Raoult's Law: The Effect of Nonvolatile Solutes on Vapor Pressure
Problems #11 - 25 (Incomplete)

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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: The vapor pressure of carbon tetrachloride (CCl4) at 50.0 °C is 0.437 atm. When 7.42 g of a pure nonvolatile substance is dissolved in 100.0 g of carbon tetrachloride, the vapor pressure of the solution is 0.411 atm. Calculate the molar mass of the solute.

Solution:

0.411 atm = (0.437 atm) (x)

x = 0.9405 (this is the mole fraction of the CCl4)

100.0 g / 153.823 g/mol = 0.6501 mol

0.9405 = 0.6501 / (0.6501 + x)

0.61142 + 0.9405x = 0.6501

x = 0.041127 mol (this is the moles of the unknown substance that was dissolved in 100.0 g of CCl4)

7.42 g / 0.041127 mol = 180. g/mol


Problem #16: At 27.0 °C, the vapor pressure of pure water is 23.76 mmHg and that of an aqueous solution of urea is 22.97 mmHg. Calculate the molality of urea in this solution.

Solution:

Psolution = Psolvent times mole fraction of the solvent

22.97 = (23.76) (x)

x = 0.96675

Let's dissolve some urea in 1.00 mole of water. The mole fraction of the water is:

0.96675 = 1 / (1+x)

where x is the number of moles of urea and 1+x is the total moles in the solution.

x = 0.0344 mole of urea

0.0344 mol / 0.0180 kg = 1.91 molal

Notice that 1 mole of water equals 18.0 g = 0.0180 kg


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