In a mixture of a solid and liquid, or two liquids, the major component represents the solvent, and the minor component represents the solute. The presence of the solute induces the phenomenon of a freezing-point depression in the solvent, where the freezing point of the solvent in the mixture becomes lower than that of the pure solvent. The freezing-point depression is calculated according to delta(T) = Km, where K represents the freezing-point depression constant of the solvent, and m represents the molality of the solution. Molality, in this case, represents the moles of solute particles per kilogram of solvent. Chemists determine the moles of solute particles by dividing the mass of the solute by its molecular weight, as determined by adding together the atomic masses of all of the atoms in its chemical formula.

Identify the solute and solvent in the mixture. By definition, the solute represents the compound present in lesser amount. For example, for a mixture of 10 grams of sodium chloride (salt) dissolved in 100 grams of water, the sodium chloride represents the solute.

Determine the formula weight or molecular weight of the solute by adding together the atomic weights of all of the atoms in the solute’s chemical formula. Sodium chloride contains one sodium atom and one chlorine atom, and the atomic weights from the periodic table of the elements for sodium and chlorine are 22.99 and 35.45, respectively. Its formula weight is therefore (1 x 22.99) + (1 x 35.45), which is 58.44.

Calculate the moles of solute by dividing the grams of solute by its formula weight. Continuing the previous example of sodium chloride, 10 grams/58.44, or 0.171 moles of sodium chloride .

Determine the moles of particles by multiplying the moles of solute by the number of particles created when the solute dissolves. For molecular substances with covalent bonds, such as sugar, each formula represents one molecule or particle in the solution. However, ionic compounds such as sodium chloride produce two or more particles per formula unit. You can identify ionic compounds easily because they always consist of a metal and a nonmetal, whereas molecular compounds such as sugar contain nonmetals only. A compound such as calcium chloride would produce three particles. For the example of 10 grams of sodium chloride (0.171 moles of NaCl) x (2 particles per formula), or 0.342 moles of particles.

Determine the molality of the solution by dividing the moles of particles by the mass of the solvent in kilograms. In the previous example, the prepared solution contained 10 grams of sodium chloride dissolved in 100 grams of water. Because 1 kilogram contains 1000 grams, 100 grams of water represents 0.100 kilograms of water. Use the online conversion tool to convert the mass of solvent to kilograms, if necessary. The particle molality of 10 grams of sodium chloride in 100 grams of water is therefore 0.342 / 0.100, or 3.42 moles per kilogram.

Refer to a table of freezing-point depression constants to determine the freezing-point depression constant, K, of the solvent. The K of water, for example, is 1.86 degrees C per molal.

Calculate the freezing-point depression, delta(T), of the solvent by multiplying its K value by the molality of the solute: delta(T) = Km. Continuing the previous example, delta(T) = 3.42 x 1.86, or 6.36 degrees C.

Determine the freezing point of the mixture by subtracting delta(T) from the freezing point of the pure solvent. Most tables of freezing-point depression constants will also provide the freezing point - sometimes listed as the melting point - of the pure solvent. In the case of water, the freezing point is 0 degrees C. The freezing point of 100 grams of water containing 10 grams of sodium chloride is therefore 0 - 6.36, or -6.36 degrees C.

For mixing rules and freezing, you may consult the book Heat and Thermodynamics, 7th edition, by MW Zemansky, and RH Dittman (McGraw-Hill, New York, 1997).

Thanks a lot Dear Dr. Tahery and Dr. Farid

The answers are very helpful.

I calculated "Solid Formation Temperature" for the binary systems of 1-propanethiol + Water and 1-butanethiol + water as a function of water contents of mixture using a commercial process simulation software.

As you can see, by increasing water content from 1 ppb to 1 % (on a molar basis), "Solid Formation Temperature" almost varies between freezing point of 1-propanethiol or 1-butanethiol to that of water.

You are welcome Amir.

Your calculated results are reminiscent of the phase diagram of the mixture of kitchen salt and water, for concentrations of salt above approximately 30% -- see Fig. 17-2, p. 441, of the book by Zemansky & Dittman, 7th edition. Can you verify your calculated results experimentally?

Dear Dr Farid,

Based on the theory of "freezing-point depression", which fully described by Dr. Tahery, freezing point temperature of a solution, must be less than that of solvent. However, in the case I described before, "Solid Formation Temperature" increases with increasing water contents of the system.

It seems that concept of "Solid Formation Temperature" is differ from the "freezing point temperature". Solid formation temperature is the highest temperature at which at least one of the species of the system became solid. For example, in the above case is the temperature at which water freezes and ice particles are form. This is somehow like the hydrate formation temperature.

Unfortunately, I cannot verify the simulation results, as I haven't an experimental setup for it.

Thanks again for your kind response.

Dear Amir,

What you describe is perfectly reflected in the phase diagram in the book by Zemansky & Dittman to which I referred earlier.

As for experiment, all you need, besides the materials you wish to mix, is a vessel and a thermocouple. There are various types of thermocouples, some of which capable of measuring very wide ranges of temperatures. Please consult the attached entry of Wikipedia for details.

https://en.wikipedia.org/wiki/Thermocouple

Being a colligative property, the freezing point depression for aqueous solutions ― θf = Tfº - Tf, where Tfº is the freezing point of water and Tf is that for the diluted solution  ― is often rather insensitive to the actual solute, while considering a set of comparable (weak or strong) electrolyte solutions. Quite possibly, it might be taken as primarily dependent on the molality of the solute (m) and on its van't Hoff factor υ (*); regardless of the actual solute, which identity may have a lesser significant role for coarse estimation purposes.

Therefore, in the absence of freezing point experimental data, I would propose the following approximate correlation, for a first-guess estimate (prediction) of the freezing point depression (ºC) for such a generic dilute aqueous solution (**): 

θf ≈ 1.86·υ·m·(1 - 0.283(kg/mol)1/2 √m) ºC·kg/mol

(*) https://en.wikipedia.org/wiki/Freezing-point_depression; https://en.wikipedia.org/wiki/Van_%27t_Hoff_factor

(**) Cf. my next post.

I have derived the correlation proposed at the above post as follows:

Concerning to the activity coefficient of the solute at the dilute aqueous solution, we may write: ln γf± ≈ ϕf - 1 - ∫ [(1- ϕf)/m]dm, where ϕf = θf / (1.86υ·m) (*), with symbols defined at the above post and omitting units for simplicity. Accepting Debye-Hückel ‘second approximation’ (*): log10 γf± ≈ - 0.492√m / (1 + √m), or: ln γf± ≈ - 1.133√m / (1 + √m). From both expressions, we have: ϕf  ≈ 1 + ∫ [(1- ϕf)/m]dm - 1.133√m / (1 + √m); what can be differentiated as: dϕf / dm ≈ (1- ϕf)/m - 1.133 d[√m / (1 + √m)]/dm; or: dϕf / dm ≈ (1- ϕf)/m - 1.133[1/(1+√m)]·[1 - √m/(1+√m)] / (2√m). As long as √m

Dear Dr. Queiroz

Thanks for your comprehensive response.

Dear Hosein

Thanks for your response.

To complete my earlier posts to this query, I would like to add that for a mixture of two or more solutes in aqueous solution ― identified by the i-index ― we should substitute ∑mi for m and ∑(υi·mi) for υ·m at the correlation proposed at my first post to this query, which then could be given as:

θf ≈ 1.86·(1 - 0.283(kg/mol)1/2√∑mi)·∑(υi·mi) ºC·kg/mol

while Blagden's Law would rewrite itself as:

θf ≈ 1.86·∑(υi·mi) ºC·kg/mol