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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?
1) We know these statements are true:
PA = P°A χA
PB = P°B χB
2) Equal moles of A and B in the vapor means PA = PB. Therefore:
P°A χA = P°B χB
3) We set χA = x and χB = 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 (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.
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):
Ptotal = P°P χP + P°H χH
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 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.
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:
Ptotal = P°Br2 χBr2 + P°CCl4 χCCl4
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?
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:
Ptotal = P°CP χCP + P°CH χCH
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?
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.
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?
Ptotal = P°H χH + P°P χP
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.
1) Find partial pressure of B in vapor:
Ppartial, B = Ptotal, vapor χB
x = (745) (0.59)
x = 439.55 torr
2) Determine mole fraction of B in solution that gives above partial pressure:
PB = P°B χB
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?
1) For part a, Teh Google yields:
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 vapor
also, 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 0.1592
x = 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.
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 times 760 mmHg/atm = 18.9 mmHg
3) We look up the vapor pressure for water at 25 °C:
4) Now, we use Raoult's Law:
18.9 = (23.8) (χsolvent)
χsolvent = 0.794
χsolute = 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 mol
mole 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 = (χsolvent) (26.02 torr)
χsolvent = 0.903536
χsolute = 0.096464
3) divide #2 by #1:
0.096464 / 0.0241155 = 4
The electrolyte dissociated into 4 ions per formula unit. An example of such a substance would be FeCl3.
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.
1) State Raoult's Law:
Let Cl = 1,1-dichloroethane and F = 1,1-dichlorotetrafluoroethane
Ptotal = P°Cl χCl + P°F χF
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.
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
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