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Solutions question

2025 · 23 Jan · Shift 2 · Q2
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Solutions question

2025 · 23 Jan · Shift 2 · Q2

JEE MainChemistrySolutionsMCQ+4 / −1
When a non-volatile solute is added to the solvent, the vapour pressure of the solvent decreases by 10 mm of Hg . The mole fraction of the solute in the solution is 0.2 . What would be the mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg ?
  1. A
    0.2
  2. B
    0.4
  3. C
    0.8
  4. D
    0.6
View written solutionFree

Correct answer: D

  1. Use Raoult’s law for relative lowering of vapour pressure

For a non-volatile solute,

ΔPP0=Xsolute\frac{\Delta P}{P^0} = X_{\text{solute}}P0ΔP​=Xsolute​

where:

  • ΔP\Delta PΔP = lowering in vapour pressure
  • P0P^0P0 = vapour pressure of pure solvent
  • XsoluteX_{\text{solute}}Xsolute​ = mole fraction of solute
  1. From the first condition

Given:

  • ΔP1=10\Delta P_1 = 10ΔP1​=10 mm Hg
  • Xsolute,1=0.2X_{\text{solute,1}} = 0.2Xsolute,1​=0.2

So,

10P0=0.2\frac{10}{P^0} = 0.2P010​=0.2

Hence,

P0=100.2=50 mm HgP^0 = \frac{10}{0.2} = 50 \text{ mm Hg}P0=0.210​=50 mm Hg

  1. For the second condition

Given:

  • ΔP2=20\Delta P_2 = 20ΔP2​=20 mm Hg

Again using,

ΔP2P0=Xsolute,2\frac{\Delta P_2}{P^0} = X_{\text{solute,2}}P0ΔP2​​=Xsolute,2​

2050=Xsolute,2=0.4\frac{20}{50} = X_{\text{solute,2}} = 0.45020​=Xsolute,2​=0.4

  1. Find mole fraction of solvent

Xsolvent=1−Xsolute=1−0.4=0.6X_{\text{solvent}} = 1 - X_{\text{solute}} = 1 - 0.4 = 0.6Xsolvent​=1−Xsolute​=1−0.4=0.6

  1. Match with options

0.60.60.6 corresponds to Option D.

Final Answer: 0.60.60.6

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