Benzene–nitrogen mixture at 900 mmHg: If the absolute humidity is 0.2 kg benzene per kg nitrogen, what is the benzene partial pressure (mmHg)?

Introduction / Context:
Gas–vapor mixtures are often described by mass ratios (absolute humidity) as well as partial pressures. Converting between these requires molecular weights and total pressure. This skill is vital in drying, stripping, and environmental emissions calculations.

Given Data / Assumptions:

• Total pressure P_total = 900 mmHg.
• Absolute humidity w = 0.2 kg benzene per kg nitrogen.
• Molecular weights: MW_benzene = 78 kg/kmol; MW_N2 = 28 kg/kmol.

Concept / Approach:
Absolute humidity w relates to mole fraction y via w = (y * MW_vapor)/[(1 − y) * MW_gas]. Solve for y, then compute the benzene partial pressure p_b = y * P_total.

Step-by-Step Solution:

Verification / Alternative check:
Convert the benzene mole fraction back to mass ratio to ensure consistency: w ≈ (0.06699*78)/((0.93301)*28) ≈ 0.2 kg/kg, confirming the calculation.

Why Other Options Are Wrong:

• 180 or 200 mmHg: would imply much higher mass ratios than 0.2 kg/kg.
• 720 mmHg: near total pressure; impossible given the nitrogen carrier gas.
• 35.0 mmHg: underestimates the required benzene content.

Common Pitfalls:
Using air molecular weight instead of nitrogen; forgetting that w is a mass ratio, not mole ratio; mixing up mmHg and atm units during conversions.

Final Answer:
60.3