Question

200 cm³ of an aqueous solution of protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300k is 2.57×10³ bar. Calculate the molar mass of the protein.​

Answer 1

Given :

Volume of aqueous solution : 200 cm³ = 0.200L.

Temperature : 300K

Proteins : 1.26 g

Osmotic pressure : 2.57 × 10¯³ bar

Real constant : 0.083 L bar K¯¹ mol¯¹

To find :

The molar mass of the protein.

Solution :

we know that,

The relationship between molar mass of solute M₂ and osmotic pressure.

$\boxed{\bf M_2 = \dfrac{ w_2RT}{ \pi v}}$

By substituting the values in the formula,

$\dashrightarrow\sf M_2 = \dfrac{ w_2RT}{ \pi v}$

$\dashrightarrow\sf M_2 = \dfrac{1.26 \times 0.083 \times 300}{2.57 \times {10}^{ - 3} \times 0.200}$

$\dashrightarrow\sf M_2 = 61038.9 \: g \: mol^{ - 1}$

Thus, the molar mass of the protein is 61038.9 g/mol.