## Raoult's Law: Vapor Pressure and Volatile Solutes (in Ideal Solutions)

The key point to remember about solutions with two (or more) volatile components? All the components are represented in the vapor that is in contact with the solution.

This is important:

ALL the components.

In a solution with a nonvolatile solute, only the pure vapor of the solvent is present above the solution. 100% of the nonvolatile solute stays in solution, none of it enters the vapor above the solution.

By the way, at this introductory level, we will only discuss solutions with two volatile components. More complex solutions (with three or more volatile components) are discussed at a level beyond the scope of the ChemTeam's mission. Also, note the presence of the word 'ideal' in the title. We will discuss the real behavior of solutions in a different tutorial.

The key equation to use is Raoult's Law, but in a slightly expanded form from how it is first presented:

Psolution = P°A χA + P°B χB

The subscripts A and B stand for the two different volatile substances in the solution.

Example #1: A solution is composed of 1.40 mol cyclohexane (P°cy = 97.6 torr) and 2.50 mol acetone (P°ac = 229.5 torr). What is Ptotal, the total vapor pressure, above this solution?

Solution:

1) Calculate the mole fraction of each substance:

cyclohexane ⇒ 1.40 mol / (1.40 mol + 2.50 mol) = 0.358974359
acetone ⇒ 2.50 mol / (1.40 mol + 2.50 mol) = 0.641025641

2) Use Raoult's Law:

Ptotal = P°cy χcy + P°ac χac

x = (97.6 torr) (0.358974359) + (229.5) (0.641025641)

x = 35.03589744 + 147.1153846 = 182.15 torr = 182 torr (to 3 sf)

There is something very important about the above calculation for the vapor pressure of the solution. It is that we also know the vapor pressures of the two components of the vapor.

This is important because it allows us to calculate the composition (using mole fractions) of the vapor. How? Divide each component's vapor pressure by the total vapor pressure. Here's the calculation for the example:

χ for cyclohexane ⇒ 35.0359 torr / 182.15 torr = 0.19
χ for acetone ⇒ 147.1154 torr / 182.15 torr = 0.81

(By the way, we know that pressure is directly proportional to the number of moles from consideration of the Ideal Gas Law.)

A key point about the above result is this: the vapor is richer in the component with the higher vapor pressure than the solution. In our example, the acetone in the solution had χ = 0.64, but acetone's mole fraction in the vapor equals 0.81.

The component with the higher vapor pressure? There's more of it in the vapor (when using mole fractions) than in the solution.

If we were to constantly sweep away the vapor from above the solution, then more of the solution would vaporize. However, what happens then is that we would remove (from the solution) more and more of the component with the higher vapor pressure. This fact allows us the means to separate two volatile components in a solution and obtain each substance in a (relatively) pure form.

The general name of the separation process that exploits volatility differences is distillation. It is an extremely important research and industrial process.

However, please be aware that the reality is much more complex than the discussion just above. That discussion is for another time, another class.

You might be interested to know that there are some solutions that resist distillation. These solutions are called azeotropes. Here is the begining of the Wikipedia entry:

"An azeotrope is a mixture of two or more liquids (chemicals) in such a ratio that its composition cannot be changed by simple distillation. This occurs because, when an azeotrope is boiled, the resulting vapor has the same ratio of constituents as the original mixture."

Notice how the composition of the solution and the vapor are the same. This means that it is extremely difficult (although not impossible in some cases) to distill an azeotrope.

Example #2: A solution is prepared by mixing 0.0400 mol CH2Cl2 and 0.0800 mol of CH2Br2 at 25 °C. Assuming the solution is ideal, calculate the composition of the vapor (in terms of mole fractions) at 25 °C. At 25 °C, the vapor pressures of pure CH2Cl2 and pure CH2Br2 are 133 and 11.4 torr, respectively.

Solution:

1) Calculate the mole fraction of each substance:

CH2Cl2 ⇒ 0.0400 mol / (0.0400 mol + 0.0800 mol) = 0.333
CH2Br2 ⇒ 0.0800 mol / (0.0400 mol + 0.0800 mol) = 0.667

2) Use Raoult's Law:

Psolution = P°Cl χCl + P°Br χBr

Psolution = (133 torr) (0.333) + (11.4 torr) (0.667) = 44.289 torr + 7.604 torr = 51.893 torr

I won't round off here because I want to go to the mole fractions in the vapor.

