The Effect of Nonvolatile Solutes on Vapor Pressure


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To make a solution, you need a solvent and a solute. Usually (but not always), liquid water is the solvent and some solid substance (sugar, for example) is the solute.

Notice the word "nonvolatile" in the title. The volatility of a substance refers to the readiness with which it vaporizes. Generally speaking, substances with a boiling point below 100 °C are considered volatile and all others are called nonvolatile. Ethyl alcohol and pentane are examples of volatile substances; sugar and sodium chloride are considered nonvolatile.

The presence of a solute leads to a colligative property called the "lowering of the vapor pressure of the solution" when compared to the vapor pressure of the pure solvent. This is not a nice, tidy name like osmosis, but then again, life itself is not always nice and tidy either. There is enough of a difference between the types (volatile and nonvolatile) of solutes to merit treating them separately.

Here is the difference:

(a) solutions with a nonvolatile solute produce a vapor which is pure solvent
(b) solutions with a volatile solute produce a vapor which is a mixture of both solvent and solute.

Another way to express this difference:

(a) a nonvolatile solute does not appear as a component of the vapor above the solution.
(b) a volatile solute does appear as a component of the vapor above the solution.

You will discover, if you have not already, that the second type of solution (composed of two volatile components) is more complicated. Thus, teachers like to ask the second type on the test, especially as a calculational problem as opposed to a multiple-guess choice question.

One last point: in all discussions concerning vapor pressure, the vapor is always in contact with the solution. ALWAYS! The whole basis to the concepts presented here is that the vapor and the solution are in equiibrium with each other. To do this, they must be in contact.

Often, to a teacher, this is such as obvious point that they forget to state it. If you read this and your teacher has not yet stated the above in class, you might consider asking in class (in a very nice tone of voice) if these problems always assume that the vapor is in contact with the solution.

The goal to the above is to get the teacher to tell the whole class about this important, but forgotten, point. You don't need to tell your classmates, your goal is to (in a nice way) get the teacher to do so.


In the mid-1800s, it was discovered that the vapor pressure of a solution was lowered and that the amount was more-or-less proportional to the amount of solute. In the early 1880s, Francios Marie Raoult was able to determine the equation which governs this property:

Psolution = χsolventsolvent

The capital P stands for the vapor pressure and χ is the lower case Greek letter chi. It stands for the mole fraction, in this case of the solvent. The P° is the vapor pressure of the pure solvent, which is usually (but not always) water.

Not unsuprisingly, the above law is called Raoult's Law.

By the way, you might ask if there are any "real world" applications for the lowering of the vapor pressure of a solution when compared to the pure solvent. I'm glad you asked:

"Solutions of certain salts are in equilibrium with a particular relative humidity less than 100%. So saturated solutions of salts like sodium chloride and nitrite are typically used to calibrate RH meters."

and

"Vapor pressure osmometry for molecular weight determination. Commercial equipment gets by with a few microliters of solvent."

I copied both those statements from the USENET group sci.chem many years ago.


Example #1: If 0.340 mol of a nonvolatile nonelectrolyte are dissolved in 3.00 mol of water, what is the vapor pressure of the resulting solution? (The vapor pressure of pure water is 23.8 torr at 25.0 °C.)

Solution:

1) Calculate the mole fraction of the solvent (NOT the solute):

χsolvent = 3.00 mol / (0.340 mol + 3.00 mol)

χsolvent = 0.8982

2) Calculate the vapor pressure:

Psolution = χsolventsolvent

Psolution = (0.8982) (23.8)

Psolution = 21.4 torr


There's a key word in the above problem and it is 'nonelectrolyte.' It turns out that many substances (called electrolytes) ionize in solution. This creates more particles in solution than would be if the substance did not ionize, thus lowering the vapor pressure more than would be expected.

However, we'll do two more nonelectrolyte problems first. Sucrose is a very popular nonelectrolyte. It does not ionize in water and its van 't Hoff factor is equal to one. (The van 't Hoff factor comes into play with eletrolytes. Make sure you go learn about it if you are not aware of its existence.)

