Vapor–liquid equilibrium (Raoult’s law):\nCalculate the total pressure of a liquid mixture (benzene, toluene, o-xylene) at 100 °C given their pure-component vapor pressures.

Introduction / Context:
This problem tests application of Raoult’s law for an ideal liquid solution at a fixed temperature. The total pressure above the mixture equals the sum of each component’s partial pressure, which is the product of its liquid mole fraction and its pure-component vapor pressure at that temperature.

Given Data / Assumptions:

• Liquid moles: benzene = 0.45 kmol, toluene = 0.44 kmol, o-xylene = 0.23 kmol.
• Total moles in liquid = 0.45 + 0.44 + 0.23 = 1.12 kmol.
• P^sat at 100 °C (mmHg): benzene = 1340, toluene = 560, o-xylene = 210.
• Assume ideal solution behavior (Raoult’s law) and negligible gas-phase non-ideality.

Concept / Approach:
For each component i: p_i = x_i * P_i^sat, where x_i is the liquid mole fraction. The total pressure P_total = Σ p_i across all components.

Step-by-Step Solution:

Verification / Alternative check:
Recalculate with rounded x_i values to two or three decimals; the total remains ≈ 801.5 mmHg, confirming robustness to small rounding differences.

Why Other Options Are Wrong:

• 756.2 / 780.5 mmHg: Underestimates; likely from arithmetic or mole-fraction rounding errors.
• 880.5 / 900.0 mmHg: Overestimates; would require higher P^sat weighting or incorrect x_i.

Common Pitfalls:
Forgetting to normalize mole fractions, mixing activity coefficients into an ideal law problem, or confusing mass fractions with mole fractions.

Final Answer:
801.5 mmHg