If a 5050 mixture of cyclohexane-toluene is boiled what will the relative ratios of the vapor be?

Answer 1

Answer

Determine the bp of clycohexane and that of toluene. The lower bp chemical will come off first. A simple distillation rig will provide a means of monitoring when the first chemical has been removed. The remaining chemical will likely have some amount of the first remaining, depending upon distillation rig used.

The more complicated answer is that the ratios of vapor will change over time as the solution boils.

When the solution is boiling the total vapor pressure of the solution is equal to atmospheric pressure. That total vapor pressure is the sum of the two partial pressures of toluene and cyclohexane (the components). So what your asking is what is the partial pressure of each; then you can compare them to make a ratio. Well, the partial pressure of each component in the vapor is the vapor pressure of that component at the current temperature of the system times the mole fraction of that component in the liquid.

Toluene has a lower vapor pressure than cyclohexane (it is less volatile) so if you start with a 50/50 mixture of cyclohexane and toluene (50/50 by mole fraction, that is, but 50/50 by volume is pretty close), the partial pressure of toluene is (lower number)x50% and the partial pressure of cyclohexane is (higher number)x50%, so there's more cyclohexane in the vapor than there is toluene. However...

Those vapors tend to float away and leave the system. Since there's initially more cyclohexane in the vapor, more cyclohexane is leaving the system and you no longer have a 50/50 mixture. Eventually it gets to the point where the partial pressure of toluene is (lower number)x(larger fraction) and the partial pressure of cyclohexane is (higher number)x(smaller fraction) so there is actually more toluene in the vapor.