10 Examples

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:

P_{solution}= ($\text{P}{\text{}}_{\mathrm{A}}^{\mathrm{o}}$) ($\text{\chi}{\text{}}_{\mathrm{A}}^{\mathrm{}}$) + ($\text{P}{\text{}}_{\mathrm{A}}^{\mathrm{o}}$) ($\text{\chi}{\text{}}_{\mathrm{B}}^{\mathrm{}}$)

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 ($\text{P}{\text{}}_{\mathrm{cy}}^{\mathrm{o}}$ = 97.6 torr) and 2.50 mol acetone ($\text{P}{\text{}}_{\mathrm{ac}}^{\mathrm{o}}$
= 229.5 torr). What is P_{solution}, 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_{solution}= ($\text{P}{\text{}}_{\mathrm{cy}}^{\mathrm{o}}$) ($\text{\chi}{\text{}}_{\mathrm{cy}}^{\mathrm{}}$) + ($\text{P}{\text{}}_{\mathrm{ac}}^{\mathrm{o}}$) ($\text{\chi}{\text{}}_{\mathrm{ac}}^{\mathrm{}}$)P

_{solution}= (97.6 torr) (0.358974359) + (229.5 torr) (0.641025641)P

_{solution}= 35.03589744 + 147.1153846 = 182.15 torrTo three sig figs, this is 182 torr.

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 (expressed using mole fractions) of the vapor. This is done by dividing 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 (expressed in 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 (I'll use a mid-dot rather than parenthesis to show the multiplication):

P_{solution}= $\text{P}{\text{}}_{\mathrm{Cl}}^{\mathrm{o}}$·$\text{\chi}{\text{}}_{\mathrm{Cl}}^{\mathrm{}}$ + $\text{P}{\text{}}_{\mathrm{Br}}^{\mathrm{o}}$·$\text{\chi}{\text{}}_{\mathrm{Br}}^{\mathrm{}}$P

_{solution}= (133 torr) (0.333) + (11.4 torr) (0.667)P

_{solution}= 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 − 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 mol$\text{\chi}{\text{}}_{\mathrm{C10H16}}^{\mathrm{}}$ = 0.36701 mol / (0.36701 mol + 0.32415 mol) = 0.531

$\text{\chi}{\text{}}_{\mathrm{C10H18O}}^{\mathrm{}}$ = 0.32415 mol / (0.36701 mol + 0.32415 mol) = 0.469

2) Use Raoult's Law:

P_{solution}= $\text{P}{\text{}}_{\mathrm{C10H16}}^{\mathrm{o}}$ $\text{\chi}{\text{}}_{\mathrm{C10H16}}^{\mathrm{}}$ + $\text{P}{\text{}}_{\mathrm{C10H18O}}^{\mathrm{o}}$ $\text{\chi}{\text{}}_{\mathrm{C10H18O}}^{\mathrm{}}$P

_{solution}= (100.3 torr) (0.531) + (9.8 torr) (0.469)P

_{solution}= 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

**Example #4:** At a certain temperature, the pure vapor pressures of benzene (C_{6}H_{6}) and toluene (C_{7}H_{8}) are P_{ben} = 94.6 torr and P_{tol} = 29.1 torr. What are the mole fractions of benzene and toluene, $\text{\chi}{\text{}}_{\mathrm{benz}}^{\mathrm{}}$
and $\text{\chi}{\text{}}_{\mathrm{tolu}}^{\mathrm{}}$
in both the liquid and vapor phases above a mixture where the total vapor pressure is P_{solution} = 82.0 torr?

**Solution:**

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

P_{solution}= $\text{P}{\text{}}_{\mathrm{benz}}^{\mathrm{o}}$ $\text{\chi}{\text{}}_{\mathrm{benz}}^{\mathrm{}}$ + $\text{P}{\text{}}_{\mathrm{tolu}}^{\mathrm{o}}$ $\text{\chi}{\text{}}_{\mathrm{tolu}}^{\mathrm{}}$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 vapor76.4 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 vapor5.59786 torr / 82.0 torr = 0.06827

The fact that the two calculated mole fractions add up to 0.9997 is due to rounding along the way. Three sig figs gives 1.00.

