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JEE Advanced 2020 · Paper 2, Chemistry question 14

Liquids A and B form ideal solution for all compositions of A and B at 25 °C. Two such solutions with 0.25 and 0.50 mole fractions of A have the total vapor pressures of 0.3 and 0.4 bar, respectively. What is the vapor pressure of pure liquid B in bar?

Answer and worked solution

Official answer: 0.2

Raoult's law for an ideal solution gives p=xApA∘+(1−xA)pB∘p=x_Ap_A^\circ+(1-x_A)p_B^\circ.

Two equations.

  • 0.25pA∘+0.75pB∘=0.30.25p_A^\circ+0.75p_B^\circ=0.3.
  • 0.50pA∘+0.50pB∘=0.40.50p_A^\circ+0.50p_B^\circ=0.4, so pA∘+pB∘=0.8p_A^\circ+p_B^\circ=0.8.

Solving. Substituting into the first:
0.25(0.8−pB∘)+0.75pB∘=0.3  ⟹  0.2+0.5pB∘=0.3  ⟹  pB∘=0.20 bar.0.25(0.8-p_B^\circ)+0.75p_B^\circ=0.3\implies0.2+0.5p_B^\circ=0.3\implies p_B^\circ=0.20\text{ bar}.
Answer: 0.20

Topics:Raoult's law and vapour pressureIdeal and non-ideal solutions
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