Freezing Point Depression

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A solution will solidfy (freeze) at a lower temperature than the pure solvent. This is the colligative property called freezing point depression.

The more solute dissolved, the greater the effect. An equation has been developed for this behavior. It is:

ΔT = i Kf m

The temperature change from the pure to the solution is equal to two constants times the molality of the solution. The constant Kf is actually derived from several other constants and its derivation is covered in textbooks of introductory thermodynamics. Its technical name is the cryoscopic constant. The Greek prefix cryo- means "cold" or "freezing." In a more generic way, it is called the "molal freezing point depression constant."

These are some sample cryoscopic constants:

Substance Kf
benzene 5.12
camphor 40.
carbon tetrachloride 30.
ethyl ether 1.79
water 1.86

The units on the constant are degrees Celsius per molal (°C m¯1). There are some variations on the theme you should also know:

1) K m¯1 - the distance between a single Celsius degree and a Kelvin are the same.
2) °C kg mol¯1 - this one takes molal (mol/kg) and brings the kg (which is in the denominator of the denominator) and brings it to the numerator.

This last one is very useful because it splits out the mol unit. We will be using this equation (or the freezing point) to calculate molecular weights. Keep in mind that the molecular weight unit is grams / mol.

Another reminder: molal is moles solute over kg solvent.


Go below the example problems for some discussion about the van 't Hoff factor.

Example #1: Pure benzene freezes at 5.50 °C. A solution prepared by dissolving 0.450 g of an uknown substance in 27.3 g of benzene is found to freeze at 4.18 °C. Determine the molecular weight of the unknown substance. The freezing point constant for benzene is 5.12 °C/m.

Solution:

Δt = i Kf m

1.32 °C = (1) (5.12 °C kg mol-1) (x / 0.0273 kg)

1.32 °C = (187.5458 °C mol-1) (x)

x = 0.007038 mol

0.450 g / 0.007038 mol = 63.9 g/mol

Note the assumption that the substance does not ionize. This is a fairly safe assumption when benzene is the solvent. Also, note the assumption that the solute is nonvolatile.


Example #2: How many grams of ethylene glycol, C2H4(OH)2, must be added to 400.0 g of water to yield a solution that will freeze at -8.35 °C?

Solution:

Δt = i Kf m

8.35 °C = (1) (1.86 °C kg mol-1) (x / 0.4000 kg)

8.35 °C = (4.65 °C mol-1) (x)

x = 1.7957 mol

1.7957 mol times 62.07 g/mol = 111 g (to three sig figs)

The solution freezes at -1.37 °C.


Example #3: A 33.7 g sample of a nonelectrolyte was dissolved is 750. g of water. The solution's freezing point was -2.86 °C. What is the molar mass of the compound? Kf = 1.86 °C/m.

Solution:

Δt = i Kf m

2.86 °C = (1) (1.86 °C kg mol-1) (x / 0.750 kg)

2.86 °C = (2.48 °C mol-1) (x)

x = 1.1532 mol

33.7 g / 1.1532 mol = 29.2 g/mol


Example #4: A 1.60 g sample of napthalene (a non-electrolyte with a formula of C10H8) is dissolved in 20.0 g of benzene. The freezing point of benzene is 5.5 °C and Kf = 5.12 kg/mol. What is the freezing point of the solution?

Solution:

1) Determine the molality of napthalene:

(1.60 g / 128.1732 g/mol) / 0.0200 kg = 0.624155 m

2) Determine the freezing point depression:

Δt = i Kf m

x = (1) (5.12 °C kg mol-1) (0.624155 mol/kg)

x = 3.2 °C

3) Determine the freezing point:

5.5 - 3.2 = 2.3 °C

Example #5: Camphor (C6H16O) melts at 179.8 °C, and it has a particularly large freezing point depression constant, Kf = 40.0 °C/m. When 0.186 g of an organic substance of unknown molar mass is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7 °C. What is the molar mass of the solute?

Solution:

179.8 - 176.7 = 3.1 °C

Δt = i Kf m

3.1 °C = (1) (40.0 °C kg mol-1) (x / 0.02201 kg)

3.1 °C = (1) (1817.356 °C mol-1) (x)

x = 0.001705775 mol

0.186 g / 0.001705775 mol = 109 g/mol


Bonus Example: What volume of ethylene glycol (C2H6O2), a nonelectrolyte, must be added to 20.0 L of water to produce an antifreeze solution with a freezing point of -34.0 °C? (The density of ethylene glycol is 1.11 g/cm3, and the density of water is 1.00 g/cm3.)

