Logo
Jeeify
MCQEasyJEE 2025Raoult's Law & Vapour Pressure

JEE Chemistry 2025 Question with Solution

When a non-volatile solute is added to the solvent, the vapour pressure of the solvent decreases by 10 mm Hg10 \, \text{mm Hg}. The mole fraction of the solute in the solution is 0.20.2. What would be the mole fraction of the solvent if the decrease in vapour pressure is 20 mm Hg20 \, \text{mm Hg}?

  • A

    0.60.6

  • B

    0.40.4

  • C

    0.20.2

  • D

    0.80.8

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The decrease in vapour pressure is 10 mm Hg10 \, \text{mm Hg} when the mole fraction of solute is 0.20.2.

Find: The mole fraction of the solvent when the decrease in vapour pressure becomes 20 mm Hg20 \, \text{mm Hg}.

For a non-volatile solute, Raoult's law gives the lowering of vapour pressure as directly proportional to the mole fraction of solute.

ΔP=P0xsolute\Delta P = P^0 x_{\text{solute}}

Initially,

ΔP=10 mm Hg,xsolute=0.2\Delta P = 10 \, \text{mm Hg}, \qquad x_{\text{solute}} = 0.2

So if the lowering of vapour pressure doubles from 10 mm Hg10 \, \text{mm Hg} to 20 mm Hg20 \, \text{mm Hg}, the mole fraction of solute also doubles:

xsolute′=0.4x'_{\text{solute}} = 0.4

Now use the relation between mole fractions of solute and solvent:

xsolvent′=1−xsolute′x'_{\text{solvent}} = 1 - x'_{\text{solute}} xsolvent′=1−0.4=0.6x'_{\text{solvent}} = 1 - 0.4 = 0.6

Therefore, the mole fraction of the solvent is 0.60.6. The correct option is A.

Using proportionality of lowering in vapour pressure

Given: At one condition, the lowering in vapour pressure is 10 mm Hg10 \, \text{mm Hg} and the solute mole fraction is 0.20.2.

Find: The solvent mole fraction when the lowering becomes 20 mm Hg20 \, \text{mm Hg}.

Raoult's law for a non-volatile solute can be written as:

Psolution=xsolventP0P_{\text{solution}} = x_{\text{solvent}} P^0

Hence,

P0−Psolution=ΔP=P0xsoluteP^0 - P_{\text{solution}} = \Delta P = P^0 x_{\text{solute}}

This shows that ΔP\Delta P is proportional to xsolutex_{\text{solute}}. So,

xsolute′xsolute=ΔP′ΔP=2010=2\frac{x'_{\text{solute}}}{x_{\text{solute}}} = \frac{\Delta P'}{\Delta P} = \frac{20}{10} = 2

Therefore,

xsolute′=2×0.2=0.4x'_{\text{solute}} = 2 \times 0.2 = 0.4

Now,

xsolvent′=1−xsolute′=1−0.4=0.6x'_{\text{solvent}} = 1 - x'_{\text{solute}} = 1 - 0.4 = 0.6

Therefore, the mole fraction of the solvent in the new condition is 0.60.6.

Common mistakes

  • Using the given solute mole fraction 0.20.2 directly as the new solvent mole fraction is incorrect because the vapour pressure lowering has changed. First update the solute mole fraction using Raoult's law, then subtract from 11.

  • Taking the initial solvent mole fraction as the final answer is wrong. The initial solvent mole fraction is 1−0.2=0.81 - 0.2 = 0.8, but the problem asks for the new condition when the lowering in vapour pressure is 20 mm Hg20 \, \text{mm Hg}.

  • Assuming the lowering in vapour pressure is proportional to the solvent mole fraction is incorrect here. For a non-volatile solute, the lowering ΔP\Delta P is proportional to the solute mole fraction, so doubling ΔP\Delta P doubles xsolutex_{\text{solute}}.

Practice more Raoult's Law & Vapour Pressure questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Raoult's Law & Vapour Pressure

Related questions