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.

Got that?

__ALL__ the components.

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:

P_{solution}= P°_{A}χ_{A}+ P°_{B}χ_{B}

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 P_{total}, 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:

P_{total}= P°_{cy}χ_{cy}+ P°_{ac}χ_{ac}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 CH_{2}Cl_{2} and 0.0800 mol of CH_{2}Br_{2} 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 CH_{2}Cl_{2} and pure CH_{2}Br_{2} are 133 and 11.4 torr, respectively.

**Solution:**

1) Calculate the mole fraction of each substance:

CH_{2}Cl_{2}⇒ 0.0400 mol / (0.0400 mol + 0.0800 mol) = 0.333

CH_{2}Br_{2}⇒ 0.0800 mol / (0.0400 mol + 0.0800 mol) = 0.667

2) Use Raoult's Law:

P_{solution}= P°_{Cl}χ_{Cl}+ P°_{Br}χ_{Br}P

_{solution}= (133 torr) (0.333) + (11.4 torr) (0.667) = 44.289 torr + 7.604 torr = 51.893 torrI won't round off here because I want to go to the mole fractions in the vapor.

3) Calculate mole fractions in vapor:

CH_{2}Cl_{2}⇒ 44.289 / 51.893 = 0.8535

CH_{2}Br_{2}⇒ 7.604 / 51.893 = 0.1465I actually did the second mole fraction by subtraction (1 minus 0.8535).

**Example #3:** C_{10}H_{16} and C_{10}H_{18}O 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.)C

_{10}H_{16}⇒ 50.00 g / 136.2364 g/mol = 0.36701 mol

C_{10}H_{18}O ⇒ 50.00 g / 154.2512 g/mol = 0.32415 molχ

_{C10H16}= 0.36701 mol / (0.36701 mol + 0.32415 mol) = 0.531

χ_{C10H18O}= 0.32415 mol / (0.36701 mol + 0.32415 mol) = 0.469

2) Use Raoult's Law:

P_{solution}= P°_{C10H16}χ_{C10H16}+ P°_{C10H18O}χ_{C10H18O}P

_{solution}= (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:

C_{10}H_{16}⇒ 53.2593 / 57.8555 = 0.9206

C_{10}H_{18}O ⇒ 4.5962 / 57.8555 = 0.0794

Go to more problems about solutions of two volatile substances