Boiling point elevation tutorial

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 K_{f}m

Δt is the temperature change from the pure solvent's freezing point to the freezing point of the solution. It is equal to two constants times the molality of the solution. The constant K_{f} 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."

The constant called the van 't Hoff factor is symbolized with the letter 'i' and is discussed below the example problems.

These are some sample cryoscopic constants:

Substance K _{f}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 two variations on the units of the constant you should also know:

1) K m¯^{1}: Kelvin is used rather than degrees Celsius. However, the "distance" between a single Celsius degree and a Kelvin are the same, the numerical value is unaffected. It's seldom seen and I will tend to ignore it.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 the above equation 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 K_{f}m1.32 °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, C_{2}H_{4}(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 K_{f}m8.35 °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) (62.07 g/mol) = 111 g (to three sig figs)

The solution freezes at −1.37 °C.

Note the van 't Hoff factor of 1. This value is used for substances that do not ionize in solution. Practically all, if not all, organic substances do not ionize in solution. each one of them has a van 't Hoff factor of 1 and they are called nonelectrolytes.

**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? K_{f} = 1.86 °C/m.

**Solution:**

Δt = i K_{f}m2.86 °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 C_{10}H_{8}) is dissolved in 20.0 g of benzene. The freezing point of benzene is 5.5 °C and K_{f} = 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 K_{f}mx = (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 (C_{6}H_{16}O) melts at 179.8 °C, and it has a particularly large freezing point depression constant, K_{f} = 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 °CΔt = i K

_{f}m3.1 °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

**Example #6:** The freezing point of a solution prepared by dissolving 150. mg of caffeine in 10.0 g of camphor is 3.07 Celsius degrees lower than that of pure camphor (K_{f} = 40.0 °C/m). What is the molar mass of caffeine?

**Solution:**

1) Use the freezing point change to calculate the molality of the solution:

Change in FP = K_{f}(m) <--- assume van 't Hoff factor is equal to 13.07 °C = (40.0 °C/m) (m)

m = 0.07675 molal

Remember: molality is moles of solute per kilogram of solvent

2) Convert the concentration of the solution into grams of solute per 1000 g of solvent:

150. mg 1 g 1000 g ––––––– x ––––––– x ––––––– = 15.0 g solute / 1 kg solvent 10.0 g 1000 mg 1 kg

3) Dividing that concentration by the molality of the solution will give you the molar mass:

15.0 g / kg –––––––––––––– = 195 g/mol 0.07675 mol / kg

**Example #7:** What is the freezing point of a water solution made by dissolving 10.90 g MgCl_{2} in 88.41 g H_{2}O?

**Solution:**

10.90 g / 95.211 g/mol = 0.1144826 mol0.1144826 mol / 0.08841 kg = 1.2949 m

Δt = i K

_{f}mΔt = (3) (1.86 °C/m) (1.2949 m)

Δt = 7.226 °C (to 4 sig figs)

The freezing point of the solution is −7.226 °C

In reality, the freezing point may be closer to −6.5 °C due to ion pairing between Mg

^{2+}and Cl¯ ions. The van 't Hoff factor is closer to 2.7 for a concentrated solution of MgCl_{2}(I don't have a source on that, I've just seen it mentioned a few times over the years.). Ion pairs are briefly formed as oppositely charge particles attract and reduce the apparent number of particles. Remember, the colligative properties depend on the total number of particles, reduce those and you will reduce the effect.

**Example #8:** A 29.3%(w/w) solution of strontium fluoride will freeze at what temperature?

**Solution:**

1) We need the molality of the SrF_{2} solution. To do that, we first assume 100. g of the solution is present. Therefore:

29.3%(w/w) means 29.3 g of SrF_{2}in the 100. g of solution as well as 70.7 g of water.

2) Since molality involves moles of solute, we calculate moles of SrF_{2}:

29.3 g / 125.62 g/mol = 0.233243 mol

3) Now, the molality:

0.233243 mol / 0.0707 kg = 3.29905 m

4) We are now ready for the freezing point calculation:

Δt = i K_{f}mΔt = (3) (1.86) (3.29905) <--- the 3 is the van 't Hoff factor

Δt = 18.4087 °C

The solution will freeze at −18.4 °C

**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 MgCl_{2}, 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 MgCl_{2}, 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 MgCl_{2}-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.

