For a binary liquid solution of ' A ' and ' B '. P A 0 = vapour pressure of pure ' A '. P B 0 = vapour pressure of pure B . X A = mole fraction of A in liquid phase

Question

For a binary liquid solution of ' A ' and ' B '. $P_{A}^{0} =$ vapour pressure of pure ' A '. $P_{B}^{0} =$ vapour pressure of pure B . $X_{A} =$ mole fraction of $A$ in liquid phase $Y_{A} =$ mole fraction of ' $A$ ' vapour phase.

LIST-I	LIST-II
(P) $P_{A}^{0} > P_{B}^{0}$ (ideal liquid solution)	(1) $Y_{A} > X_{A}$
(Q) Azeotropic mixture	(2) $Y_{A} = X_{A}$
(R) Equilmolar ideal mixture of 'A' and 'B' with $P_{A}^{0} < P_{B}^{0}$	(3) $Y_{B} > Y_{A}$
(S) Equimolar mixture of 'A' and 'B', $P_{A}^{0} = 100 t orr, P_{B}^{0} = 200 t orr$ has vapor pressure 140 torr	(4) Neagtive Deviation

TestHub · Accepted Answer

P. If P A 0 > P B 0 , component A is more volatile, so Y A > X A . Q. For an azeotropic mixture, the composition in liquid and vapor phases is the same, so Y A = X A . R. If P A 0 < P B 0 , component B is more volatile, so Y B > Y A . S. For an equimolar ideal mixture, P T = X A P A 0 + X B P B 0 = 0.5 ( 100 ) + 0.5 ( 200 ) = 150 torr. Since P T = 140 < 150 torr, it shows negative deviation.

Chemistry - Liquid Solution Question with Solution | TestHub

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JEE Advanced 2027 (Class 12)