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← JEE Advanced 2018 · Paper 1
ChemistryNumerical+3Question 30 of 54

JEE Advanced 2018 · Paper 1, Chemistry question 12

Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar binary solution of liquids A and B has vapour pressure 45 Torr. At the same temperature, a new solution of A and B having mole fractions xAx_A and xBx_B, respectively, has vapour pressure of 22.5 Torr. The value of xA/xBx_A/x_B in the new solution is ______.

(given that the vapour pressure of pure liquid A is 20 Torr at temperature T)

Answer and worked solution

Official answer: 18.5

Find PB∘P_B^\circ. For the equimolar solution,
45=12(20)+12PB∘  ⟹  PB∘=70 Torr.45=\tfrac12(20)+\tfrac12P_B^\circ\implies P_B^\circ=70\text{ Torr}.

New solution.
22.5=20xA+70(1−xA)  ⟹  50xA=47.5  ⟹  xA=0.95, xB=0.05.22.5=20x_A+70(1-x_A)\implies50x_A=47.5\implies x_A=0.95,\ x_B=0.05.

xAxB=19.\frac{x_A}{x_B}=19.

Answer: 19