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NEET CHEMISTRYMedium

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)

A

61038 g mol⁻¹

B

51022 g mol⁻¹

C

122044 g mol⁻¹

D

31011 g mol⁻¹

Step-by-Step Solution

Osmotic pressure (Π\Pi) is a colligative property used to determine the molar mass of solutes, especially biomolecules like proteins. The relationship is given by the formula: M2=w2RTΠVM_2 = \frac{w_2 R T}{\Pi V}

Given: Mass of protein (w2w_2) = 1.26 g1.26 \text{ g} Volume of solution (VV) = 200 mL=0.200 L200 \text{ mL} = 0.200 \text{ L} Temperature (TT) = 300 K300 \text{ K} Osmotic Pressure (Π\Pi) = 2.57×103 bar2.57 \times 10^{-3} \text{ bar}

  • Gas Constant (RR) = 0.083 L bar mol1 K10.083 \text{ L bar mol}^{-1} \text{ K}^{-1}

Calculation: Substitute the values into the equation : M2=1.26 g×0.083 L bar K1 mol1×300 K2.57×103 bar×0.200 LM_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}} M2=31.3740.000514 g mol1M_2 = \frac{31.374}{0.000514} \text{ g mol}^{-1} M261038.9 g mol1M_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).

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