Q1: Assume that benzene is insoluble in water. The normal boiling points of benzene and water are 80.1 and 100°C respectively. At a pressure of 1 atm, the boiling point of a mixture of benzene and water is

ANS:D - greater than 80.1°C but less than 100°C.

In a mixture of benzene and water, the boiling point will be higher than the boiling point of benzene but lower than the boiling point of water due to the colligative property of boiling point elevation. Benzene is insoluble in water, so the solute concentration is effectively zero. According to Raoult's law, the vapor pressure of the solution is proportional to the mole fraction of the solvent (water). Since water has a higher boiling point than benzene, adding benzene to water will decrease the vapor pressure of the solution compared to pure water, resulting in an increase in the boiling point. Therefore, the boiling point of the mixture of benzene and water at 1 atm pressure will be greater than 80.1°C but less than 100°C.