Talk:Boiling-point elevation

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Contents 1 Error in formula 2 Molarity --> Molality 3 Boiling-point depression? 4 Why?

[edit] 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)

[edit] 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.

[edit] 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)

[edit] 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)