200 cm^3 of an aqueous protein solution contains 1.26 g of protein and has an osmotic pressure of 2.57 x 10^-3 bar at 300 K (R = 0.083 L bar/K/mol). The molar mass of the protein is approximately:

1. $M_2 = (w_2 R T)/(\Pi V)$, with $V = 0.200$ L. 2. Numerator = $1.26 \times 0.083 \times 300 = 31.37$. 3. Denominator = $2.57 \times 10^{-3} \times 0.200 = 5.14 \times 10^{-4}$. 4. $M_2 = 31.37 / 5.14 \times 10^{-4} \approx 61{,}022$ g/mol. (Option C wrongly uses V = 2 L.) _Source: NCERT Class 12 Chemistry Ch 1 "Solutions", p.21_