Q1: The boiling points for pure water and pure toluene are 100°C and 110.6°C respectively. Toluene and water are completely immiscible in each other. A well agitated equimolar mixture of toluene and water are prepared. The temperature at which the above mixture will exert a pressure of one standard atm. is

ANS:C - between 100 and 110°C

When two immiscible liquids are mixed together, they behave as if they are in separate layers, each exerting their own vapor pressure independent of the other. In this case, the vapor pressure of the mixture will be determined by the vapor pressure of each component at a given temperature. Given:

• Boiling point of water (𝐻2𝑂H2​O) = 100°C
• Boiling point of toluene (𝐶7𝐻8C7​H8​) = 110.6°C
• The mixture is equimolar (equal moles) of toluene and water.

At the boiling point of a liquid, its vapor pressure is equal to the external pressure (in this case, one standard atmosphere). Since the boiling point of water is lower than the boiling point of toluene, at the boiling point of water (100°C), the vapor pressure of water will be equal to one standard atmosphere. Therefore, the mixture will exert a pressure of one standard atmosphere at a temperature less than 100°C. So, the correct answer is: less than 100°Cless than 100°C