At T(K), 2 moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of ideal solution formed is 320 mm Hg. At this stage, one mole of A and one mole of B are added to the solution. The vapour pressure is now measured as 328.6 mm Hg. The vapour pressure (in mm Hg) of A and B are respectively:

At T(K), 2 moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of ideal solution formed is 320 mm Hg. At this stage, one mole of A and one mole of B are added to the solution. The vapour pressure is now measured as 328.6 mm Hg. The vapour pressure (in mm Hg) of A and B are respectively

Q. At T(K), 2 moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of ideal solution formed is 320 mm Hg. At this stage, one mole of A and one mole of B are added to the solution. The vapour pressure is now measured as 328.6 mm Hg. The vapour pressure (in mm Hg) of A and B are respectively:

(A) 400,300

(B) 600,400

(D) 300,200

Correct Answer: 500,200

Explanation

According to Raoult's law, the total vapour pressure of an ideal solution is given by:

P = x_AP_A + x_BP_B

Initially, moles of A = 2 and moles of B = 3.

x_A = 2 / 5 , x_B = 3 / 5

So:

(2/5)P_A + (3/5)P_B = 320

After adding one mole each of A and B:

moles of A = 3 , moles of B = 4

x_A = 3 / 7 , x_B = 4 / 7

Now:

(3/7)P_A + (4/7)P_B = 328.6

Solving the two equations:

2P_A + 3P_B = 1600

3P_A + 4P_B = 2300

P_A = 500 mm Hg , P_B = 200 mm Hg

Hence, the vapour pressures of A and B respectively are:

500 mm Hg and 200 mm Hg

Explanation

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