**Example #5:** Calculate the vapor pressure of a solution of 74.0 g of benzene (C_{6}H_{6}) in 48.8 g of toluene (C_{7}H_{8}) 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 moltoluene ---> 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) = 0.6414toluene ---> 1 − 0.6414 = 0.3586

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

P_{solution}= $\text{P}{\text{}}_{\mathrm{benz}}^{\mathrm{o}}$ χ_{benz}+ $\text{P}{\text{}}_{\mathrm{tolu}}^{\mathrm{o}}$ χ_{tolu}P

_{solution}= (95.1) (0.6414) + (28.4) (0.3586)P

_{solution}= 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.

**Example #6:** What is the ideal vapor pressure of a solution which contains 4.401 moles of C_{6}H_{6} and 1.623 moles of CCl_{4} at 20.00 °C? The vapor pressure of pure CCl_{4} at 20.00 °C is 91.32 mmHg and C_{6}H_{6} is 74.61 mmHg.

**Solution:**

1) Mole fractions:

4.401 + 1.623 = 6.024$\text{\chi}{\text{}}_{\mathrm{C6H6}}^{\mathrm{}}$ ---> 4.401 / 6.024 = 0.7305777

$\text{\chi}{\text{}}_{\mathrm{CCl4}}^{\mathrm{}}$ ---> 1.623 / 6.024 = 0.2694223

2) Raoult's Law

P_{solution}= $\text{P}{\text{}}_{\mathrm{C6H6}}^{\mathrm{o}}$ $\text{\chi}{\text{}}_{\mathrm{C6H6}}^{\mathrm{}}$ + $\text{P}{\text{}}_{\mathrm{CCl4}}^{\mathrm{o}}$ $\text{\chi}{\text{}}_{\mathrm{CCl4}}^{\mathrm{}}$P

_{solution}= (74.61 mmHg) (0.7305777) + (91.32 mmHg) (0.2694223)P

_{solution}= 54.508402197 mmHg + 24.603644436 mmHgP

_{solution}= 79.11 mmHg (to four sig figs)

3) A reminder about a possible mistake: inadvertently switching the two vapor pressures. And, yes, the ChemTeam has made this mistake. So, double- and triple-check. It's easy to make a mistake, even if you're already experienced.

**Example #7:** Acetone and methanol form an ideal solution. At 25 °C, the vapor pressure of pure acetone is 30.784 kPa and that of methanol is 16.958 kPa. Calculate the mole fraction of methanol in a solution that boils at 25 °C and 25.130 kPa.

**Solution:**

1) The mole fractions must add up to 1:

$\text{\chi}{\text{}}_{\mathrm{acet}}^{\mathrm{}}$ + $\text{\chi}{\text{}}_{\mathrm{meth}}^{\mathrm{}}$ = 1

2) Therefore:

$\text{\chi}{\text{}}_{\mathrm{acet}}^{\mathrm{}}$ = 1 − $\text{\chi}{\text{}}_{\mathrm{meth}}^{\mathrm{}}$

3) State Raoult's Law and solve:

P_{solution}= $\text{\chi}{\text{}}_{\mathrm{acet}}^{\mathrm{}}$ $\text{P}{\text{}}_{\mathrm{acet}}^{\mathrm{o}}$ + $\text{\chi}{\text{}}_{\mathrm{meth}}^{\mathrm{}}$ $\text{P}{\text{}}_{\mathrm{meth}}^{\mathrm{o}}$25.130 kPa = (1 − $\text{\chi}{\text{}}_{\mathrm{meth}}^{\mathrm{}}$) (30.784 kPa) + ($\text{\chi}{\text{}}_{\mathrm{meth}}^{\mathrm{}}$) (16.958 kPa)

$\text{\chi}{\text{}}_{\mathrm{meth}}^{\mathrm{}}$ = 0.11843

**Example #8:** Liquids A and B form an ideal solution. The vapor pressure of pure A is 0.700 atm at the normal boiling point of a solution prepared from 0.250 mole of B and 0.650 mole of A.What is the vapor pressure of pure B at this temperature?

