A mixture contains 1 mol of volatile liquid A (p°A= 100 mmHg) and 3 moles of volatile liquid B (p°b = 80 mmHg). If solution is ideally behaved, total vapour pressure of distillate is (approx)

Question

A mixture contains 1 mol of volatile liquid A (p°A= 100 mmHg) and 3 moles of volatile liquid B (p°b = 80 mmHg). If solution is ideally behaved, total vapour pressure of distillate is (approx) (A) 85 mm Hg (B) 86 mm Hg (C) 90 mm Hg (D) 92 mm Hg

Tardigrade Admin · Accepted Answer

pA = p° A xA = 100 × (1/4) = 25 mm Hg pB = p° B xB = 80 × (3/4) = 60 mmHg Mole fraction of A in vapour x'A = (pA/pA + pB) = (25/85) Mole fraction of B in vapour x'B = 1 - (25/85)= (60/85) V.P. of distillate = x'A p° A + x' B p° B = (25/85) × 100 + (60/85) × 80 = (1/85) ( 2500 + 4800) = (7300/85) = 85.88 ≈ 86 mm Hg.

Q. A mixture contains 1 mol of volatile liquid A (p∘A= 100 mmHg) and 3 moles of volatile liquid B (p∘b = 80 mmHg). If solution is ideally behaved, total vapour pressure of distillate is (approx)

85 mm Hg

86 mm Hg

90 mm Hg

92 mm Hg

Q. A mixture contains 1 mol of volatile liquid A (p $^{\circ}$ A= 100 mmHg) and 3 moles of volatile liquid B (p $^{\circ}$ b = 80 mmHg). If solution is ideally behaved, total vapour pressure of distillate is (approx)