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Chemistry Question 55 – JEE-MAIN 2026

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Solution A is prepared by dissolving 1 g of a protein (molar mass =50000 g 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 B is y bar. Entire solution of A is mixed with entire solution of B at same temperature. The osmotic pressure of resultant solution is z bar. x,y and z respectively are : (R=0.083 L bar mol−1 K−1)

Osmotic pressure is a colligative property that depends on the molar concentration of the solute particles, not their identity.

Step 1: Calculate Molar Concentrations for Solutions A and B✦ Active

First, determine the number of moles of protein (n) for Solution A and Solution B using the given mass (w) and molar mass (M=50000 g mol−1). Then, calculate their respective molar concentrations (C=n/V). For Solution A:

nA=1 g50000 g mol−1=2×10−5 mol CA=2×10−5 mol0.5 L=4×10−5 mol L−1

For Solution B:

nB=2 g50000 g mol−1=4×10−5 mol CB=4×10−5 mol1 L=4×10−5 mol L−1
Step 2: Calculate Osmotic Pressures x and y○ Expand

Use the van't Hoff equation for osmotic pressure, Π=CRT, where R=0.083 L bar mol−1 K−1 and T=300 K. For Solution A (osmotic pressure x):

x=CART=(4×10−5 mol L−1)×(0.083 L bar mol−1 K−1)×(300 K) x=9.96×10−4 bar

For Solution B (osmotic pressure y):

y=CBRT=(4×10−5 mol L−1)×(0.083 L bar mol−1 K−1)×(300 K) y=9.96×10−4 bar
💡 Teacher's Secret Hint

Note that since CA=CB and R,T are constant, x and y must be equal.

Step 3: Calculate Osmotic Pressure z for the Resultant Solution○ Expand

When solutions A and B are mixed, the total moles of protein and total volume are summed. Then, calculate the total concentration (Ctotal) and the resultant osmotic pressure (z). Total moles:

ntotal=nA+nB=2×10−5 mol+4×10−5 mol=6×10−5 mol

Total volume:

Vtotal=VA+VB=0.5 L+1 L=1.5 L

Total concentration:

Ctotal=ntotalVtotal=6×10−5 mol1.5 L=4×10−5 mol L−1

Resultant osmotic pressure (z):

z=CtotalRT=(4×10−5 mol L−1)×(0.083 L bar mol−1 K−1)×(300 K) z=9.96×10−4 bar

Therefore, x=9.96×10−4, y=9.96×10−4, and z=9.96×10−4.

💡 Teacher's Secret Hint

Notice that the final concentration of the mixed solution is the same as the initial concentrations of solutions A and B. This is because the ratio of moles to volume remained constant.

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