Also, notice that the word 'nonvolatile' is used below. Make sure you know what it means. If you're not sure, go back and review the discussion above.


Example #2: 210.0 g of the nonvolatile solute sucrose (C12H22O11) is added to 485.0 g of water at 25.0 °C. What will be the pressure of the water vapor over this solution? (The vapor pressure of pure water is 23.8 torr at 25.0 °C.)

Solution:

1) Determine moles of water and sucrose:

water: 485.0 g / 18.0152 g/mol = 26.92171 mol
sucrose: 210.0 g / 342.3014 g/mol = 0.61349 mol

2) Determine the mole fraction of the solvent:

χsolvent = 26.92171 mol / (26.92171 mol + 0.61349 mol) = 0.9777

3) Using Raoult's Law, determine the vapor pressure:

Psolution = χsolventsolvent

Psolution = (0.9777) (23.8)

Psolution = 23.3 torr


Example #3: The vapor pressure of water above a solution of water and a nonvolatile solute at 25.0 °C is 19.3 mm Hg. What is the mole fraction of the solute? (The vapor pressure of pure water is 23.8 torr at 25.0 °C.)

Solution:

1) Use Raoult's Law:

Psolution = χsolventsolvent

19.3 = (x) (23.8)

x = 0.811


The following problem uses the concept of the van 't Hoff factor.

Example #4: A solution is prepared by dissolving 3.423 g of aluminium sulfate in 25.00 mL water at 45.0 °C. At 45.0 °C, the vapor pressure of water is 26.47 torr and its density is 0.985 g cm-3. What is the vapor pressure of the solution?

Solution:

1) Determine moles of water and aluminium sulfate:

water:

25.00 mL x 0.985 g cm-3 = 24.625 g
24.625 g / 18.0152 g/mol = 1.3669 mol

Al2(SO4)3:

3.423 g / 342.3 g/mol = 0.0100 mol

2) However, (this is important) . . .

Al2(SO4)3 ionizes in solution to produce 5 ions for every formula unit. (Those ions are two Al3+ and three SO43-.)

Since all colligative properties depend on the number of particles in solution, the effective number of moles that Al2(SO4)3 produces is:

0.0100 mol x 5 = 0.0500 mol

3) Determine the mole fraction of the solvent:

χwater = 1.3669 mol / (1.3669 mol + 0.0500 mol)

χwater = 0.96471

4) Use Raoult's Law:

Pwater = (0.96471) (26.47 torr)

Pwater = 25.54 torr


Example #5: Calculate the vapor pressure of water above a solution at 35.0 °C that is 1.600 m fructose, C6H12O6. (The vapor pressure of pure water at 35.0 °C is 42.2 mmHg.)

Solution:

1) From the molality, we know this:

1.600 mol of fructose is dissolved in 1.000 kg of water.

2) Determine moles of water:

1000 g / 18.015 g/mol = 55.5093 mol of H2O

3) Determine mole fraction of solvent:

55.5093 / (55.5093 + 1.600) = 0.97198

4) Using Raoult's Law, the vapor pressure above the solution is:

(42.2 mmHg) (0.97198) = 41.0 mmHg

Example #6: The vapor pressure of pure methanol, CH3OH, at 30. °C is 160. torr. How many grams of the nonvolatile solute glycerol, C3H5(OH)3, must be added to 116. g of methanol to obtain a solution with a vapor pressure of 127. torr?

Solution:

1) Write an expression for the mole fraction of the solvent (methanol):

116. g divided by 32.0416 g/mol = 3.6203 mol

set mol of glycerol equal to x

χsolvent = 3.6203/(3.6203 + x)

2) Solve for 'x' with Raoult's Law:

127 torr = (χsolvent) (P°solvent)

127 = (3.6203/(3.6203 + x)) times 160. torr

Algebra!

x = 0.940708 mol

3) Convert moles to grams:

0.940708 mol is the amount of glycerol, so use its molar mass

0.940708 mol times 92.0932 g/mol = 86.6 g (to three sig figs)


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