Return to volatile solutions tutorial

**Problem #1:** At 333 K, substance A has a vapor pressure of 1.0 atm and substance B has a vapor pressure of 0.20 atm. A solution of A and B is prepared and allowed to equilibrate with its vapor. The vapor is found to have equal moles of A and B. What was the mole fraction of A in the original solution?

**Solution:**

1) We know these statements are true:

P_{A}= P°_{A}χ_{A}

and

P_{B}= P°_{B}χ_{B}

2) Equal moles of A and B in the vapor means P_{A} = P_{B}. Therefore:

P°_{A}χ_{A}= P°_{B}χ_{B}

3) We set χ_{A} = x and χ_{B} = 1 - x. Substituting, we obtain:

(1.0) (x) = (0.20) (1-x)x = 0.17

If the question had asked for the mole fraction of B, it would be 1 - 0.17 = 0.83 atm.

**Problem #2:** 30.0 mL of pentane (C_{5}H_{12}, d = 0.626 g/mL, v.p. = 511 torr) and 45.0 mL of hexane (C_{6}H_{14}, d = 0.655 g/mL, v.p. = 150. torr) are mixed at 25.0 ° C to form an ideal solution.

a) Calculate the vapor pressure of this solution.

b) Calculate the composition (in mole fractions) of the vapor in contact with this solution.

**Solution:**

1) Calculate (then add) moles of pentane and hexane:

pentane:(0.626 g/mL) (30.0 mL) = 18.78 g

18.78 g / 72.15 g/mol = 0.26029 molhexane:

(0.655 g/mL) (45.0 mL) = 29.475 g

29.475 g / 87.1766 g/mol = 0.338107 moltotal moles = 0.26029 mol + 0.338107 mol = 0.598397 mol

2) Calculate mole fractions:

pentane ⇒ 0.26029 mol / 0.598397 mol = 0.435

hexane ⇒ 0.338107 mol / 0.598397 mol = 0.565

3) Calculate total pressure (the answer to part a):

P_{total}= P°_{P}χ_{P}+ P°_{H}χ_{H}x = (511 torr) (0.435) + (150. torr) (0.565)

x = 222.285 torr + 84.75 torr

x = 307.035 torr (to three sf, it is 307 torr)

4) Calculate composition of the vapor (the answer to part b)

pentane ⇒ 222.285 torr / 307.035 torr = 0.724

hexane ⇒ 84.75 torr / 307.035 torr = 0.276

The substance with the higher vapor pressure (because of the weaker intermolecular forces) is present in the vapor to a larger mole fraction than it is present in the solution.

**Problem #3:** What is the vapor pressure (in mmHg) of a solution of 4.40 g of Br_{2} in 101.0 g of CCl_{4} at 300 K? The vapor pressure of pure bromine at 300 K is 30.5 kPa and the vapor pressure of CCl_{4} is 16.5 kPa.

**Solution:**

1) Calculate moles, then mole fraction of each substance:

bromine ⇒ 4.40 g / 159.808 g/mol = 0.027533 mol

CCl_{4}⇒ 101.0 g / 153.823 g/mol = 0.6566 molχ

_{Br2}⇒ 0.027533 mol / 0.684133 mol = 0.040245

χ_{CCl4}⇒ 0.6566 mol / 0.684133 mol = 0.959755

2) Calculate total pressure:

P_{total}= P°_{Br2}χ_{Br2}+ P°_{CCl4}χ_{CCl4}x = (30.5 kPa) (0.040245) + (16.5 kPa) (0.959755)

x = 1.2275 + 15.8360 = 17.0635 kPa

3) Convert to mmHg:

17.0635 kPa x (760.0 mmHg / 101.325 kPa) = 128 mmHg (to three sig fig)

**Problem #4:** A solution has a 1:3 ratio of cyclopentane to cyclohexane. The vapor pressures of the pure compounds at 25 °C are 331 mmHg for cyclopentane and 113 mmHg for cyclohexane. What is the mole fraction of cyclopentane in the vapor above the solution?

**Solution:**

1) mole fractions for each substance:

cyclopentane: 1/4 = 0.25

cyclohexane: 3/4 = 0.75

Note: one part cyclopentane and three parts cyclohexane means four total parts to the solution, hence four in the denominator.

2) total pressure above the solution is:

P_{total}= P°_{CP}χ_{CP}+ P°_{CH}χ_{CH}x = (331) (0.25) + (113) (0.75)

x = 82.75 + 84.75 = 167.5 mmHg

3) mole fraction of cyclopentane in the vapor:

82.75 mmHg / 167.6 mmHg = 0.494To 2 sig figs, write 0.49

**Problem #5:** Acetone and ethyl acetate are organic liquids often used as solvents. At 30.0 °C, the vapor pressure of acetone is 285 mmHg and the vapor pressure of ethyl acetate is 118 mmHg. What is the vapor pressure at 30.0 °C of a solution prepared by dissolving 25.0 g of acetone in 22.5 g of ethyl acetate?

