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JEE Advanced 2026 · Paper 2, Chemistry question 15

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Question Stem for Question Nos. 15 and 16

Two volatile liquids A and B form an ideal solution. Consider a 5 molal solution of B in A inside a closed container having a total vapour pressure of 100 mm Hg at 300 K. The vapour pressure of pure A at 300 K is 105 mm Hg. Assume that A and B behave as ideal gases in the vapour phase.

Given:
The gas constant R=0.08R=0.08 L atm K−1^{-1} mol−1^{-1}
Molar mass of A is 50 g mol−1^{-1}
Molar mass of B is 57 g mol−1^{-1}
Density of liquid B at 300 K is 0.5 g/mL
1 atm = 760 mm Hg

At 300 K, the ratio of the molar volume of pure B in vapour phase to its molar volume in liquid phase is ____.

Answer and worked solution

Official answer: 1991

Vapour pressure of pure B. A 5 molal solution has 5 mol B per 1000 g A, i.e. per 20 mol A. So xB=525=0.2x_B=\frac5{25}=0.2 and xA=0.8x_A=0.8. Raoult:
100=0.8(105)+0.2 pB∘  ⟹  pB∘=80 mm Hg.100=0.8(105)+0.2\,p_B^\circ\implies p_B^\circ=80\text{ mm Hg}.

Molar volumes of B at 300 K.

  • Vapour, at its own vapour pressure: Vg=RTp=0.08×30080/760=228V_g=\dfrac{RT}{p}=\dfrac{0.08\times300}{80/760}=228 L mol⁻¹.
  • Liquid: Vl=570.5=114V_l=\dfrac{57}{0.5}=114 mL =0.114=0.114 L mol⁻¹.

VgVl=2280.114=2000.\frac{V_g}{V_l}=\frac{228}{0.114}=2000.

Topics:Raoult's law and vapour pressureConcentration terms
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