Depression of freezing point [closed]

Question

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.

Facts and definitions

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.

My argument as to why depression in freezing point shouldn’t take place

consider this 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.

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.

Reality check

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

Where and why my "intuitive" argument is failing?

Answer 1

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.