[tvtokyo]

I have some questions on real solution..
Question 1:

First statement: Correct
2nd statement: False (contant pressure, should boil at a particular temperature?)
3rd: True
4th: Unsure ??
5th: True


Question 2:
Liquids A and B form an azeotropic mixture. Boiling point of pure liquid A is 79 °C and the boiling point of pure liquid B is 69 °C. Minimum b.p is observed for a mixture having the mole fraction of A equal to 0.3.

Can I say that A-B interaction is weaker than A-A and B-B ?

When mole fraction of A =0.5 undergoes fractional distillation?
(i) What would be collected at the top of the fractional column ? (is it vapour having mole fraction of A = 0.3 collected?)
(ii) Can we obtain pure A and pure B in the flask or neither pure A nor pure B can be obtained?

Thanks!

[mjc123]

Q1: 1.True
2.True - boiling point varies with composition. (Apply the phase rule, 1 degree of freedom if pressure is fixed.) A mixture of given composition will have a particular b.p., but as it boils the liquid will become richer in the less volatile component and the b.p. will rise.
3. True
4. False - not mixtures that form azeotropes. If 5 is true, 4 must be false.
5. True

Q2: Yes, but not too much, or A and B would be immiscible.
(i) Yes, it will be the azeotrope.
(ii) Azeotrope will distil off and pure A will be left in the flask. You can't get pure B from this experiment. (You could get it by distilling a mixture richer in B than the azeotrope.)

[tvtokyo]

Hi thanks. May I know why A would be left in the flask?

[mjc123]

Because you have distilled off all the B (as azeotrope). Of course once you've got pure A it will boil at the boiling point of A, so you could collect it as your last fraction. It all depends when you decide to stop heating. The point is that azeotrope (most volatile) is separated from A (least volatile).