Recently, I learnt that by adding a non-volatile solute to a solvent, the vapour pressure of the solution is decreased and consequently its boiling point.

Also consider the definition of the freezing point: "The temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase."

So as the solution cools down, the vapour pressure decreases and eventually equals the vapour pressure of the solid.

Now consider this intuitive argument: the atmospheric pressure is constant regardless of the temperature of the solution. As we cool down a solution, we are basically removing the kinetic energy of the solution's particles so due to less kinetic energy and due to atmospheric pressure the Van der Waal's forces come to play, hence turning the solution into a solid state.

Also the pressure exerted on the liquid by the atmosphere is the same whether you add a solute or not, so no depression of freezing point should take place.

I know that this argument is not true, as experiments proves the colligative properties are true.

Where and why my "intuitive" argument is failing?

For some of you who failing to understand my question, this is another form of it . A logical explanation for depression in freezing point , where inevitability of atmospheric pressure regardless of solution's vapor pressure is considered? If not considered , appropriate reason for the same.


closed as unclear what you're asking by Mithoron, Todd Minehardt, A.K., airhuff, Jon Custer Nov 13 '18 at 4:08

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I have edited it pl remove the hold $\endgroup$ – sudhanva b Nov 19 '18 at 2:17
  • $\begingroup$ I need an explination for depression in freezing point $\endgroup$ – sudhanva b Dec 6 '18 at 7:15

You argument is largely based on enthalpy, whereas freezing point depression and boiling point elevation is (in the situation when the solute only weakly interacts with the solvent) mainly an entropic phenomenon.

Because of the increased entropy by adding a solute (i.e. there are more configurations by having two types of molecules instead of one), the solution will liquify at a lower temperature as compared to the pure solvent and with a similar reasoning, the boiling point will increase with respect to the pure solvent.

  • $\begingroup$ but the reason given by my teacher and and textbooks is that , add solute -vapour pressure decreases-bp and mp decreases-as they are dependent on the pressure above the solution $\endgroup$ – sudhanva b Nov 11 '18 at 18:46
  • $\begingroup$ Perhaps I am overlooking something, but to the best of my knowledge the boiling point increases when you add a weak-interacting solute to a solvent. For instance, have a look at this page: en.wikipedia.org/wiki/Boiling-point_elevation Did your teacher (or textbook) by any chance mentioned some additional details? $\endgroup$ – Ivo Filot Nov 11 '18 at 18:48
  • $\begingroup$ yeah boiling of liquids takes place when vapour pressure is equal or greater than atmospheric pressure $\endgroup$ – sudhanva b Nov 11 '18 at 18:49
  • $\begingroup$ but adding of solute decreases the vapor pressure. now the elevation of boiling point completely makes sense to me whereas depression in freezing point does not $\endgroup$ – sudhanva b Nov 11 '18 at 18:51
  • 1
    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – sudhanva b Nov 11 '18 at 19:11

Not the answer you're looking for? Browse other questions tagged or ask your own question.