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 called "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.

**Example #9:** The freezing point of a 0.0925 m aqueous solution of ammonium chloride was found to be –0.325 °C. What is the actual van ’t Hoff factor for this salt at this concentration compared to the ideal one of 2? K_{f} = 1.86 °C/m

**Solution:**

Δt = i K_{f}m0.325 °C = (x) (1.86 °C/m) (0.0925 m)

x = 1.89

**Example #10:** A solution of 5.00 g of sodium chloride in 1.00 kg of water has a freezing point of −0.299 °C. What is the actual van 't Hoff factor for this salt at this concentration compared to the ideal one of 2? K_{f} of water = 1.86 °C/m

**Solution:**

1) Determine molality of the NaCl solution:

(5.00 g / 58.443 g/mol) / 1.00 kg = 0.085553m

2) Determine van 't Hoff factor:

Δt = i K_{f}m0.299 °C = (x) (1.86 °C/m) (0.085553 m)

x = 1.89

**Example #11:** A solution is prepared by dissolving 1.53 g of acetone (CH_{3}COCH_{3}) in 50.00 g of water. Its freezing point is measured to be −0.980 °C. Does acetone dissociate in solution?

**Solution:**

1) Determine moles of acetone:

1.53 g / 58.0794 g/mol = 0.02634325 mol

2) Solution path #1: assume no dissociation and calculate the expected freezing point:

Δt = i K_{f}mx = (1) (1.86 °C kg mol¯

^{1}) (0.02634325 mol / 0.0500 kg)x = 0.9799689 °C = 0.980 °C (to three sig figs)

Acetone does not dissociate in solution.

3) Solution path #2: calculate the value of the van 't Hoff factor:

Δt = i K_{f}m0.980 °C = (y) (1.86 °C kg mol¯

^{1}) (0.02634325 mol / 0.0500 kg)y = 1.00

Acetone does not dissociate in solution.

**Example #12:** An aqueous solution is 0.8402 molal in Na_{2}SO_{4}. It has a freezing point of -4.218 °C. (a) Determine the effective number of particles arising from each Na_{2}SO_{4} formula unit in this solution. (b) In comparison to the theoretical van 't Hoff factor of 3, what behavior of the sodium sulfate in solution accounts for the difference?

Δt = i K_{f}m4.218 = (x) (1.86) (0.8402)

x = 2.699 = 2.7

The Na

_{2}SO_{4}exhibits 'ion pairing.' At any given moment in the solution, it is not 100% sodium ions and sulfate ions. Some NaSO_{4}^{2}¯ (and even some Na_{2}SO_{4}) exists, forming and falling apart from instant to instant. This ion pairing reduces the number of particles in solution, thus lowering the van 't Hoff factor.

**Example #13:** The freezing point of a 5.00% CH_{3}COOH(aq) solution is -1.576 °C. (a) Determine the experimental van't Hoff factor for this solution. (b) On the basis of your understanding of intermolecular forces, account for its value.

**Solution:**

1) Let us assume the percentage is w/w (and that it is an aqueous solution) and calculate the molality:

5.00% means 5.00 g of acetic acid and 95.0 g of water.5.00 g / 60.054 g/mol = 0.0832584 mol

0.0832584 mol / 0.0950 kg = 0.8764 m

2) Calculate the van 't Hoff factor:

Δt = i K_{f}m1.576 °C = (x) (1.86 °C kg mol

^{-1}) (0.8764 mol / kg)x = 0.9668

For (b), note that the van 't Hoff factor is

lessthan one. If the dissolved CH_{3}COOH had ionized, the van 't Hoff factor would have beengreaterthan one.The explanation is that CH

_{3}COOH forms dimers (two CH_{3}COOH molecules associating into one "molecule"). This reduces the number of particles in solution, thereby reducing the van 't Hoff factor.Lots of images of acetic acid dimers can be found on the Internet. Here are some.

