### Raoult's Law: The Effect of Nonvolatile Solutes on Vapor PressureProblems #1 - 10

Problem #1: What is the vapor pressure of an aqueous solution that has a solute mole fraction of 0.1000? The vapor pressure of water is 25.756 mmHg at 25 °C.

Solution:

χsolvent = 1.0000 - 0.1000 = 0.9000

Use Raoult's Law:

Psolution = (χsolvent) (P°solvent)

x = (0.900) (25.756)

x = 23.18 mmHg (to four sig figs)

Problem #2: The vapor pressure of an aqueous solution is found to be 24.90 mmHg at 25 °C. What is the mole fraction of solute in this solution? The vapor pressure of water is 25.756 mm Hg at 25 °C.

Solution:

Use Raoult's Law:

Psolution = (χsolvent) (P°solvent)

24.90 = (x) (25.756)

x = 0.966765 (this is the solvent mole fraction)

χsolute = 1 - 0.966765 = 0.033235

χsolute = 0.03324 (to four sig figs)

Problem #3: How many grams of nonvolatile compound B (molar mass= 97.80 g/mol) would need to be added to 250.0 g of water to produce a solution with a vapor pressure of 23.756 torr? The vapor pressure of water at this temperature is 42.362 torr.

Solution:

We will assume that B does not ionize in solution.

1) Determine mole fraction of solvent that produces a solution vapor pressure of 23.756 torr:

Psolution = (χsolvent) (P°solvent)

23.756 torr = (x) (42.362 torr)

x = 0.5608

2) Determine moles of compound B needed to produce the above solvent mole fraction:

0.5608 = 13.88 / (13.88 + B)

7.7839 + 0.5608B = 13.88

0.5608B = 6.0961

B = 10.87 mol

3) Determine mass of B

10.87 mol times 97.80 g/mol = 1063 g

Comment: this is a completely ridiculous amount to dissolve in 250.0 g of water, but that's not the point. The point is to solve the problem.

Problem #4: At 29.6 °C, pure water has a vapor pressure of 31.1 torr. A solution is prepared by adding 86.8 g of "Y", a nonvolatile non-electrolyte to 350. g of water. The vapor pressure of the resulting solution is 28.6 torr. Calculate the molar mass of Y.

Solution:

1) Use Raoult's Law to determine mole fraction of the solvent:

Psolution = (χsolvent) (P°solvent)

28.6 torr = (χsolvent) (31.1 torr)

χsolvent = 28.6 torr / 31.1 torr

χsolvent = 0.91961415

2) Use the mole fraction and the moles of water to determine the moles of Y:

350. g / 18.015 g/mol = 19.428254 mol

0.91961415 = 19.428254 / (19.428254 + x) <--- x equals the moles of Y dissolved

(0.91961415) (19.428254 + x) = 19.428254

17.8665 + 0.91961415x = 19.428254

0.91961415x = 1.561754

x = 1.69827 mol

3) Calculate molar mass of Y:

86.8 g / 1.69827 mol = 51.1 g/mol (to three sig figs)

Problem #5: The vapor pressure of pure water is 23.8 mmHg at 25.0 °C. What is the vapor pressure of 2.50 molal C6H12O6

Solution:

1) Covert the molality to a mole fraction. First, calculate total moles:

2.50 m C6H12O6 = 2.50 mol / 1.00 kg H2O

1000 g / 18.015 g/mol = 55.51 mol

55.51 mol + 2.50 mol = 58.01 mol

2) We need the mole fraction of water:

55.51 / 58.01 = 0.9569

3) Use Raoult's law:

Psolution = (χsolvent) (P°solvent)

Psolution = (0.9569) (23.8 mmHg) = 22.8 mmHg

Problem #6: How many grams of testosterone , C19H28O2, a nonvolatile, nonelectrolyte (MW = 288.4 g/mol), must be added to 207.8 grams of benzene to reduce the vapor pressure to 71.41 mm Hg? (Benzene = C6H6 = 78.12 g/mol. The vapor pressure of benzene is 73.03 mm Hg at 25.0 °C.)

