Vapor Pressure and Volatile Solutes: Ideal Solution
Problems #1 - 15


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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:

PA = P Ao· χ A

and

PB = P Bo· χ B

2) Equal moles of A and B in the vapor means PA = PB. Therefore:

P Ao· χ A = P Bo· 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

4) 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 (C5H12, d = 0.626 g/mL, v.p. = 511 torr) and 45.0 mL of hexane (C6H14, 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 mol

hexane:

(0.655 g/mL) (45.0 mL) = 29.475 g
29.475 g / 87.1766 g/mol = 0.338107 mol

total 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):

Psolution = P pento· χ pent + P hexo· χ hex

x = (511 torr) (0.435) + (150. torr) (0.565)

x = 222.285 torr + 84.75 torr

x = 307 torr (to three sig figs)

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 Br2 in 101.0 g of CCl4 at 300 K? The vapor pressure of pure bromine at 300 K is 30.5 kPa and the vapor pressure of CCl4 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
CCl4 ⇒ 101.0 g / 153.823 g/mol = 0.6566 mol

χBr2 ⇒ 0.027533 mol / 0.684133 mol = 0.040245
χCCl4 ⇒ 0.6566 mol / 0.684133 mol = 0.959755

2) Calculate total pressure:

Psolution = P Br2o· χ Br2 + P CCl4o· χ CCl4

Psolution = (30.5 kPa) (0.040245) + (16.5 kPa) (0.959755)

Psolution = 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:

Psolution = P pento· χ pent + P hexo· χ hex

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 = 0.494

To 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:

Psolution = (0.62762) (285 mmHg) + (0.37238) (118 mmHg)

Psolution = (178.872) + (43.941) = 222.813 mmHg

Psolution = 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:

Psolution = P hexo· χ hex + P pento· χ pent

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 (χB) 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:

PB = P vaporo· χ B

x = (745 torr) (0.59)

x = 439.55 torr

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

PB = P Bo· χ B

439.55 torr = (637 torr) (y)

y = 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: 157.02) 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 20.0 grams of water?

Solution:

1) For (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:

20.0 g/18.015 g/mol = 1.11018 moles of water represents 0.8408 of the vapor

also, the mole fraction of the bromobenzene is:

1 − 0.8408 = 0.1592

6) Set up a ratio and proportion:

1.11018   x
–––––––  =  –––––––
0.8408   0.1592

x = 0.210205 mol

(157.02 g/mol) (0.210205 mol) = 33.0 g


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 mol

PV = 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) (760. mmHg/atm) = 18.9 mmHg

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

23.8 mmHg

4) Now, we use Raoult's Law:

18.9 = (23.8) (χsolvent)

χsolvent = 0.794

χsolute = 1 − 0.794 = 0.106


Problem #10:

1,1-Dichloroethane (CH3CHCl2) has a vapor pressure of 228 torr at 25.0 °C; at the same temperature, 1,1-dichlorotetrafluoroethane (CF3CCl2F) 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

Psolution = P Clo· χ Cl + P Fo· 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 + 0.6690963

0.0048152x = 0.7350637

x = 152.654864 g

Rounded to four significant figures would be 152.6 g


Problem #11: The vapor pressure of pure benzene (C6H6, symbolized by B) and toluene (C7H8, 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:

Psolution = P Bo· χ B + P To· χ T

Psolution = (95.1) (0.25) + (28.4) (0.75)

Psolution = 23.775 torr + 21.3 torr

Psolution = 45.075 torr

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

21.3 torr / 45.075 torr = 0.472545757

Rounded to three sig figs, the answer is 0.472.


Problem #12: 1-propanol (P 1o = 20.9 torr at 25.0 °C) and 2-propanol (P 2o = 45.2 torr at 25.0 °C) form ideal solutions in all proportions. Let χ1 and χ2 represent the mole fractions of 1-propanol and 2-propanol in a liquid mixture, respectively. For a solution of these liquids with χ1 = 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) (0.520) = 10.868 torr
vapor pressure of 2-propanol: (45.2 torr) (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 torr

mole fraction 1-propanol in vapor: 10.868 / 32.564 = 0.334
mole fraction 2-propanol in vapor: 1 − 0.334 = 0.666


Problem #13: Butanone (CH3CH2COCH3) has a vapor pressure of 100. torr at 25 °C. At the same temperature, propanone (CH3COCH3) has a vapor pressure of 222 torr. What mass of propanone must be mixed with 190. g of butanone to give a solution with a vapor pressure of 135 torr at 25 °C? Assume ideal behavior.

Solution:

1) Moles of each component:

butanone ---> 190. g/ 72.107 g/mol = 2.634973 mol
let z = moles of propanone

2) Mole fraction of each component:

butanone ---> 2.63 / (2.63 + z)
propanone ---> z / (2.63 + z)

3) Insert values into Raoult's Law for two volatile components:

   2.63   z
135 = 100. x ––––––– + 222 x ––––––
   2.63 + z   2.63 + z

You might see it formatted like this:

135 = [(100.) (2.63 / 2.63 + z)] + [(222) (z / (2.63 + z)]

4) Solve for z:

 263 222z
135 = ––––––– + ––––––
 2.63 + z 2.63 + z

 263 + 222z
135 = –––––––––
 2.63 + z

355.05 + 135z = 263 + 222z

92.05 = 87z

z = 1.058 mol of propanone

5) Determine mass of propanone required:

(1.058 mol) (58.0794 g/mol = 61.4 g (to three sig figs)

6) We can check the answer:

Mole fraction of each component:

butanone ---> 2.63 / (2.63 + 1.058) = 0.713124
propanone ---> 1.058 / (2.63 + 1.058) = 0.286876

Insert into Raoult's Law:

Psolution = (100.) (0.713124) + (222) (0.286876)

Psolution = 135 torr

Yay!


