Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE Main 20262 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
2 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Solutions

JEE Main 2026 · 2 April · Evening Shift · Chemistry · Solutions

Solution A is prepared by dissolving 1 g of a protein (molar mass =...

Solution A is prepared by dissolving 11 g of a protein (molar mass =50000= 50000 g mol−1^{-1}) in 0.50.5 L of water at 300300 K. Its osmotic pressure is xx bar. Solution B is made by dissolving 22 g of same protein in 11 L of water at 300300 K. Osmotic pressure of solution B is yy bar. Entire solution of A is mixed with entire solution of B at same temperature. The osmotic pressure of resultant solution is zz bar. x,yx, y and zz respectively are:
(R=0.083(R = 0.083 L bar mol−1^{-1} K−1)^{-1})

Options

  1. A9.96×10−4;9.96×10−4;9.96×10−49.96 \times 10^{-4}; 9.96 \times 10^{-4}; 9.96 \times 10^{-4}
  2. B9.96×10−4;9.96×10−4;19.92×10−49.96 \times 10^{-4}; 9.96 \times 10^{-4}; 19.92 \times 10^{-4}
  3. C4.98×10−4;4.98×10−4;9.96×10−44.98 \times 10^{-4}; 4.98 \times 10^{-4}; 9.96 \times 10^{-4}
  4. D4.98×10−4;4.98×10−4;4.98×10−44.98 \times 10^{-4}; 4.98 \times 10^{-4}; 4.98 \times 10^{-4}

Correct answer

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}

Step-by-step solution

Osmotic pressure is given by π=CRT=wMVRT\pi = CRT = \dfrac{w}{MV}RT

For solution A:
x=150000×0.5×0.083×300x = \dfrac{1}{50000 \times 0.5} \times 0.083 \times 300
x=125000×24.9=9.96×10−4x = \dfrac{1}{25000} \times 24.9 = 9.96 \times 10^{-4} bar

For solution B:
y=250000×1×0.083×300y = \dfrac{2}{50000 \times 1} \times 0.083 \times 300
y=250000×24.9=9.96×10−4y = \dfrac{2}{50000} \times 24.9 = 9.96 \times 10^{-4} bar

For the resultant solution (mixture of A and B):
Total mass of protein =1+2=3= 1 + 2 = 3 g
Total volume of solution =0.5+1=1.5= 0.5 + 1 = 1.5 L
z=350000×1.5×0.083×300z = \dfrac{3}{50000 \times 1.5} \times 0.083 \times 300
z=375000×24.9=9.96×10−4z = \dfrac{3}{75000} \times 24.9 = 9.96 \times 10^{-4} bar

Therefore, x=y=z=9.96×10−4x = y = z = 9.96 \times 10^{-4} bar.

Answer: 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}
Practice Solutions on Quantrex Academy →

More from Solutions

In a solvent S , a compound B is partially dissociated into C and D as given below: B… JEE Advanced 2026Passage: Two volatile liquids A and B form an ideal solution. Consider a 5 molal solution… JEE Advanced 2026Passage: Two volatile liquids A and B form an ideal solution. Consider a 5 molal solution… JEE Advanced 2026Given below are two statements: Given: Molar mass of C, H, O, Cl are 12, 1, 16 and 35.5 g… JEE Main 2026Which of the following statements are not correct? A. For water, magnitude of K_b is more… JEE Main 2026Given below are two statements : Statement I : H _2 O molecules move from the chamber 1… JEE Main 2026When 0.25 moles of a non-volatile, non-ionizable solute was dissolved in 1 mole of a… JEE Main 202620 g hemoglobin in a 1 L aqueous solution (A) at 300 K is separated from pure water by… JEE Main 2026 Full Solutions list All JEE Main PYQs