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. If, at a total pressure of one standard atm. exerted by the vapours of water and toluene, the mole fraction of water X^w in the vapour phase satisfies

ANS:C - 0.5 < X_w < 1.0

To solve this problem, we need to consider Raoult's law, which describes the vapor pressure of an ideal solution. According to Raoult's law: 𝑃total=𝑃A⋅𝑋A+𝑃B⋅𝑋BPtotal​=PA​⋅XA​+PB​⋅XB​ Where:

• 𝑃totalPtotal​ is the total vapor pressure of the mixture.
• 𝑃APA​ and 𝑃BPB​ are the vapor pressures of the pure components A and B, respectively.
• 𝑋AXA​ and 𝑋BXB​ are the mole fractions of components A and B in the mixture.

Since toluene and water are completely immiscible in each other, the vapor above the mixture will contain only toluene vapor and water vapor, each exerting partial pressures according to their mole fractions. Given that the mole fraction of toluene and water in the equimolar mixture is 0.5 each, we can substitute these values into Raoult's law. Let's denote:

• 𝑃APA​ as the vapor pressure of toluene (𝐶7𝐻8C7​H8​).
• 𝑃BPB​ as the vapor pressure of water (𝐻2𝑂H2​O).
• 𝑋AXA​ as the mole fraction of toluene.
• 𝑋BXB​ as the mole fraction of water.

Given:

• Vapor pressure of toluene (𝑃APA​) = 110.6°C (since toluene is the major component)
• Vapor pressure of water (𝑃BPB​) = 100°C

Now, we can plug these values into Raoult's law: 𝑃total=(110.6 mmHg)×(0.5)+(100 mmHg)×(0.5)Ptotal​=(110.6mmHg)×(0.5)+(100mmHg)×(0.5) 𝑃total=55.3 mmHg+50 mmHgPtotal​=55.3mmHg+50mmHg 𝑃total=105.3 mmHgPtotal​=105.3mmHg At one standard atmosphere (1 atm), the total pressure is ≈760 mmHg≈760mmHg. Therefore, the mole fraction of water (𝑋wXw​) in the vapor phase is: 𝑋w=𝑃B⋅𝑋B𝑃total=50105.3≈0.474Xw​=Ptotal​PB​⋅XB​​=105.350​≈0.474 Since 0<𝑋w<0.50<Xw​<0.5, the correct option is 0<𝑋w<0.50<Xw​<0.5.