Calculate the boiling point of a 8.25 m aqueous solution of ethanol.

Question

I originally attempted to solve this problem (as shown below), but it is not right. How can I correctly calculate the boiling point?

My attempt:

The normal boiling point for ethanol is 78.4° C and the boiling point elevation is 1.22° C per molal. 

So the boiling point will be elevation by 8.25 * 1.22° C = 10.1° C. So the boiling point will be 78.4+ 10.1 = 88.5° C.

Answer 1

The answer to this will depend to a large extent on the level of chemistry you are studying. There is a simplified answer (which will be incorrect), and a more complicated solution (the correct one) that deals with azeotropes and non-ideal solutions. One key in this question is that ethanol is a volatile liquid, so the normal boiling point elevation formula cannot be used. Following is the simplified approach.

If you consider the ethanol to be the solute (not the solvent) and if you treat it as a non volatile component, then you are really looking at the boiling point elevation of water, where the K_b (ebullioscopic constant) is 0.512ºC/m. Since you have 8.25 m ethanol, we can use the same approach you used but looking at it as a change in boiling point of water, not ethanol.

∆T = imK_b where ∆T is change in temp; i = van't Hoff factor = 1 for ethanol; m= molality = 8.25 and K = 0.512

∆T = (1)(8.25)(0.512) = 4.22ºC

Boiling point = 100º + 4.22º = 104ºC

Again, please note that this is the simplified, incorrect approach. The boiling point of this solution will actually be lower than the boiling point of pure water.