### Boiling Point Elevation

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

Δt is the temperature change from the pure solvent's boiling point to the boiling point of the solution. It is equal to two constants times the molality of the solution. The constant Kb is 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 called the van 't Hoff factor is symbolized with the letter 'i' and is discussed below the example problems.

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 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: 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.

Solution

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

Sometimes, the boiling point is asked for. In this example, it would be 101.74 °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?

Solution

1) Determine number of moles of adrenaline in solution:

ΔT = i Kb m

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

0.49 °C = (137.5 °C mol¯1) (x)

x = 0.0035636 mol

2) 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.

Notice also, the unit used on the boiling point constant. In problems of this type, you should not use the °C/m unit.

This is a popular test question.

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

Solution:

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) (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?

Solution:

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.

Solution:

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)

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 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 #6: The freezing point depression of a 0.10 m solution of HF(aq) solution is -0.201 °C. Calculate the percent dissociation of HF(aq).

Solution

The freezing point depression equation is:

ΔT = i Kf m

The only value we do not know is i:

0.201 °C = (x) (1.86 °C m¯1) (0.10 m)

x = 1.08

The HF is about 8% dissociated.

Example #7: A 1.00 m solution of acetic acid (CH3COOH) in benzene has a freezing point depression of 2.6 K. Calculate the value for i and suggest an explanation for its value.

Solution

The freezing point depression equation is:

ΔT = i Kf m

The only value we do not know is i:

2.6 K = (x) (5.10 K m¯1 ) (1.00m)

x = 0.51

Here's an explanation: If i = 1, then one item dissolving yields one item in solution. If i = 2, then one item dissolving yields two items in solution. What about i = 0.5?

Think of it in the reverse fashion to i = 2 and one item (like a molecule) into solution yielded two items (the dissociated molecule) in solution. This is the reverse. When one item goes into solution, only one-half is present in the solution. Sort of a reverse dissociation - call it an association.

Maybe if I went to whole numbers? If two solute items go into solution, only one item is present in the solvated state.

The acetic acid dimerized in the benzene and formed (CH3COOH)2. Two molecules went into making the dimer, but only one item (the dimer) is present in solution.

Think how confused poor J.H. van't Hoff must have been. And then, when he got to the right answer, I think the joy of figuring it out must have been mixed with some relief.

Example #8: Calculate the boiling point of a 4.81% by mass solution of magnesium chloride (MgCl2, MW = 95.211 g/mol). The van 't Hoff factor of MgCl2 is 2.7, and the boiling point elevation constant, Kb, of water is 0.512 °C/molal.

Solution:

1) 4.81% (w/w) means that, in 100. g of solution, there are:

4.81 g of MgCl2
95.19 g of H2O

mole of MgCl2 ---> 4.81 g / 95.211 g/mol = 0.0505194 mol

2) Boiling point elevation equation:

Δt = i Kb m

x = (2.7) (0.512 °C kg mol¯1) (0.0505194 mol / 0.09519 kg)

Note how I modified the unit on the boiling point elevation constant.

x = 0.73 °C (to two sig figs)

3) The boiling point of the solution:

100.73 °C

Example #9: Arrange the five solutions in the order of increasing boiling point elevation and state your reason:

(a) 0.25 m sucrose(aq)
(b) 0.15 m KNO3(aq)
(c) 0.048 m C10H8 (naphthalene) in benzene
(d) 0.15 m CH3COOH(aq)
(e) 0.15 m H2SO4(aq)

Solution:

1) Examine each solution in terms of its effective molality, based on total particles dissolved.

(a) 0.25 m x 1 = 0.25 m
(b) 0.15 m x 2 = 0.30 m
(c) 0.048 m x 1 = 0.048 m
(d) 0.15 m x 1 = 0.15 m <--- a weak acid is considered to be unionized in solution
(e) 0.15 m x 3 = 0.45 m <--- see note below

The value to the right of each above calculation is the van 't Hoff factor.

2) The order, based on effective molality:

C <--- least bp elevation
D
A
B
E <--- most bp elevation

Reason: the more particles in solution, the greater the bp elevation

3) Note: I treated the H2SO4 as if both hydrogen ions ionize 100%. The reality is that only the first hydrogen ion is strong (ionized 100%) whereas the second hydrogen ion has a Ka equal to 0.012. Thus, the actual van 't Hoff factor for H2SO4 is probably some value slightly greater than 2, but less than 3. I ignored all that for the purposes of this question.

Example #10: The following information has been determined for an unknown organic compound:

(i) 4.98 g of the compound was dissolved in 100. g of benzene. The boiling temperature of the resulting solution was 82.3 °C.
(ii) A separate experiment found the compound to be 46.65% (by mass) nitrogen, 6.71% hydrogen, 26.64% oxygen and the remainder was carbon. (iii) The density of the solution is 0.8989 g/mL

(a) Determine the molecular mass of the compound.
(b) Determine the molecular formula of the compound.
(c) Determine the mole fraction of the compound in the solution.
(d) Calculate the molarity of the compound in the solution.

Solution to (a):

These values are not provided in the problem: bp of benzene = 82.3 °C and Kb for benzene = 2.53 °C/m

Δt = i Kb m

2.1 °C = (1) (2.53 °C/m) (x)

x = 0.83 m

0.83 mol/kg = y / 0.1 kg

y = 0.083 mol

molecular mass ---> 4.98 g / 0.083 mol = 60.0 g/mol

Solution to (b):

1) Assume 100. g of the compound is present. Therefore:

N ---> 46.65 g
H ---> 6.71 g
O ---> 26.64 g
C ---> 20.00 g

2) Determine moles of each:

N ---> 46.65 g / 14.007 g/mol = 3.330 mol
H ---> 6.71 g / 1.008 g/mol = 6.657 mol
O ---> 26.64 g / 16.00 g/mol = 1.665 mol
C ---> 20.00 g / 12.011 g/mol = 1.665 mol

3) Divide through by smallest:

N ---> 3.330 mol / 1.665 mol = 2
H ---> 6.657 mol / 1.665 mol = 3.998 = 4
O ---> 1.665 mol / 1.665 mol = 1
C ---> 1.665 mol / 1.665 mol = 1

4) The empirical formula is CH4N2O. Determine the molecular formula:

The "empirical formula weight" equals 60.05 g.

The molecular weight equals 60.0 g.

Since the two values are the same, we can determine the molecular formula to be CH4N2O, same as the empirical formula.

Solution to (c):

1) Determine moles present in 100. g of benzene:

100. g / 78.1118 g/mol = 1.280 mol

2) The moles of the organic compound dissolved is known to be 0.083 mol.

3) Determine the requested mole fraction:

 0.083 mol –––––––––––––––––– = 0.061 1.280 mol + 0.083 mol

Solution to (d):

1) Determine the volume of the solution:

104.98 g / 0.8989 g/mL = 116.787 mL = 0.116787 L

2) Determine the molarity:

0.083 mol / 0.116787 L = 0.7107 M

Two sig figs would give a final answer of 0.71 M

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 Percentby 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.