### Raoult's Law: The Effect of Nonvolatile Solutes on Vapor PressureProblems #1 - 15

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:

χsolv = 1.0000 − 0.1000 = 0.9000 <--- solv means solvent

Use Raoult's Law:

Psolution = (χsolv) ($\text{P}{\text{}}_{solv}^{o}$)

x = (0.9000) (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 = (χsolv) ($\text{P}{\text{}}_{solv}^{o}$)

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 = (χsolv) ($\text{P}{\text{}}_{solv}^{o}$)

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) (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 = (χsolv) ($\text{P}{\text{}}_{solv}^{o}$)

28.6 torr = (χsolv) (31.1 torr)

χsolv = 28.6 torr / 31.1 torr

χsolv = 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 = (χsolv) ($\text{P}{\text{}}_{solv}^{o}$)

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 = ($\text{P}{\text{}}_{solv}^{o}$) (χsolv)

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) (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 = ($\text{P}{\text{}}_{solv}^{o}$) (χsolv)

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

Psolution = ($\text{P}{\text{}}_{benz}^{o}$) (χbenz)

χbenz = Psolution / $\text{P}{\text{}}_{benz}^{o}$

= 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:

χbenz + χsolu = 1 <--- the 'solu' means solute

Hence:

χsolu = 1 − χbenz

= 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) (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) (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 = (χsolv) ($\text{P}{\text{}}_{solv}^{o}$)

754.5 torr = (χsolv) (760.0 torr)

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

χsolv = 754.5 torr / 760.0 torr

χsolv = 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:

Psolution = ($\text{P}{\text{}}_{solv}^{o}$) (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:

(0.960 mol) (18.015 g/mol) = 17.2944 g

Calculate molality:

0.040 mol / 0.0172944 kg = 2.31 m

Problem #11: Calculate the mass of propylene glycol (C3H8O2) that must be added to 500. grams of water to reduce the vapor pressure by 4.75 mmHg at 40.0 °C.

Solution:

Look up the vapor pressure of water at 40.0 °C. It is 55.3 mmHg.

Use Raoult's Law to get the mole fraction of the solvent:

50.55 mmHg = (55.3 mmHg) (x)

x = 0.9141

The 50.55 comes from 55.3 minus 4.75.

The mole fraction of the solute is 0.0859

Let's set up a mole fraction calculation:

(x / 76.0942) divided by [(x / 76.0942) + (500. / 18.0152)] = 0.0859

The x / 76.0942 is the mole of C3H8O2, the 500. / 18.0152 is the mol of water and the sum of the two (inside the square brackets) is the total mol in the solution.

Solve for x.

Problem #12: What is the vapor pressure of water above a solution in which 32.5 g of glycerin (C3H8O3) are dissolved in 125. g of water at 343 K? The vapor pressure of pure water at 343 K is 233.7 torr.

Solution:

The vapor pressure is proportional to the mole fraction in the solution.

moles glycerin = 32.5 g / 92.19 g/mol = 0.3525 moles
moles water = 125 g / 18.0 g/mol = 6.944

total moles = 7.2965 mol
mole fraction of water = 6.944 mol / 7.2965 mol = 0.9516

(0.9516) (233.7 torr) = 222.4 torr

Problem #13: A solution is prepared by dissolving 396 g of sucrose in 624 g of water at 30.0 °C. What is the vapor pressure of this solution? (The vapor pressure of water is 31.82 mmHg at 30.0 °C.)

Solution:

1) Determine moles of solute and solvent:

396 g / 342.2948 g/mol = 1.1569 mol
624 g / 18.015 g/mol = 34.6378 mol

2) Determine mole fraction of the solvent:

34.6378 mol / (34.6378 mol + 1.1569 mol) = 0.96768

3) Determine vapor pressure:

x = (31.82 mmHg) (0.96768) = 30.8 mmHg (to three sig figs)

Problem #14: Calculate the vapor pressure of a solution made by dissolving 21.80 g of glucose (molar mass = 180.155 g/mol) in 460.0 g of H2O at 30.0 °C. (The vapor pressure of the pure solvent is 31.82 mmHg at 30.0 °C.)

Solution:

1) Determine moles of solute and solvent:

21.80 g / 180.155 g/mol = 0.1210 mol
460.0 g / 18.015 g/mol = 25.5343 mol

2) Determine mole fraction of the solvent:

25.5343 mol / (25.5343 mol + 0.1210 mol) = 0.9952836

3) Determine vapor pressure:

x = (31.82 mmHg) (0.9952836) = 31.67 mmHg (to four sig figs)

Problem #15: A solution of sodium chloride in water has a vapor pressure of 18.5 torr at 25 °C. What would be the vapor pressure of this solution at 45 °C? The vapor pressure of pure water is 23.8 torr at 25 °C and 71.9 torr at 45 °C

Solution:

Raoult's law states that the vapor pressure of a solvent above a solution is equal to the vapor pressure of the pure solvent at the same temperature multiplied by the mole fraction of the solvent present:

vapor pressure solution = (mole fraction of solvent) (vapor pressure solvent)

18.5 torr = (mole fraction) (23.8 torr)

mole fraction solvent = 0.77731

At 45 °C,

vapor pressure solution = (0.77731)(71.9 torr) = 55.9 torr (to three SF)

Bonus Problem: At 300 °C, the vapor pressure of Hg is 32.97 torr. What mass of Au would have to be dissolved in 5.000 g of Hg to lower its vapor pressure to 25.00 torr?

Solution #1:

Au dissolves in Hg to form a solution. At 300 °C, gold is still a solid and has a vapor pressure of Au is much lower than that of Hg (which is liquid at 300 °C. Hence, we will treat the solute, Au, as non-volatile.

Psolution = $\text{P}{\text{}}_{Hg}^{o}$· $\text{χ}{\text{}}_{Hg}^{}$ <--- mole fraction calculated using vapor pressures

25.00 torr = (32.97 torr) ($\text{χ}{\text{}}_{Hg}^{}$)

$\text{χ}{\text{}}_{Hg}^{}$ = 0.758265

moles of Hg present ---> 5.00 g / 200.59 g/mol = 0.0249265 mol

Let M = moles of Au that dissolved.

 mol Hg $\text{χ}{\text{}}_{Hg}^{}$ = –––––––––– <--- mole fraction calculated using moles mol Hg + M

 0.0249265 0.758265 = ––––––––––––– 0.0249265 + M

(0.758265) (0.0249265 + M) = 0.0249265

0.0189009 + 0.758265M = 0.0249265

0.758265M = 0.0060256

M = 0.00794656 mol of Au

(196.96657 g/mol) (0.00794656 mol) = 1.565 g (to 4 sig figs)

Solution #2:

This method sets the two calculation methods of mole fraction equal to each other.

$\text{χ}{\text{}}_{Hg, using vp}^{}$ = $\text{χ}{\text{}}_{Hg, using mol}^{}$

 Psolution mol Hg ––––––– = –––––––––– $\text{P}{\text{}}_{Hg}^{o}$ mol Hg + M

Let m = mass of Au

25.00
 5.00 ––––– 200.6
–––––  =  ––––––––––––––
32.97
 5.00 m ––––– + ––––– 200.6 197.0

Note: molar masses have been rounded off and all units have been dropped.

0.0249265
0.758265  =  –––––––––––––––––

 m 0.0249265 + ––––– 197.0

 0.758265m 0.01890089 + ––––––––– = 0.0249265 197.0

 0.758265m ––––––––– = 0.00602561 197.0

0.758265m = 1.18704517

m = 1.565 g (to four sig figs)