Question

Starting at a solution that is 50% A and 50% B (at X), if we were to heat this solution to boiling point T1, the vapour that would be in equilibrium would have a composition of ______, which practicaly speaking means the distillate will become ______ in the ______ boiling component of the mixture, while the liquid that remains in the original RBF will be_________ in the __________ boiling component. (pg 62).

Answer 1

When heating a solution that is 50% A and 50% B to boiling point T1, the vapor in equilibrium will have a composition that is enriched in the more volatile component (A). As a result, the distillate, or collected vapor, will become richer in A, while the liquid in the original round-bottom flask (RBF) will become relatively richer in the less volatile component (B).

Step-by-step explanation:

When heating a solution that is 50% A and 50% B to boiling point T1, the vapor in equilibrium will have a composition that is enriched in the more volatile component, which we can denote as A. This means that the vapor will have a higher mole fraction of A compared to the liquid.

Consequently, the distillate, which is the collected vapor, will become richer in A. On the other hand, the remaining liquid in the original round-bottom flask (RBF) will become relatively richer in B. Therefore, the distillate will become more concentrated in the lower boiling component (A), while the liquid in the RBF will be more concentrated in the higher boiling component (B).