Solution:

1) We need to determine the moles of ethylene glycol:

Δt = i Kf m

34.0 °C = (1) (1.86 °C kg mol-1) (x / 20.0 kg)

34.0 °C = (0.093 °C mol-1) (x)

x = 365.59 mol

2) How many grams is this?

365.59 mol * 62.0674 g/mol = 22691.22 g

3) What volume does this occupy?

22691.22 g / 1.11 g/cm3 = 20442 cm3

This is 20.44 L

An approximate 50/50 (by volume) mixing of ethylene glycol and water gives us the freezing point depression we require. Note that I did not bother to show how 20.0 L of water was converted into 20.0 kg.


Go to freezing point depression problems #1-10

Go to freezing point depression problems #11-25


The van 't Hoff Factor

The van 't Hoff factor is symbolized by the lower-case letter i. It is a unitless constant directly associated with the degree of dissociation of the solute in the solvent.

Substances which do not ionize in solution, like sugar, have i = 1.
Substances which ionize into two ions, like NaCl, have i = 2.
Substances which ionize into three ions, like MgCl2, have i = 3.
And so on. . . .

That's the modern explanation. In the 1880's, when van 't Hoff was compiling and examining boiling point and freezing point data, he did not understand what i meant. His use of i was strictly to try and make the data fit together. Essentially, this is what he had:

Take a 1.0-molal solution of sugar and measure its bp elevation. Now examine a 1.0-molal solution of NaCl. Its bp elevation is twice the sugar's value. When he did MgCl2, he got a value three times that of sugar.

All his values begain to group together, one groups with sugar-like values, another with NaCl-like values and a third with MgCl2-like values.

This is how each group got its i value and he had no idea why. That is, until he learned of Svante Arrhenius' theory of electrolytic dissociation. Then, the modern explanation above became very clear. >P>Substances that ionize partially insolution will have i values between 1 and 2 usually. I will do an example problem in osmosis that involves i = 1.17. Also, i values can be lowered by a concept calle "ion pairing" For example, NaCl has an actual i = 1.8 because of ion pairing. I will leave it to you to find out what ion pairing is.

Two van 't Hoff factor problems


Some additional comments about the boiling point and freezing point of a Solution

Pure substances have true boiling points and freezing points, but solutions do not. For example, pure water has a boiling point of 100 °C and a freezing point of 0 °C. In boiling for example, as pure water vapor leaves the liquid, only pure water is left behind. Not so with a solution.

As a solution boils, if the solute is non-volatile, then only pure solvent enters the vapor phase. The solute stays behind (this is the meaning of non-volatile). However, the consequence is that the solution becomes more concentrated, hence its boiling point increases. If you were to plot the temperature change of a pure substance boiling versus time, the line would stay flat. With a solution, the line would tend to drift upward as the solution became more concentrated.


A non-volatile solute is one which stays in solution. The vapor that boils away is the pure solvent only. A volatile solute, on the other hand, boils away with the solvent.

Salt in water is an example of a non-volatile solute. Only water will boil away and, when dry, a white solid (the NaCl) remains. Hexane dissolved in pentane is an example of a volatile solute. The vapor will be a hexane-pentane mixture. However, here is something very interesting. The hexane-pentane percentages in the vapor will be DIFFERENT that the percentages of each in the solution. We will get into that in a different tutorial.


One last thing that deserves a small mention is the concept of an azeotrope. This is a constant boiling mixture. What this means is that the mixture of the vapor coming from the boiling solution is the same as the mixture of the solution. The first occurence was reported by Dalton in 1802, but the word was not coined until 1911.

One example of a binary azeotrope is 4% (by weight) water and 96% ethyl alcohol. By the way, what this means is that you cannot produce pure, 100% alcohol (called absolute alcohol) by boiling. You must use some other means to get the last 4% out. It also means that absolute alcohol is hygroscopic - it absorbs water from the atmosphere.

The Handbook of Chemistry and Physics for 1992 lists the following:

Azeotrope Number
Binary 1743
Ternary 177
Quaternary 21
Quinary 2

Here is the composition of one quinary system. It boils at 76.5 ° C

Substance Percent
by Weight
Water 9.45
Nitromethane 37.30
Tetrachloroethylene 21.15
n-Propyl alcohol 10.58
n-Octane 21.52

Pretty exciting, eh?

Oh, by the way, the same lowering of the freezing (sometimes called solidification) point also happens with metal alloys such as solders. An alloy actually has a melting point below that of either of its parent metals. The ratio with the lowest point is called a "eutectic" alloy; a 63 parts tin to 37 parts lead electrical solder is one such eutectic mixture.


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