Stronger intermolecular forces lower vapor pressure in mixture Z Steps:

• X and Y have identical boiling points, so similar intermolecular forces (IMFs) and vapor pressures.
• They mix without reaction to form homogeneous Z.
• Average IMFs in Z exceed the average of pure X and Y IMFs.
• Stronger IMFs in Z reduce molecular escape to vapor phase.

Why B is correct:

• Vapor pressure decreases with stronger IMFs, as per the definition that IMFs hold molecules in liquid against evaporation; Z has lower vapor pressure than X or Y at same temperature.

Why the others are wrong:

• A: Stronger average IMFs imply exothermic mixing via enhanced X-Y attractions, but identical boiling points provide no direct enthalpy data.
• C: Stronger IMFs raise boiling point by requiring more heat to overcome attractions, opposite of lower boiling point claimed.

Final answer: B