Boiling Point Elevation

Go to boiling point elevation problems #1-10

Return to Solutions Menu

Go to the freezing point depression tutorial

A solution will boil at a higher temperature than the pure solvent. This is the colligative property called boiling point elevation.

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

ΔT = i Kb m

The temperature change from the pure to the solution is equal to two constants times the molality of the solution. The constant Kbis actually derived from several other constants and its derivation is covered in textbooks of introductory thermodynamics. Its technical name is the ebullioscopic constant. The Latin prefix ebulli- means "to bubble" or "to boil." In a more generic way, it is called the "molal boiling point elevation constant." The constant i (called the van 't Hoff factor) will be discussed below.

These are some sample ebullioscopic constants:

Substance Kb
benzene 2.53
camphor 5.95
carbon tetrachloride 5.03
ethyl ether 2.02
water 0.52

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: What is the boiling point elevation when 11.4 g of ammonia (NH3) is dissolved in 200. g of water? Kb for water is 0.52 °C/m.


1) Determine molality of 11.4 g of ammonia in 200. g of water:

11.4 g / 17.031 g/mol = 0.6693676 mol

0.6693676 mol / 0.200 kg = 3.3468 m

2) Determine bp elevation:

Δt = i Kb m

Δt = (1) (0.52 °C/m) (3.3468 m)

Δt = 1.74 °C

Example #2: 0.64 g of adrenaline in 36.0 g of CCl4 produces a bp elevation of 0.49 °C. What is adrenaline's molecular weight?


1) determine number of moles of adrenaline in solution:

ΔT = i Kb m

0.49 °C = (1) (4.95 °C / m) (x / 0.0360 kg)

x = 0.0035636 mol

From here it is a simple step: grams divided by moles equals the desired answer.

0.64 g / 0.0035636 mol = 180 g/mol (to two sig figs)

Notice that the Kb value for CCl4 was not given in the problem. In a textbook, the value would be in a chapter table or in an appendix. On ye olde Intertubes, one must unleash the Great Googlizer.

By the way, the textbook value for adrenaline's molecular weight is 183.204 g/mol

Example #3: How many grams of fructose (C6H12O6) must be dissolved in 937 g of acetic acid to raise the boiling point by 9.1 °C? The boiling point constant for acetic acid is 3.08 °C/m


1) We utilize this formula:

Δt = i Kb m

9.1 °C = (1) (3.08 °C kg mol-1) (x / 0.937 kg)

9.1 °C = (3.287 °C mol-1) (x)

x = 2.7684 mol

2) Determine grams of fructose:

2.7684 mol times 180.1548 g/mol = 499 g (to three sig figs)

Example #4: A 166.5 mg sample of the compound eugenol was dissolved in 1.00 g of chloroform (Kb = 3.63 °C/m), increasing the boiling point of chloroform by 3.68 °C. What is the molar mass of eugenol?


Δt = i Kb m

3.68 °C = (1) (3.63 °C kg mol¯1) (x / 0.00100 kg)

x = 0.00368 kg °C / 3.63 °C kg mol¯1 = 0.0010137741 mol

0.1665 g / 0.0010137741 mol = 164.2 g/mol

Note: °C / m equals °C kg mol¯1

Example #5: A 5.00 g sample of a large biomolecule was dissolved in 16.0 g of carbon tetrachloride. The boiling point of this solution was determined to be 77.85 °C. Calculate the molar mass of the biomolecule. For carbon tetrachloride, the boiling point constant is 5.03 °C/m, and the boiling point of pure carbon tetrachloride is 76.50 °C.


1) Determine moles of the compound dissolved:

Δt = i Kb m

1.35 °C = (1) (5.03 °C kg mol-1) (x / 0.0160 kg)

1.35 °C = (314.375 °C mol-1) (x)

x = 0.00429423 mol

2) Determine the molar mass:

5.00 g / 0.00429423 mol = 1160 g/mol (to three sig figs)

Go to boiling point elevation problems #1-10

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.

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.

Return to Solutions Menu