200 mL of an aqueous solution contains 1.26 g of protein. The osmotic pressure of this solution at 300 K is found to be 2.57 × 10⁻³ bar. The molar mass of protein will be: (Use: R = 0.083 L bar mol⁻¹ K⁻¹)

Osmotic pressure ($\Pi$) is a colligative property used to determine the molar mass of solutes, especially biomolecules like proteins. The relationship is given by the formula: $M_2 = \frac{w_2 R T}{\Pi V}$

Given: Mass of protein ($w_2$) = $1.26 \text{ g}$ Volume of solution ($V$) = $200 \text{ mL} = 0.200 \text{ L}$ Temperature ($T$) = $300 \text{ K}$ Osmotic Pressure ($\Pi$) = $2.57 \times 10^{-3} \text{ bar}$

• Gas Constant ($R$) = $0.083 \text{ L bar mol}^{-1} \text{ K}^{-1}$

Calculation: Substitute the values into the equation : $M_2 = \frac{1.26 \text{ g} \times 0.083 \text{ L bar K}^{-1} \text{ mol}^{-1} \times 300 \text{ K}}{2.57 \times 10^{-3} \text{ bar} \times 0.200 \text{ L}}$ $M_2 = \frac{31.374}{0.000514} \text{ g mol}^{-1}$ $M_2 \approx 61038.9 \text{ g mol}^{-1}$

The calculated molar mass is approximately 61039 g mol⁻¹. (Note: NCERT Example 1.11 lists the result as 61,022 g mol⁻¹, likely due to internal rounding, but the calculation with the provided values matches Option A closely).