**Solution:**

1) Determine moles of each compound in solution:

acetone: 25.0 g / 58.08 g/mol = 0.43044 mol

ethyl acetate: 22.5 g / 88.10 g/mol = 0.25539 mol

2) Determine mole fraction for each compound in solution:

acetone: 0.43044 mol / 0.68583 mol = 0.62762

ethyl acetate: 1 - 0.62762 = 0.37238

3) Determine vapor pressure of vapor above solution:

P = (0.62762) (285 mmHg) + (0.37238) (118 mmHg)P = (178.872) + (43.941) = 222.813 mmHg

P = 223 mmHg (to three sig figs)

Special bonus question: determine the composition (expressed in mole fraction) of the vapor above this solution.

**Solution:**

acetone: 178.872 / 222.813 = 0.8028

ethyl acetate: 1 - 0.8028 = 0.1972

Note how the vapor is richer than the solution in the component with the higher vapor pressure. This is the basis for fractional distillation.

**Problem #6:** A solution containing hexane and pentane has a pressure of 252.0 torr. Hexane has a pressure at 151.0 torr and pentane has a pressure of 425.0 torr. What is the mole fraction of pentane?

**Solution:**

P_{total}= P°_{H}χ_{H}+ P°_{P}χ_{P}252 = (151) (1 - x) + (425) (x)

x = 0.3686

**Problem #7:** The vapor pressure above a solution of two volatile components is 745 torr and the mole fraction of component B (χ_{B}) in the vapor is 0.59. Calculate the mole fraction of B in the liquid if the vapor pressure of pure B is 637 torr.

**Solution:**

1) Find partial pressure of B in vapor:

P_{partial, B}= P_{total, vapor}χ_{B}x = (745) (0.59)

x = 439.55 torr

2) Determine mole fraction of B in solution that gives above partial pressure:

P_{B}= P°_{B}χ_{B}439.55 = (637) (x)

x = 0.69

We could calculate the vapor pressure of pure A, if we so desired. The solution is left to the reader. The answer is 1454.5 torr.

Notice also, that the vapor is richer than the solution in A, the more volatile component. In the solution, the mole fraction of A is 0.21 and in the vapor it is 0.41.

**Problem #8:** Bromobenzene (MW: 157) steam distills at 95 °C. Its vapor pressure at 95 °C is 120 mmHg.

a. What is the vapor pressure of water at 95 °C?

b. How many grams of bromobenzene would steam distill with 20 grams of water?

**Solution:**

1) For part a, Teh Google™ yields:

633.9 mmHg

2) The Wikipedia page for steam distillation says:

"When a mixture of two practically immiscible liquids is heated while being agitated to expose the surfaces of both the liquids to the vapor phase, each constituent independently exerts its own vapor pressure as a function of temperature as if the other constituent were not present."

3) The total pressure of the vapor phase is:

633.9 + 120 = 753.9 mmHg

4) The mole fraction of the water vapor:

633.9 / 753.9 = 0.8408

5) This means:

20/18.0 = 1.11 moles of water represents 0.8408 of the vaporalso, please note that the mole fraction of the bromobenzene is:

1 minus 0.8408 = 0.1592

6) Set up a ratio and proportion:

1.11 is to 0.8408 as x is 0.1592x = 0.148728 mol

23.35 g of bromobenzene

**Problem #9:** Given that the vapor above an aqueous solution contains 18.3 mg water per liter at 25.0 °C, what is the concentration of the solute within the solution in mole percent? Please assume ideal behavior.

**Solution:**

1) We need to know the pressure exerted by the vapor:

0.0183 g / 18.015 g/mol = 1.01582 x 10^{-3}molPV = nRT

(x) (1.00 L) = (1.01582 x 10

^{-3}mol) (0.08206) (298 K)x = 2.4841 x 10

^{-2}atm

2) Let's convert the pressure to mmHg:

2.4841 x 10^{-2}atm times 760 mmHg/atm = 18.9 mmHg

3) We look up the vapor pressure for water at 25 °C:

23.8 mmHg

4) Now, we use Raoult's Law:

18.9 = (23.8) (χ_{solvent})χ

_{solvent}= 0.794χ

_{solute}= 1 minus 0.794 = 0.106

**Problem #10:** An 18.2% by mass aqueous solution of an electrolyte is prepared (molar mass = 162.2 g/mol). If the vapor pressure of the solution is 23.51 torr, into how many ions does the electrolyte dissociate? The vapor pressure of water at this temperature is 26.02 torr.