Solution:

Use Raoult's law:

Psolution = (P°solvent) (χsolvent)

71.41 = (73.03) (2.66 / (2.66+x))

71.41 = 194.2598 / (2.66+x)

71.41 = 194.2598 / (2.66+x)

194.2598 = 189.9506 + 71.41x

71.41x = 4.3092

x = 0.0603445 mol

288.4 g/mol times 0.0603445 mol = 17.4 g (to three sig figs)

Problem #7: At 25.0 °C, the vapor pressure of benzene (C6H6) is 0.1252 atm. When 10.00 g of an unknown non-volatile substance is dissolved in 100.0 g of benzene, the vapor pressure of the solution at 25.0 °C is 0.1199 atm. Calculate the mole fraction of solute in the solution, assuming no dissociation by the solute.

Solution:

1) Because solute is non-volatile, the vapor of the solution contains only benzene. All of the unknown substance remains in solution. Assuming an ideal mixture, the vapor pressure of the solution is given by Raoult's Law:

Psolution = (P°solvent) (χsolvent)

2) The mole fraction of benzene in this mixture is:

Psolution = (P°benzene) (χbenzene)

χbenzene = Psolution / P°benzene

= 0.1199 atm / 0.1252 atm

= 0.9576677

3) The mole fractions of the components in any mixture sum up to unity. So, for this solution:

χbenzene + χsolute = 1

Hence:

χsolute = 1 - χbenzene

= 1 - 0.9576677 = 0.04233 (to four sig figs)

Problem #8: What is the vapor pressure at 25.0 °C of a solution composed of 42.71 g of naphthalene (a non-volatile compound, MW = 128 g/mol) and 40.65 g of ethanol (MW = 46.02 g/mol). (The vapor pressure of pure ethanol at 25.0 °C is 96 torr. )

Solution:

1) The vapor pressure of this kind of solvent is related to the mole fraction of the solvent and its pure vapor pressure:

vapor pressure of the solution = mole fraction of solvent times vapor pressure of the pure solvent

This is known as Raoult's Law.

2) Calculating:

moles naphthalene ---> 42.71 g / 128 g/mol = 0.334 mol

moles ethanol ---> 40.65 g / 46.02 g/mol = 0.883 mol

mole fraction ethanol ---> 0.883 / (0.883 + 0.334) = 0.726

vapor pressure of solution ---> 96 torr times 0.726 = 70. torr

Problem #9: A nonvolatile organic compound Z was used to make up a solution. Solution A contains 5.00 g of Z dissolved in 100. g of water and has a vapor pressure of 754.5 mmHg at the normal boiling point of water. Calculate the molar mass of Z.

Solution:

1) Use Raoult's Law to determine mole fraction of the solvent:

Psolution = (χsolvent) (P°solvent)

754.5 torr = (χsolvent) (760.0 torr)

Note: 760.0 torr is the vapor pressure of water at its normal boiling point, 100 °C

χsolvent = 754.5 torr / 760.0 torr

χsolvent = 0.99276316

2) Use the mole fraction and the moles of water to determine the moles of Z:

100. g / 18.015 g/mol = 5.55093 mol

0.99276316 = 5.55093 / (5.55093 + x) <--- x equals the moles of Z dissolved

x = 0.040464 mol

3) Calculate molar mass of Z:

5.00 g / 0.040464 mol = 124 g/mol (to three sig figs)

Problem #10: What is the molality of an aqueous solution of urea, CO(NH2)2, if the vapor pressure above the solution is 22.83 mmHg at 25 °C? Assume that urea is non-volatile. The vapor pressure of pure water is 23.77 mmHg at 25 °C

Solution:

Psoln = (Psolvent) (mole fraction of solvent)

22.83 = (23.77) (x)

x = 0.960 (this is the mole fraction of water)

mole fraction of urea = 0.040

Let us assume a total of 1.000 mole of solvent and solute is present.

Change 0.960 mole to grams of water:

0.960mol times 18.015 = 17.2944 g

Calculate molality:

0.040 mol / 0.0172944 kg = 2.31 m