Talk:Boiling-point elevation

Error in formula
Not sure of this, so not changing it myself, but if "n is the number of dissolved particles the solute will create when dissolved", then isn't that redundant, because that's what the van't Hoff factor is for? EvilStorm 16:02, 6 February 2006 (UTC)

Yes, that is redundant. The formula is ^T = iKM, so I'm changing it. 24.125.117.4 03:05, 23 February 2006 (UTC)

Molarity --> Molality
This page mistakenly states that "m" in the formula ΔT = Kb x m stands for "molarity." In actuality, "m" stands for "molality," a completely separate measurement. This is confusing and inncorect, and ought to be changed.

Boiling-point depression?
What about Boiling-point depression, such as with ethanol and water?
 * In that case, we're not talking about a solution of a non-volatile solute in a volatile solvent. Rather, we're talking of a mixture of two volatile liquids, which is another chapter in the book. In this specific case an azeotrope is also formed, complicating things.Tomas e 21:57, 20 October 2007 (UTC)

Why?
Why does it happen? A.Z. 20:28, 9 November 2006 (UTC)
 * Solvation involves the formation of weak bonds between solute and solvent particles. Boiling essentially breaks these bonds, since gases do not form solutions the way liquids do. As an inherently endothermic process, boiling already requires a certain amount of energy (i.e., a certain temperature) to occur. The presence of a solute simply increases this energy barrier, and thus, more energy (corresponding to a higher temperature) must be added in order to achieve vaporization. This explanation takes into account both the strength of the bonds (included in Kb) and the number of bonds (included in i), and the boiling point elevation is proportional to each. -128.101.53.232 08:56, 13 February 2007 (UTC)
 * This explanation is quite simply wrong. Since this is a collagative phenomenon, it does not depend on any specific interactions between solute and solvent, and there are no additional energy barriers involved. The solvent is quite simply diluted, and must therefore reach a higher temperature before its vapour pressure is equal to the pressure of the surroundings, as can be read from the explanation I've now added. Tomas e 21:57, 20 October 2007 (UTC)


 * Well ultimately energetics is still important. Only not in the way the original post described, nor how you described it Tomas.  As water becomes more disordered due to being diluted, its entropy increases.  This decreases the change in entropy gained as a result of evaporating.  In other words, it takes more energy to vaporize pure water vs water in a solution, because pure water is lower in entropy.  Or simply, that water in a solution has a lower chemical potential than pure water.  Your reply, and the current version of the wiki is very misleading, implying that it is the number of surface water molecules present that dictates the vapor pressure; since because the water is diluted, there are less water molecules available to turn to gas at any given time.  However, vapor pressure is INDEPENDENT of surface area.  The equilibrium is established based on chemical potentials, not on surface area, or exposed water.  Therefore, the dilution bit really should be clarified.

=== You say "it takes more energy to vaporize pure water", However with boiling point elevation (of solutions) it seem to take more energy to vaporize un-pure water, and not pure water! What's wrong here? Also, in the Wiki it says: "A non-volatile solute has a vapor pressure of zero, so the vapor pressure of the solution is the same as the vapor pressure of the solvent. Thus, a higher temperature is needed for the vapor pressure to reach the surrounding pressure, and the boiling point is elevated." If the vapor pressure of the solution is the same as that of the solvent then why does it require MORE energy (or higher T) to boil it? Isn't the point that the vapor pressure of the solution is LOWERED and therefore one needs higher T to reach the point of vapor pressure being equal to surrounding (gas) pressure, i.e. bring it to boiling? —Preceding unsigned comment added by 75.80.129.136 (talk) 00:50, 11 October 2010 (UTC)

Salt water example
It would be nice if someone more knowledgeable than me could add more to the "10 g of salt per 1 kg water => 0.17deg increase" sentence. More detailed maths would be useful. It would look nice as an example. janto (talk) —Preceding undated comment added 10:21, 3 February 2011 (UTC).