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 m

Question

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: [2026]

Smart Achievers · Accepted Answer

(1)

$P_{s} = X_{A} P_{A} ° + X_{B} P_{B} °$

$320 = P_{A} ° (\frac{2}{5}) + P_{B} ° (\frac{3}{5})$

$2 P_{A} ° + 3 P_{B} ° = 1600 . . . . (I)$

$Now 1 mole of A & 1 mole of B is added$

$X_{A}' = \frac{3}{7}, X_{B}' = \frac{4}{7}$

$P_{s}' = 328.6 = P_{A} ° (\frac{3}{7}) + P_{B} ° (\frac{4}{7})$

$3 P_{A} ° + 4 P_{B} ° = 2300.2 . . . . (I I)$

$Now eq (I) \times 3 - eq (II) \times 2$

$6 P_{A} ° + 9 P_{B} ° = 4800$

$6 P_{A} ° + 8 P_{B} ° = 4600.4$

$P_{B} ° ≃ 200 mm of Hg$

$P_{A} ° ≃ 500 mm of Hg$

Solution

1 500, 200

2 400, 300

3 300, 200

4 600, 400

Want Us to Email you About Special Offers & Updates?