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ChemMedMain 2026 • 2nd Apr - Shift 2

Calculate Osmotic Pressure of Protein Solutions and Mixture

Solution A is prepared by dissolving 1 g1\text{ g} of a protein (molar mass =50000 g mol−1= 50000\text{ g mol}^{-1}) in 0.5 L0.5\text{ L} of water at 300 K300\text{ K}. Its osmotic pressure is x barx\text{ bar}. Solution B is made by dissolving 2 g2\text{ g} of same protein in 1 L1\text{ L} of water at 300 K300\text{ K}. Osmotic pressure of solution B is y bary\text{ bar}. Entire solution of A is mixed with entire solution of B at same temperature. The osmotic pressure of resultant solution is z barz\text{ bar}. xx, yy and zz respectively are : (R=0.083 L bar mol−1 K−1)(\text{R} = 0.083\text{ L bar mol}^{-1}\text{ K}^{-1})

Options

A

9.96×10−4;9.96×10−4;9.96×10−49.96 \times 10^{-4}; 9.96 \times 10^{-4}; 9.96 \times 10^{-4}

Correct
B

9.96×10−4;9.96×10−4;19.92×10−49.96 \times 10^{-4}; 9.96 \times 10^{-4}; 19.92 \times 10^{-4}

C

4.98×10−4;4.98×10−4;9.96×10−44.98 \times 10^{-4}; 4.98 \times 10^{-4}; 9.96 \times 10^{-4}

D

4.98×10−4;4.98×10−4;4.98×10−44.98 \times 10^{-4}; 4.98 \times 10^{-4}; 4.98 \times 10^{-4}

Topics & Concepts

SolutionsConcentration Terms

Step-by-Step Solution

To find the osmotic pressures xx, yy, and zz, we use the formula for osmotic pressure: π=CRT\pi = C R T where:

  • CC is the molar concentration of the solution (C=nVC = \frac{n}{V})
  • nn is the number of moles of solute (n=wMn = \frac{w}{M})
  • R=0.083 L bar mol−1 K−1R = 0.083\text{ L bar mol}^{-1}\text{ K}^{-1}
  • T=300 KT = 300\text{ K}

First, calculate the product R×TR \times T: R⋅T=0.083×300=24.9 L bar mol−1R \cdot T = 0.083 \times 300 = 24.9\text{ L bar mol}^{-1}

1. Calculation of Osmotic Pressure xx for Solution A

  • Mass of protein, wA=1 gw_A = 1\text{ g}
  • Molar mass of protein, M=50000 g mol−1M = 50000\text{ g mol}^{-1}
  • Moles of protein in A, nA=150000=2×10−5 moln_A = \frac{1}{50000} = 2 \times 10^{-5}\text{ mol}
  • Volume of solution A, VA=0.5 LV_A = 0.5\text{ L}

Molarity of Solution A: CA=nAVA=2×10−5 mol0.5 L=4×10−5 MC_A = \frac{n_A}{V_A} = \frac{2 \times 10^{-5}\text{ mol}}{0.5\text{ L}} = 4 \times 10^{-5}\text{ M}

Osmotic pressure xx: x=CART=(4×10−5)×24.9=9.96×10−4 barx = C_A R T = (4 \times 10^{-5}) \times 24.9 = 9.96 \times 10^{-4}\text{ bar}

2. Calculation of Osmotic Pressure yy for Solution B

  • Mass of protein, wB=2 gw_B = 2\text{ g}
  • Moles of protein in B, nB=250000=4×10−5 moln_B = \frac{2}{50000} = 4 \times 10^{-5}\text{ mol}
  • Volume of solution B, VB=1 LV_B = 1\text{ L}

Molarity of Solution B: CB=nBVB=4×10−5 mol1 L=4×10−5 MC_B = \frac{n_B}{V_B} = \frac{4 \times 10^{-5}\text{ mol}}{1\text{ L}} = 4 \times 10^{-5}\text{ M}

Osmotic pressure yy: y=CBRT=(4×10−5)×24.9=9.96×10−4 bary = C_B R T = (4 \times 10^{-5}) \times 24.9 = 9.96 \times 10^{-4}\text{ bar}

3. Calculation of Osmotic Pressure zz for the Resultant Mixture

When solution A and solution B are mixed:

  • Total moles of protein, ntotal=nA+nB=2×10−5+4×10−5=6×10−5 moln_{\text{total}} = n_A + n_B = 2 \times 10^{-5} + 4 \times 10^{-5} = 6 \times 10^{-5}\text{ mol}
  • Total volume of mixture, Vtotal=VA+VB=0.5 L+1.0 L=1.5 LV_{\text{total}} = V_A + V_B = 0.5\text{ L} + 1.0\text{ L} = 1.5\text{ L}

Molarity of the mixture: Cmix=ntotalVtotal=6×10−5 mol1.5 L=4×10−5 MC_{\text{mix}} = \frac{n_{\text{total}}}{V_{\text{total}}} = \frac{6 \times 10^{-5}\text{ mol}}{1.5\text{ L}} = 4 \times 10^{-5}\text{ M}

Osmotic pressure zz: z=CmixRT=(4×10−5)×24.9=9.96×10−4 barz = C_{\text{mix}} R T = (4 \times 10^{-5}) \times 24.9 = 9.96 \times 10^{-4}\text{ bar}

Conclusion:

x=9.96×10−4 bar,y=9.96×10−4 bar,z=9.96×10−4 barx = 9.96 \times 10^{-4}\text{ bar}, \quad y = 9.96 \times 10^{-4}\text{ bar}, \quad z = 9.96 \times 10^{-4}\text{ bar}

This corresponds to Option A.

Calculate Osmotic Pressure of Protein Solutions and Mixture | Chemistry PYQ Solution - JEE Challenger