**Solution Overview:** (1) we will calculate the mole fraction of the solute in the 18.2% solution (2) we will calculate the mole fraction of the solute in the solution via Raoult's Law (3) will divide #2 by #1

1) 18.2 % means this:

18.2 g solute in 100 g of solution. Therefore, 81.8 g of water.

moles solute ⇒ 18.2 g / 162.2 g/mol = 0.112207 mol

moles water ⇒ 81.8 g / 18.015 g/mol = 4.5407 molmole fraction of solute ⇒ 0.112207 mol / 4.652907 mol = 0.0241155

We might call this the 'analytical mole fraction.' It completely ignores any dissociation that might occur, looking only at how many moles of the solid solute were dissolved into the solution.

2) mole fraction via Raoult's Law:

23.51 torr = (χ_{solvent}) (26.02 torr)χ

_{solvent}= 0.903536

χ_{solute}= 0.096464

3) divide #2 by #1:

0.096464 / 0.0241155 = 4The electrolyte dissociated into 4 ions per formula unit. An example of such a substance would be FeCl

_{3}.

The problems below will eventually go into a new file.

**Problem #11:**

1,1-Dichloroethane (CH_{3}CHCl_{2}) has a vapor pressure of 228 torr at 25.0 °C; at the same temperature, 1,1-dichlorotetrafluoroethane (CF_{3}CCl_{2}F) has a vapor pressure of 79 torr. What mass of
1,1-dichloroethane must be mixed with 240.0 g of 1,1-dichlorotetrafluoroethane to give a solution with vapor pressure 157 torr at 25 °C? Assume ideal behavior.

**Solution:**

1) State Raoult's Law:

Let Cl = 1,1-dichloroethane and F = 1,1-dichlorotetrafluoroethane

P

_{total}= P°_{Cl}χ_{Cl}+ P°_{F}χ_{F}

2) Substitute values and solve:

157 torr = (228 torr) (x) + (79 torr) (1 - x)where 'x' is the mole fraction of 1,1-dichloroethane and '1 - x' is the mole fraction of 1,1-dichlorotetrafluoroethane.

157 = 228x + 79 - 79x

149x = 78

x = 0.52349 (this is the mole fraction of 1,1-dichloroethane)

3) Set up a mole fraction equation:

0.52349 = (x / 98.9596) divided by [(x / 98.9596) + (240.0 / 170.92)]where 'x' is the mass of 1,1-dichloroethane (which is our answer).

However, I will solve the other mole fraction expression. (I did solve the above equation on paper when I formatted this answer (Nov. 16, 2011) and I did get the answer below.)

4) Use the mole fraction of 1,1-dichlorotetrafluoroethane:

0.47651 = (240.0 / 170.92) divided by [(x / 98.9596) + (240.0 / 170.92)]where 'x' is still the mass of 1,1-dichloroethane (which is our answer).

The reason? One less 'x' in the above equation.

5) Algebra!

1.40416 = 0.47651x / 98.9596 + 0.66909630.0048152x = 0.7350637

x = 152.654864 g

Rounded to four significant figures would be 152.6 g

**Problem #12:** The vapor pressure of pure benzene (C_{6}H_{6}, symbolized by B) and toluene (C_{7}H_{8}, symbolized by T) at 25.0° C are 95.1 and 28.4 torr, respectively. A solution is prepared with a mole fraction of toluene of 0.75. Determine the mole fraction of toluene in the gas phase. Assume the solution to be ideal.

**Solution:**

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

P_{total}= P°_{B}χ_{B}+ P°_{T}χ_{T}P

_{total}= (95.1) (0.25) + (28.4) (0.75)P

_{total}= 23.775 torr + 21.3 torr P_{total}= 45.075 torr

2) The mole fraction of toluene in the vapor is this:

21.3 torr / 45.075 torr = 0.472545757Rounded to three sig figs, this is 0.472.

**Problem #13:** 1-propanol (P_{1}° = 20.9 torr at 25.0 °C) and 2-propanol (P_{2}° = 45.2 torr at 25.0 °C) form ideal solutions in all proportions. Let χ_{1} and χ_{2} represent the mole fractions of 1-propanol and 2-propanol in a liquid mixture, respectively. For a solution of these liquids with χ_{1} = 0.520, calculate the composition of the vapor phase at 25.0 °C.

**Solution:**

1) Use the mole fractions of each liquid to calculate the partial pressure of that component:

vapor pressure of 1-propanol: 20.9 torr x 0.520 = 10.868 torr

vapor pressure of 2-propanol: 45.2 torr x 0.480 = 21.696 torr(the 0.480 comes from 1 - 0.520)

2) Use the partial pressures to determine the composition of the vapor:

total pressure of the vapor: 10.868 + 21.696 = 32.564 torrmole fraction 1-propanol in vapor: 10.868 / 32.564 = 0.334

mole fraction 2-propanol in vapor: 1 - 0.334 = 0.666