**Example #14:** Arrange the following aqueous solutions in order of decreasing freezing points:

0.10 m KNO_{3}

0.10 m BaCl_{2}

0.10 m C_{2}H_{4}(OH)_{2}

0.10 m Na_{3}PO_{4}

**Solution:**

1) Determine the van 't Hoff factor for each substance:

0.10 m KNO_{3}---> one K^{+}ion and one nitrate ion per formula unit, van 't Hoff factor = 20.10 m BaCl

_{2}----> one Ba^{2+}ion and two chloride ions per formula unit, van 't Hoff factor = 30.10 m C

_{2}H_{4}(OH)_{2}---> ethylene glycol does not ionize in solution, van 't Hoff factor = 10.10 m Na

_{3}PO_{4}---> three Na^{+}ions and one phosphate ion per formula unit, van 't Hoff factor = 4

2) Determine the effective molality of each solution:

KNO_{3}---> 0.10 m x 2 = 0.20 mBaCl

_{2}---> 0.10 m x 3 = 0.30 mC

_{2}H_{4}(OH)_{2}---> 0.10 m x 1 = 0.10 mNa

_{3}PO_{4}---> 0.10 m x 4 = 0.40 m

3) I took this question to mean an order in which the first substance has a freezing point closest to pure water and that the last one has the lowest freezing point, the value farthest away from 0 °C.

C_{2}H_{4}(OH)_{2}<--- smallest amount of fp depression

KNO_{3}

BaCl_{2}

Na_{3}PO_{4}<--- largest amount of fp depression

**Example #15:** A solution of 1.00 grams of anhydrous aluminum chloride, AlCl_{3}, in 50.0 grams of water freezes at −1.11 °C. Does the molar mass determined from this freezing point agree with that calculated from the formula? If there is a difference, what caused it?

**Solution:**

1) Should a van 't Hoff factor be used for this problem?

Aluminum chloride is a strong electrolyte, ionizing 100% in solution, so the answer is yes.Each formula unit of aluminum chloride ionizes into 4 ions, so se a van 't Hoff factor of 4.

2) Determine the moles of solute:

Δt = i K_{f}m1.11 °C = (4) (1.86 °C kg mol¯

^{1}) (x / 0.0500 kg)x = 0.00745968 mol

3) Determine the molar mass from freezing point data:

1.00 g / 0.00745968 mol = 134.054 g/mol

4) Determine the molar mass from atomic weights:

26.981539 + (3) (35.453) = 133.34 g/mol

5) The difference is due to ion-pairing. Let us determine the van 't Hoff factor using 133.34 g/mol. Start by determining the moles of solute:

1.00 g / 133.34 g/mol = 0.007499625 mol

6) Determine the van 't Hoff factor:

Δt = i K_{f}m1.11 °C = (i) (1.86 °C kg mol¯

^{1}) (0.007499625 mol / 0.0500 kg)i = 3.98

**Bonus Example:** A certain solvent has a freezing point of −22.465 °C. Dilute (0.050 m) solutions of four common acids are prepared in this solvent and their freezing points are measured, with these results:

Acid: HCl H _{2}SO_{4}HClO _{3}HNO _{3}Freezing Point: −22.795 °C −22.788 °C −22.791 °C −22.796 °C

(a) Determine K_{f }for this solvent and (b) advance a reason why one of the acids differs so much from the others in its power to depress the freezing point.

**Solution:**

1) Determine the K_{f} for each acid using:

Δt = i K_{f}m

2) HCl (von 't Hoff = 2)

0.333 °C = (2) (K_{f}) (0.050 m)K

_{f}= 3.3 (to two sig figs)

3) H_{2}SO_{4} (von 't Hoff = 3) <--- here is the problem leading to the answer for (b)

0.323 °C = (3) (K_{f}) (0.050 m)K

_{f}= 2.2 (to two sig figs)

4) HClO_{3} (von 't Hoff = 2)

0.326 °C = (2) (K_{f}) (0.050 m)K

_{f}= 3.3 (to two sig figs)

5) HNO_{3} (von 't Hoff = 2)

0.331 °C = (2) (K_{f}) (0.050 m)K

_{f}= 3.3 (to two sig figs)

6) The answer to (b) lies in the fact that H_{2}SO_{4} does not ionize 100% in both hydrogens. Sulfuric acid is strong in only its first hydrogen:

The ionization of the second hydrogen is weak, giving rise to sulfuric acid having a van 't Hoff factor slighter greater than 2 and not the 3 used in step #3, just above.

**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, that 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 WeightWater 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.