3) Calculate mole fractions in vapor:

CH2Cl2 ⇒ 44.289 / 51.893 = 0.8535
CH2Br2 ⇒ 7.604 / 51.893 = 0.1465

I actually did the second mole fraction by subtraction (1 minus 0.8535).

Example #3: C10H16 and C10H18O are two of the many compounds used in perfumes and cosmetics to provide a fresh pine scent. At 69.0 °C the pure substances have vapor pressures of 100.3 torr and 9.8 torr, respectively. What is the composition of the vapor in terms of mole fractions above a solution containing equal masses of these compounds at 69.0 °C? Assume ideal behavior.

Solution:

1) Calculate the mole fraction of each substance:

Assume 50.00 g of each substance. (You can pick any mass of each substance, just as long as they are equal. Just like Henry Ford and the Model A: you can have any color of car, just as long as it was black.)

C10H16 ⇒ 50.00 g / 136.2364 g/mol = 0.36701 mol
C10H18O ⇒ 50.00 g / 154.2512 g/mol = 0.32415 mol

χC10H16 = 0.36701 mol / (0.36701 mol + 0.32415 mol) = 0.531
χC10H18O = 0.32415 mol / (0.36701 mol + 0.32415 mol) = 0.469

2) Use Raoult's Law:

Psolution = P°C10H16 χC10H16 + P°C10H18O χC10H18O

Psolution = (100.3 torr) (0.531) + (9.8 torr) (0.469) = 53.2593 torr + 4.5962 torr = 57.8555 torr

3) Calculate mole fractions in vapor:

C10H16 ⇒ 53.2593 / 57.8555 = 0.9206
C10H18O ⇒ 4.5962 / 57.8555 = 0.0794

Example #4: At a certain temperature, the pure vapor pressures of benzene (C6H6) and toluene (C7H8) are Pben = 94.6 torr and Ptol = 29.1 torr. What are the mole fractions of benzene and toluene, χben and χtol in both the liquid and vapor phases above a mixture where the total vapor pressure is Ptot = 82.0 torr?

Solution:

1) Raoult's Law for a solution of two volatiles is this:

Ptotal = P°ben χben + P°tol χtol

82.0 = (94.6) (x) + (29.1) (1 - x)

82.0 = 94.6x + 29.1 - 29.1x

52.9 = 65.5x

x = 0.8076336 <--- the mole fraction of benzene in the solution
1 - x = 0.1923664 <--- the mole fraction of toluene in the solution

2) The mole fractions in the vapor are as follows:

benzene:
(94.6) (0.8076336) = 76.4 torr <--- partial pressure of benzene in the vapor

76.4 torr / 82.0 torr = 0.9317

toluene (can be done by subtraction also)

(29.1) (0.1923664) = 5.59786 torr <--- partial pressure of toluene in the vapor

5.59786 torr / 82.0 torr = 0.06827

The fact that the two calculated mole fractions add up to 0.9997 is due to rounding errors along the way.

Example #5: Calculate the vapor pressure of a solution of 74.0 g of benzene (C6H6) in 48.8 g of toluene (C7H8) at 25.0 °C. The vapor pressure of benzene is 95.1 torr and of toluene is 28.4 torr at this temperature.

Solution:

1) Determine moles of benzene and toluene:

benzene ---> 74.0 g / 78.1134 g/mol = 0.94734 mol

toluene ---> 48.8 g / 92.1402 g/mol = 0.52963 mol

2) Determine the mole fraction of each substance:

benzene ---> 0.94734 mol / (0.94734 mol + 0.52963 mol) = 0.6414

toluene ---> 1 - 0.6414 = 0.3586

3) Raoult's Law for a solution of two volatiles is this:

Ptotal = P°ben χben + P°tol χtol

Ptotal = (95.1) (0.6414) + (28.4) (0.3586)

Ptotal = 61.0 torr + 10.2 torr = 71.2 torr

4) Although not asked for, you can proceed to the composition of the vapor:

benzene vapor mole fraction ---> 61.0 / 71.2 = 0.85674
toluene vapor mole fraction ---> 1 - 0.85674 = 0.14326

Notice I didn't pay too much attention to sig figs. Just remember to calculate with minimal rounding-off and then round off to the correct number of sig figs at the end.

Be aware that "the other" mole fraction is obtained by subtraction. Round off the first mole fraction to the proper number of sig figs, then perform the "one minus" operation.