Raoult's Law: 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 = (χsolvent) (P solvento )

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):

    3.00 mol  
χ solvent  =  –––––––––––––––––  = 0.8982
    0.340 mol + 3.00 mol  

2) Calculate the vapor pressure:

Psolution = (χ solvent) (P solvento )

Psolution = (0.8982) (23.8 torr)

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:

χ solvent = 26.92171 mol / (26.92171 mol + 0.61349 mol) = 0.9777

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

Psolution = (χ solvent) (P solvento )

Psolution = (0.9777) (23.8 torr)

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 = (χ solvent) (P solvento )

19.3 = (χ solvent) (23.8 torr)

χ solvent = 0.811


Example #4: 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 #5: 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 / 32.0416 g/mol = 3.6203 mol

set mol of glycerol equal to x

χ solvent = 3.6203 / (3.6203 + x)

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

127 torr = (χ solvent) (P solvento )

127 = (3.6203 / (3.6203 + x)) (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) (92.0932 g/mol) = 86.6 g (to three sig figs)


Example #6: A solution containing 8.30 g of a nonvolatile, nondissociating substance dissolved in 1.00 mol of chloroform, CHCl3, has a vapor pressure of 511 torr. The vapor pressure of pure CHCl3 at the same temperature is 526 torr. Calculate:

(a) the mole fraction of the solute
(b) the number of moles of solute in the solution
(c) the molecular mass of the solute

Solution to (a):

Psolution = (χ solvent) (P solvento )

511 torr = (χ solvent) (526 torr)

χ solvent ---> 511 / 526 = 0.9715 <--- keep one extra digit

χ solute ---> 1 − 0.9715 = 0.0285

Solution #1 to (b):

The 1.00 mole of chloroform represents a mole fraction of 0.9715.

1.00 mol is to 0.9715 as x is to 0.0285

x = 0.029336 mol <--- keep two extra digits

Solution #2 to (b):

1.00 / (1.00 + x) = 0.9715

0.9715 + 0.9715x = 1

0.9715x = 0.0285

x = 0.029336 mol

Solution to (c):

8.30 g / 0.029336 mol = 283 g/mol (to three sig figs)

Example #7: 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:

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 could 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:

Psolution = P solvento· χ solvent

23.51 torr = (26.02 torr) (χ solvent)

χ solvent = 0.903536

χ solute = 1 − 0.903536 = 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.


Example #8: What is the vapor pressure at 20.0 °C above a solution that has 85.5 g of sucrose and 75 g of urea dissolved per kg of water? (Sucrose and urea do not ionize in solution.)

Solution:

1) The vapor pressure of a solution is equal to the vapor pressure of the pure solvent times the mole fraction of the solvent in the solution. So, for this problem, first calculate moles of sucrose, moles of urea and moles of water:

moles sucrose ---> 85.5 g / 342.3 g/mol = 0.250 mol

moles urea ---> 75 g / 60.06 g/mol = 1.25 mol

moles water = 1000 g / 18.015 g/mol = 55.509 mol

2) Now, calculate the mole fraction of water:

χ water = 55.509 / (55.509 + 0.25 + 1.25) = 0.97369

3) Finally, look up the vapor pressure of water at 20.0 °C, and multiply that by the mole fraction of water that was just calculated.

Psolution = P solvento· χ solvent

Psolution = (17.5 torr) (0.97369)

Psolution = 17.0 torr (to three sig figs)


Example #9: 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


Example #10: 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


Bonus Example: How many grams of sucrose (C12H22O11) must be added to 527 g of water to give a solution with a vapor pressure 0.911 mmHg less than that of pure water at 20 °C? (The vapor pressure of water at 20 °C is 17.5 mmHg.)

Solution:

1) In the solution formed, H2O is considered the volatile component and sucrose is the non-volatile component. Therefore, we assume all the pressure of the vapor phase is due to PH2O. Raoult's Law applies:

Psolution = P H2Oo· χ H2O

where

χ H2O = mole fraction H2O in the solution

P H2Oo = vapor pressure of pure H2O (at the given temperature)

2) Using the data given in the problem statement:

16.589 mmHg = (17.5 mmHg) (χ H2O)

χ H2O = 16.589 mmHg / 17.5 mmHg = 0.94794

Note that 16.589 came from 17.5 minus 0.911.

3) The mole fraction of water can be expressed thusly:

    mol H2O
χ H2O  =  –––––––––––––––––––
    mol H2O + mol sucrose

4) Determine moles of water in this problem:

527 g / 18.015 g/mol = 29.2534 mol

5) Substitute into equation given in step 3 and solve:

    29.2534
0.94794  =  ––––––––––
    29.2534 + x

(0.94794) (29.2534 + x) = 29.2534

27.73047 + 0.94794x = 29.2534

0.94794x = 1.52293

x = 1.606568 mol of sucrose

6) Mass of sucrose:

(342.2948 g/mol) (1.606568 mol) = 550. g (to three sig figs)

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