**Solution:**

1) Determine mole fractions:

0.650 $\text{\chi}{\text{}}_{\mathrm{A}}^{\mathrm{}}$ = ––––––––––– = 0.72222 0.250 + 0.650

0.250 $\text{\chi}{\text{}}_{\mathrm{B}}^{\mathrm{}}$ = ––––––––––– = 0.27778 0.250 + 0.650 The value for $\text{\chi}{\text{}}_{\mathrm{B}}^{\mathrm{}}$could have also been arrived at by doing 1 minus the value for $\text{\chi}{\text{}}_{\mathrm{A}}^{\mathrm{}}$.

2) State Raoult's Law:

P_{solution}= P_{A}+ P_{B}P

_{solution}= $\text{\chi}{\text{}}_{\mathrm{A}}^{\mathrm{}}$·$\text{P}{\text{}}_{\mathrm{A}}^{\mathrm{o}}$ + $\text{\chi}{\text{}}_{\mathrm{B}}^{\mathrm{}}$·$\text{P}{\text{}}_{\mathrm{B}}^{\mathrm{o}}$

3) Here is the key step:

In the problem, the term 'normal boiling point' allows us to set the vapor pressure of the solution as 1.00 atm.

4) Solve for $\text{\chi}{\text{}}_{\mathrm{B}}^{\mathrm{}}$:

1.00 atm = (0.72222) (0.700 atm) + (0.27778) ($\text{\chi}{\text{}}_{\mathrm{B}}^{\mathrm{}}$)1.00 atm = (0.505554 atm) + (0.27778) ($\text{\chi}{\text{}}_{\mathrm{B}}^{\mathrm{}}$)

(0.27778) ($\text{\chi}{\text{}}_{\mathrm{B}}^{\mathrm{}}$) = 0.494446 atm

$\text{\chi}{\text{}}_{\mathrm{B}}^{\mathrm{}}$ = 1.78 atm (to three sig figs)

**Example #9:** Heptane and octane form an ideal solution. At 40 °C, the vapor pressure of heptane is 91.5 torr and the vapor pressure of octane is 31.2 torr. A solution is made of 5.32 g heptane and 8.80 g octane. Calculate the mole fraction of octane is the vapor above the solution at 40 °C.

**Solution:**

1) Determine moles of each component of the solution:

heptane ---> 5.32 g / 100.21 g/mol = 0.0530885 mol

octane ---> 8.80 g / 114.23 g/mol = 0.0770376 mol

2) Determine mole fraction of each component of the solution:

0.0530885 $\text{\chi}{\text{}}_{\mathrm{hept}}^{\mathrm{}}$ = ––––––––– = 0.4080 0.1301261

0.0770376 $\text{\chi}{\text{}}_{\mathrm{oct}}^{\mathrm{}}$ = ––––––––– = 0.5920 0.1301261 The value for $\text{\chi}{\text{}}_{\mathrm{oct}}^{\mathrm{}}$could have also been arrived at by doing 1 minus the value for $\text{\chi}{\text{}}_{\mathrm{hept}}^{\mathrm{}}$.

