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Name: Bernadette
Status: educator
Age: 40s
Location: N/A
Country: N/A
Date: 2001 - 2002

Freezing Point Depression and Vapor Pressure Lowering will freezing pt. be lowered if the solution shows positive deviation to Raoult's law?

Is this due to the unfavorable entropy change required as the solution separates as it freezes?

AP Chem 1994 Multiple Choice Question #14 asks Which of the following is lower for a 1.0 molar aqueous solution of ANY solute than it is for pure water?

The correct answer is supposedly Freezing Point
Vapor Pressure is an incorrect choice.

(Follow up Question below)

The question(s) confuse me. First, deviations from Raoult's law affect the depression of the freezing point quantitatively, but not qualitatively. Second, I am not sure what the terms "unfavorable entropy changes ... " means. Third, vapor pressure IS a correct answer to the question as the question is stated in this inquiry. Now let's address the effect of solutes on the physical properties of a solvent. Only the case of liquid-phase soluble/ solid and gas-phase insoluble, non-volatile, totally dissociated solutes is considered. These restrictions affect the magnitude of the effects, and involve more complicated computations, but they do not affect the qualitative direction of the effects. Consider the following general principles of thermodynamics:

*** The addition of a liquid-soluble substance to a liquid phase ALWAYS decreases the vapor pressure of liquid solution (P) relative to that of the pure liquid (Po). Raoult's Law: P = Po * X (where X is the mole fraction of the solvent) is the quantitative statement of that thermodynamic principle.

*** At temperatures less than the melting point of a pure solid, the vapor pressure of the pure super-cooled liquid is ALWAYS greater than that of the pure solid at the same temperature. The vapor pressure of the pure solid and liquid phases are equal AT the melting point.

Both these principles are direct consequences of the Second Law of Thermodynamics. To my knowledge there are no exceptions, given the conditions imposed above.

Since the addition of a liquid-soluble, solid-insoluble solute lowers the vapor pressure of the liquid solution, the temperature of the solid and the solution must decrease until the vapor pressure of the pure solid is equal to the vapor pressure of the solution in thermal contact with that solid. This is the freezing point depression. The freezing point of the solid + solution is less than the freezing point of the pure solid.

Now consider the vaporization process. At the normal boiling point of the pure liquid, the vapor pressure of the liquid, Po = 1 atm. (One could elect some other fixed pressure and the temperature corresponding to that pressure, but that only affects the numbers, not the direction of the effect.). The addition of a solute to the liquid again ALWAYS LOWERS the vapor pressure of the solution relative to the vapor pressure of the pure liquid at the same temperature. In order to bring the vapor pressure of the solution back up to the applied pressure Po = 1 atm. (or some other fixed pressure), the temperature of the solution must be increased by an appropriate amount so that the vapor pressure is equal to the applied pressure. This is the ELEVATION of the boiling point. In the idealized case, Raoult's Law ( P = Po * X ) applies. Positive or negative deviations from Raoult's Law only affect the quantitative "details", but any such deviations do not alter the basic principles of the lowering of the vapor pressure of the solution relative to the pure liquid and the concomitant increase in the boiling point of the solution compared to the pure liquid. This is the boiling point ELEVATION.

So, I contend that BOTH the equilibrium freezing point and vapor pressure of a liquid are decreased upon addition of a non-volatile, liquid-soluble/solid-insoluble solute to the liquid phase.

Vince Calder

Hi Bernadette

This is pretty strange. The addition of ANY solute to pure water will definitely depress the freezing point. However, it will also increase the boiling point (lower vapor pressure).

Now please review the table towards the bottom of the URL at the web site that I have attached:

** If you will scroll down to the table that shows the relative Kb's and Kf's of various solvents you will notice that water's Kf is greater than its Kb. For a 1 molal solution the melting point temperature DEPRESSION for the solution will be 1.86 C (lower than freezing point of pure water). The boiling point ELEVATION for the solution will be 0.52 C (higher than the boiling point of pure water.) A fair set of multiple choice answers would include T (boiling point) and exclude vapor pressure. Of course, Tfp would be correct, and Tbp would be incorrect to the question of "Which one is lowered?"

With this being said, the AP Chem tests two answers do not make sense since BOTH the vapor pressure and the freezing point will be decreased. Also it is a silly comparison since the units of vapor pressure are in [ pressure units, mm Hg, psi, atm, etc... ] and the units of freezing point are in temperature units C, K, F, R ]

My response does not speak to your first question dealing with the effects of the solution as it freezes. Sorry.

I hope you have made some sense of all this information. Thanks for all of the great questions.

-Darin Wagner

If a solute is volatile and has a weak attraction for water, wouldn't that show an increased vapor pressure (a positive deviation to Raoult's Law?)

When a solute is volatile, its vapor pressure must be taken into account since the total vapor pressure of the solution will be the sum of the partial pressures Ptotal = Pa + Pb = Poa*Xa + Pob*Xb = Poa*Xa + Pob*(1 - Xa) for a two component solution of 'a' and 'b' that obeys Raoult's Law. Such solutions are referred to as ideal.

Predicting deviations from ideality on the basis of "simple" forces is not reliable, because many factors enter into the picture in addition to like polarities etc. Differences in molecular size (even if all intermolecular forces are equal) result in deviations from Raoult's Law. In addition, intermolecular forces in the liquid state are very complex and sometimes very specific interactions occur that are not at all obvious from just looking at the polarity or the dipole moment of the components etc. especially when water is involved. If you have access to a chemistry library you can look up "theory of real solutions" in a thermodynamics book as see just how messy things can be. The website:

also will give you some insight into the complexity of the affairs.

Vince Calder

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