Problem #14: At −100. °C, ethane and propane are liquids. At this temperature, the vapor pressure of pure ethane is 394 torr and that of pure propane is 22 torr. What is the vapor pressure at −100. °C over a solution containing equal molar amounts of these substances?

Solution:

If they have equal molar amounts, then each will have a partial pressure of exactly one-half of its pure vapor pressure:

197 torr for ethane, and 11 torr for propane.

So the total pressure is 208 torr.


Problem #15: At 60 °C, compound A has a vapor pressure of 96 mmHg. Benzene has a vapor pressure of 395 mmHg at 60 °C. A 50:50 mixture by mass of benzene and A has a vapor pressure of 281 mmHg. What is the molar mass of A?

Solution:

1) Start with Raoult's law:

Ptot = PA + PB

where

Ptot = total pressure above the solution

PA = vapor pressure of component A above the solution

PB = vapor pressure of component B above the solution (the benzene, in our case)

2) Raoult's Law can be stated in an expanded form:

Ptot = P Ao· χ A + P Bo· χ B

where

P Ao = vapor pressure above pure A   χ A = mole fraction A in the solution
P Bo = vapor pressure above pure B   χ B = mole fraction B in the solution

3) More useful information (mw = molecular weight):

χ A = mol A / (mol A + mol B)   mol A = mass A / mwA
χ B = mol B / (mol A + mol B)   mol B = mass B / mwB

4) Putting all that together along with the rest of the data (do not include units):

   
mass A
–––––––
mwA
   
mass B
–––––––
mwB
281 = 96 x ––––––––––––––––––– + 395 x –––––––––––––––––––
   
mass A mass B
––––––– + –––––––
mwA mwB
   
mass A mass B
––––––– + –––––––
mwA mwB

Here's another way to format the above:

281 = 96 x (mass A / mwA) / ((mass A / mwA) + (mass B / mw B)) + 395 x (mass B / mwB) / ((mass A / mw A) + (mass B / mw B))

5) From the problem statement, we know that mass A = mass B. Therefore, replace all mass B:

   
mass A
–––––––
mwA
   
mass A
–––––––
mwB
281 = 96 x ––––––––––––––––––– + 395 x –––––––––––––––––––
   
mass A mass A
––––––– + –––––––
mwA mwB
   
mass A mass A
––––––– + –––––––
mwA mwB

Here's another way to format the above:

281 = 96 x (mass A / mwA) / ((mass A / mwA) + (mass A / mw B)) + 395 x (mass A / mwB) / ((mass A / mw A) + (mass A / mw B))

6) Each term has mass A in the numerator and denominator, so we can divide out all mass A:

   
1
–––––––
mwA
   
1
–––––––
mwB
281 = 96 x ––––––––––––––––––– + 395 x –––––––––––––––––––
   
1 1
––––––– + –––––––
mwA mwB
   
1 1
––––––– + –––––––
mwA mwB

Here's another way to format the above:

281 = 96 x (1/mwA) / ((1/mwA) + (1/mw B)) + 395 x (1/mwB) / ((1/mw A) + (1/mw B))

7) We know mwB to be 78.1134 g/mol:

   
1
–––––––
mwA
   
1
–––––––
78.1134
281 = 96 x ––––––––––––––––––– + 395 x –––––––––––––––––––
   
1 1
––––––– + –––––––
mwA 78.1134
   
1 1
––––––– + –––––––
mwA 78.1134

Here's another way to format the above:

281 = 96 x (1/mwA) / ((1/mw A) + (1/78.1134)) + 395 x (1/78.1134) / ((1/mw A) + (1/78.1134))

and now we have 1 equation and 1 unknown.

8) Move the denominator to the other side:

281 x [(1/mwA) + (1/78.11)] = 96 x (1/mwA) + 395 x (1/78.11)

9) Multiply through by mwA

281 x [(1) + (mwA/78.11)] = 96 x (1) + 395 x (mwA / 78.11)

10) Simplify:

281 + 3.598 mwA = 96 + 5.057 mwA

11) Which will result in:

mwA = 185 / 1.459 = 127 g/mole

12) Here's another way. Return to the equation in step 7 and do the two multiplications on the right-hand side:

 
96
–––––––
mwA
 
395
–––––––
78.1134
281 = ––––––––––––––––––– + –––––––––––––––––––
 
1 1
––––––– + –––––––
mwA 78.1134
 
1 1
––––––– + –––––––
mwA 78.1134

13) Move the right-hand side denominator to the other side:

281
–––––––
mwA
+
281 96 395
––––––– = ––––– + –––––––
78.1134 mwA 78.1134

I also distributed the 281 over the two terms that moved to the left-hand side.

14) Gather like terms:

185
–––––––
mwA
=
114
–––––––
78.1134

15) Cross-multiply and divide:

114 mwA = (185) (78.1134)

mwA = 127 g/mol


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