3) Raoult's Law for two volatile components and solve for P_{solution}:

P

_{solution}= $\text{\chi}{\text{}}_{\mathrm{hept}}^{\mathrm{}}$·$\text{P}{\text{}}_{\mathrm{hept}}^{\mathrm{o}}$ + $\text{\chi}{\text{}}_{\mathrm{oct}}^{\mathrm{}}$·$\text{P}{\text{}}_{\mathrm{oct}}^{\mathrm{o}}$P

_{solution}= (91.5 torr) (0.4080) + (31.2 torr) (0.5920)P

_{solution}= 37.332 + 18.4704 = 55.8024 torr

4) Raoult's Law for one component:

P_{oct in vapor}= $\text{\chi}{\text{}}_{\mathrm{oct\; in\; vapor}}^{\mathrm{}}$·$\text{P}{\text{}}_{\mathrm{total\; vpor}}^{\mathrm{o}}$18.4704 torr = ($\text{\chi}{\text{}}_{\mathrm{oct}}^{\mathrm{}}$) (55.8024 torr)

5) $\text{\chi}{\text{}}_{\mathrm{oct\; in\; vapor}}^{\mathrm{}}$ = 0.331

**Example #10:** Two liquids, A and B, form an ideal solution. The solution boils at 22 °C and 0.255 atm. The vapor pressure of pure A is 0.192 atm at 22 °C and the vapor pressure of B is 0.311 atm, also at 22 °C. Determine the mole fraction of each solution component.

**Solution:**

P

_{solution}= $\text{\chi}{\text{}}_{\mathrm{A}}^{\mathrm{}}$·$\text{P}{\text{}}_{\mathrm{A}}^{\mathrm{o}}$ + $\text{\chi}{\text{}}_{\mathrm{B}}^{\mathrm{}}$·$\text{P}{\text{}}_{\mathrm{B}}^{\mathrm{o}}$0.255 atm = ($\text{\chi}{\text{}}_{\mathrm{A}}^{\mathrm{}}$) (0.192 atm) + (1 − $\text{\chi}{\text{}}_{\mathrm{A}}^{\mathrm{}}$) (0.311 atm)

0.255 = 0.192$\text{\chi}{\text{}}_{\mathrm{A}}^{\mathrm{}}$ + 0.311 − 0.311$\text{\chi}{\text{}}_{\mathrm{A}}^{\mathrm{}}$

0.255 − 0.311 = 0.192$\text{\chi}{\text{}}_{\mathrm{A}}^{\mathrm{}}$− 0.311$\text{\chi}{\text{}}_{\mathrm{A}}^{\mathrm{}}$

0.056 = 0.119$\text{\chi}{\text{}}_{\mathrm{A}}^{\mathrm{}}$

$\text{\chi}{\text{}}_{\mathrm{A}}^{\mathrm{}}$ = 0.056 / 0.119 = 0.47

$\text{\chi}{\text{}}_{\mathrm{B}}^{\mathrm{}}$ = 1 − 0.47 = 0.53

**Bonus Example:** 20.772 g of a volatile solute is dissolved in 495.0 g of water. This solute does not react with water nor dissociate in solution. The pure solute displays, at 60 °C, a vapor pressure of 14.94 torr. Assume an ideal solution.

At 60 °C the vapor pressure of this solution is determined to be 147.34 torr. Calculate the molar mass of this volatile solute. (The vapor pressure of water at 60 °C is 149.40 torr.)

**Solution:**

1) Let the mole fractions be x (the volatile solute) and 1 − x (the water). Using Raoult's Law:

147.34 = (x) (14.94) + (1 − x) (149.40)147.34 = 14.94x + 149.40 − 149.40x

134.46x = 2.06

x = 0.01532 (this is mole fraction of the volatile solute)

1 − x = 0.98468 (this is the mole fraction of water)

2) Use a ratio and proportion to determine the moles of solute present:

moles of water = 495 g / 18.015 g/mol = 27.4771 mol27.4771 is to 0.98468 as y is to 0.01532

y = 0.42749845 mol of the unknown is present

3) Determine the molar mass:

20.772 g / 0.42749845 mol = 48.59 g/mol (to four sig figs)

4) You can double-check that the ratio and proportion is correct. Here is the mole fraction calculation for water:

27.4771 / (27.4771 + 0.42749845) = 0.98468