Match the column based on phase diagram (temperature v / s mole fraction) of a solution of liquid A & liquid B given below. List - I List - II (P) Solution of composition X A = 0.8 is heated upto T

Question

Match the column based on phase diagram (temperature $v / s$ mole fraction) of a solution of liquid A & liquid B given below.

List - I	List - II
(P) Solution of composition $X_{A} = 0.8$ is heated upto $T_{1}$ temperature. Then the mole fraction of 'A' in vapour form at equilibrium	(1) 0.2
(Q) Solution of composition $X_{A} = 0.8$ is heated upto $T_{2}$ temperature. Then the mole fraction of ' B ' in vapour form at equilibrium.	(2) 0.26
(R) Solution of composition $X_{A} = 0.66$ is heated upto $T_{4}$ temperature. Then the mole fraction of 'A' in vapour form at equilibrium.	(3) 0.66
(S) Solution of composition $X_{A} = 0.3$ is heated upto $T_{3}$ temperature. Then the mole fraction of ' B ' in liquid form at equilibrium.	(4) 0.7
	(5) 0.8

TestHub · Accepted Answer

For a binary liquid phase diagram:

The upper curve represents the dew point curve, and the lower curve represents the bubble point curve.

At any given temperature, draw a horizontal tie line and read the compositions from both curves.

For (P): $x_{A} = 0.8$ at $T_{1}$ .

The tie line cuts the vapor curve at $X_{A} = 0.8$ . Hence, the vapor mole fraction of A is 0.8 (5).

For (Q): $x_{A} = 0.8$ at $T_{2}$ .

From the tie line, the vapor composition of A is approximately 0.74, so the vapor composition of B is $1 - 0.74 = 0.26$ (2).

For (R): $x_{A} = 0.66$ at $T_{4}$ .

At the azeotropic point, the vapor and liquid compositions are the same. Therefore, the mole fraction of A in the vapor is 0.66 (3).

For (S): $x_{A} = 0.3$ at $T_{3}$ .

The liquid mole fraction of B is $1 - 0.3 = 0.7$ (4).

Therefore, (P)-5; (Q)-2; (R)-3; (S)-4.

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