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Chemistry · Solutions and Colligative Properties

JEE Main 2026 — 2 April, Evening Shift — Question 52

Solution A is prepared by dissolving 1 g of a protein (molar mass =50000 g mol−1=50000 \mathrm{~g} \mathrm{~mol}^{-1} ) in 0.5 L of water at 300 K . Its osmotic pressure is x bar. Solution B is made by dissolving 2 g of same protein in 1 L of water at 300 K . Osmotic pressure of solution BB is y bar. Entire solution of AA is mixed with entire solution of BB at same temperature. The osmotic pressure of resultant solution is zz bar. x,yx, y and z respectively are :( R=0.083 L\mathrm{R}=0.083 \mathrm{~L} bar mol−1 K−1\mathrm{mol}^{-1} \mathrm{~K}^{-1} )

  1. Option 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
  2. Option 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}

  3. Option 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}

  4. Option 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}

Answer: A

Step-by-step solution

′x′=150,000×0.5×R×300=9.96×10−4\quad{ }^{\prime} \mathrm{x}^{\prime}=\frac{1}{50,000 \times 0.5} \times \mathrm{R} \times 300=9.96 \times 10^{-4} bar y′=250,000×1×R×300=9.96×10−4\mathbf{y}^{\prime} \boldsymbol{=} \frac{2}{50,000 \times 1} \times \mathrm{R} \times 300=9.96 \times 10^{-4} bar ′z′=′x′×12+′y′×132=9.96×10−4{ }^{\prime} \mathbf{z}^{\prime}=\frac{{ }^{\prime} x^{\prime} \times \frac{1}{2}+{ }^{\prime} y^{\prime} \times 1}{\frac{3}{2}}=9.96 \times 10^{-4} bar

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
2 April, Evening Shift
Subject
Chemistry
Chapter
Solutions and Colligative Properties
Topic
Solid in Liquid Solutions (Colligative Properties)

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