200 mL of an aqueous solution of a protein contain 1.26 g of...

Given:
Volume of solution, $V = 200$ mL $= 0.200$ L
Mass of protein, $W = 1.26$ g
Osmotic pressure, $\pi = 2.57 \times 10^{-3}$ bar
Temperature, $T = 300$ K
Gas constant, $R = 0.083$ L bar K^{-1}$mol^{-1}$

Using the formula for osmotic pressure:
$\pi = \dfrac{W}{M V} R T$

Rearranging for molar mass $M$:
$M = \dfrac{W R T}{\pi V}$

Substituting the values:
$M = \dfrac{1.26 \times 0.083 \times 300}{2.57 \times 10^{-3} \times 0.200}$

$M = \dfrac{31.374}{5.14 \times 10^{-4}}$

$M = 61038.9$ g mol^{-1} \approx 61039$g mol^{-1}$

Answer: $61039$ g mol^{-1}