200 mL of an aqueous solution contains 1.26 g of protein. The osmotic pressure of this solution at 300 K is fo

Question

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)

Accepted Answer

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 ($w_2$) = $1.26 	ext{ g}$
*   Volume of solution ($V$) = $200 	ext{ mL} = 0.200 	ext{ L}$
*   Temperature ($T$) = $300 	ext{ K}$
*   Osmotic Pressure ($\Pi$) = $2.57 	imes 10^{-3} 	ext{ bar}$
*   Gas Constant ($R$) = $0.083 	ext{ L bar mol}^{-1} 	ext{ K}^{-1}$

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

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).

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