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Chemistry Question 55 – JEE-MAIN 2025

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A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapour pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapour pressure of pure B and the least volatile component of the solution, respectively, are:

The volatility of a component is directly related to its pure vapor pressure; a higher vapor pressure indicates higher volatility.

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Ninja Strategy•Estimate PB0 and Compare Volatility

Quickly estimate PB0 using Raoult's Law to eliminate options with incorrect vapor pressure, then compare the pure vapor pressures to identify the least volatile component.

● Video Walkthrough
Step 1: Calculate Mole Fractions✦ Active

Determine the mole fractions of liquid A and liquid B in the solution.

nA=1 mol,nB=3 mol Total moles (ntotal)=nA+nB=1+3=4 mol xA=nAntotal=14=0.25 xB=nBntotal=34=0.75
Step 2: Apply Raoult's Law to find PB0○ Expand

Use Raoult's Law for the total vapor pressure of the solution.

Ptotal=xAPA0+xBPB0 500 mm Hg=(0.25)(200 mm Hg)+(0.75)PB0 500=50+0.75PB0 450=0.75PB0 PB0=4500.75=600 mm Hg
💡 Teacher's Secret Hint

Ensure correct substitution of mole fractions and pure vapor pressures.

Step 3: Identify the Least Volatile Component○ Expand

Compare the pure vapor pressures of A and B. The component with the lower pure vapor pressure is the least volatile.

PA0=200 mm Hg PB0=600 mm Hg

Since PA0<PB0, component A has a lower vapor pressure and is therefore the least volatile component. Thus, the vapor pressure of pure B is 600 mm Hg and the least volatile component is A.

💡 Teacher's Secret Hint

Remember that lower vapor pressure implies lower volatility.

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