expression.

$\pi =\frac{wt\times 1000}{GMM\times V}\times RT$

$2.57\times {10}^{-3}=\frac{1.26\times 1000}{GMM\times 200}\times 0.083\times 300$

$GMM=61038g$

Thus, the molar mass of protein is $61038g{mol}^{-1}$

Hence, the correct option is $\text{D}$