For the same binary solution (P°₁ = 450 mm Hg, P°₂ = 700 mm Hg, total P = 600 mm Hg, $x_1 = 0.4$, $x_2 = 0.6$), the mole fraction of B (component 2) in the **vapour phase** is:
A0.3
B0.6
C0.5
D0.7
Answer & Solution
Correct answer: D. 0.7
$P_2 = x_2 P_2^0 = 0.6 \times 700 = 420$ mm Hg. By Dalton's law $y_2 = P_2/P = 420/600 = 0.70$. (Note: vapour is **enriched** in the more volatile component B — its mole fraction rises from 0.6 in liquid to 0.7 in vapour.)
Related questions
The mass of NaOH (M = 40 g/mol) required to prepare 500 mL of a 0.20 M NaOH solution isWhich of these solutions will boil at the highest temperature at 1 atm?The molality of a 1 M aqueous NaCl solution (density = 1.04 g/mL, NaCl M = 58.5 g/mol) is Osmotic pressure π of a dilute solution obeys π = C R T. Here C representsVan't Hoff factor i > 1 for a solute indicatesA negative deviation from Raoult law is expected whenIf a solute dimerises in solution, its observed molar mass compared with the calculated vaWhich of the following is NOT a colligative property?