Ideal solution flash (Raoult’s law):\nA liquid mixture contains 0.4 mol fraction benzene (p^sat = 3 atm) and 0.6 mol fraction toluene (p^sat = 4/3 atm) at a given temperature. Assuming ideal behavior and equilibrium, what is the benzene mole fraction in the vapor phase?

Introduction / Context:
For ideal binary mixtures, Raoult’s law and Dalton’s law provide a rapid way to compute equilibrium vapor compositions from known liquid compositions and saturation pressures at a fixed temperature. Benzene–toluene is a classic nearly ideal pair used in teaching VLE basics.

Given Data / Assumptions:

• x_B = 0.4, x_T = 0.6 in the liquid.
• p_B^sat = 3 atm, p_T^sat = 4/3 atm at the same temperature.
• Ideal behavior: p_i = x_i p_i^sat and y_i = p_i / P_total (proportional to partial pressures).

Concept / Approach:
The vapor composition y_B equals the benzene partial pressure divided by the total pressure. Under Raoult’s law, total pressure is p_total = x_B p_B^sat + x_T p_T^sat, and the benzene partial pressure is x_B p_B^sat. Hence y_B = (x_B p_B^sat) / (x_B p_B^sat + x_T p_T^sat).

Step-by-Step Solution:
Compute numerator: x_B p_B^sat = 0.4 × 3 = 1.2 atm.Compute denominator: 1.2 + (0.6 × 4/3) = 1.2 + 0.8 = 2.0 atm.Therefore y_B = 1.2 / 2.0 = 0.6.

Verification / Alternative check:
Because benzene is more volatile (higher p^sat), its vapor-phase fraction must exceed its liquid fraction: y_B (0.6) > x_B (0.4), confirming the direction of enrichment.

Why Other Options Are Wrong:
0.4: equals the liquid fraction, ignoring volatility differences.0.8 or 0.2: over- or underestimates enrichment relative to the saturation pressure ratio.

Common Pitfalls:
Dividing by total pressure from a different state; forgetting to normalize y_B + y_T = 1; mixing activity coefficients into an ideal problem.

Final Answer:
0.6