plz solve this in detailed way..
Q no.1&2
Answers
1) Vapour pressure of a liquid X, P° = 80 m bar
When solid Y is put in liquid X, then the vapour pressure P is "P° x mole fraction"
But, given that P = 64 m bar
P° x mole fraction of X = 64
80 x mole fraction of X = 64
Thus, mole fraction of X = 64/80 = 0.8
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2) Vapour pressure of pure liquid A is P°₁ = 450 mm Hg
Vapour pressure of pure liquid B is P°₂ = 700 mm Hg
Let mole fraction of Liquid A be 'x', and the mole fraction of liquid B is '1 - x'
Total vapour pressure, P = 600 mm Hg
→ P°₁ (x) + P°₂ (1 - x) = P
→ P°₁ (x) + P°₂ (1 - x) = 600
→ 450x + 700(1 - x) = 600
→ 450x - 700x = 600 - 700
→ -250x = -100
→ x = 0.4 & 1 - x = 0.6
In Liquid Phase, mole fraction of A is 0.4 and B is 0.6
Now,
Mole fraction of A in Vapour phase = P°₁ (x) / P = 450 (0.4)/600 = 0.3
Mole fraction of B in Vapour phase = P°₂ (1 - x) / P = 700 (0.6)/600 = 0.7
∴In Vapour phase, mole fraction of A is 0.3 and B is 0.7
Hope it helps!!