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>>
View the other Key Equations and Concepts
in this chapter
Colligative Properties of Solutions
Parts of this equation/concept include:
| A. Osmotic Pressure,
Equation 5.5 |
 = MRT
The most difficult part of working with this equation is to keep
the appropriate units. The units of the gas constant R
are helpful for this. You may be required to memorize the gas constant
at this time. If not now, you probably will need to do so in a later
chapter. The version of R most useful for this chapter
is
R = 0.0821 L•atm/mol•K
With the units, notice the L/mol is the reverse of M =
mol/L (M in the equation). As a colligative property, it
is the molarity of the total number of particles, not the molarity
of the solute that is generally used. (See previous section.) Since
molecular compounds do not ionize, the molarity of the compound
is the same as the total number of particles. Insoluble salts do
not dissolve; therefore they will not have a concentration unit,
so you don't have to worry about solubility rules in this type of
problem.
K is the temperature unit of Kelvin (T in the equation).
Temperature is normally measured in units of °C; it will be
your job to convert to Kelvin using the formula from Chapter 1.
C + 273.15 = K
Using these values will result in osmotic pressure, , in
units of atmospheres (atm). You must use the appropriate values
and units for M and T to obtain the correct answer.
>> Calculating Osmotic Pressure
>> Example 1
What is the osmotic pressure of a solution of 1.455 g of BeSO4
in 500.0 mL aqueous solution at 21°C?
Solution:
= MRT
To determine M, recall M = mol particles/L
solution
| mol solute |
= |
mol BeSO4 |
= |
1.455 g |
 |
|
= |
0.01385 mol BeSO4 |
but M = moles of particles and there are 2 moles of
ions for each mole of BeSO4, so 0.02770 mol ions
| liters solution |
= |
500.0 mL |
|
|
= |
0.5000 L |
R is the gas constant, 0.0821 L•atm/mol•K
T is temperature in Kelvin, so 21 °C + 273.15
= 294 K
Note: The temperature limited the value to three significant
figures. Although it started with two, it gained one in the conversion
to Kelvin. The significant figure rule for addition is to use
the value with the fewest decimal places rather than the fewest
significant figures.
>> Example 2
What is the osmotic pressure of a solution of 3.47 mg of PCl3
in 750.0 mL of ethanol solution at 23 °C?
Solution:
= MRT
| M |
= |
|
= |
| mol PCl3 |
|
| L ethanol solution |
|
| 3.47 mg PCl3 |
|
|
|
|
= |
2.5310–5 mol |
| 750.0 mL |
|
|
= |
0.7500 L |
| M |
= |
| 2.5310–5 mol |
|
| 0.7500 L |
|
= |
3.3710–5 M |
R = 0.0821 L•atm/mol•K
T = 23 °C + 273.15 = 296 K
|
= |
| 3.3710–5 mol |
|
| L |
|
|
|
|
296 K |
= |
8.1810–4 atm |
>> Calculating the van't Hoff Factor
from Osmotic Pressure
When calculating the van't Hoff factor in any colligative property
problem, including osmotic pressure, it is often easiest to include
i in the equation and let the concentration represent moles of solute
rather than moles of particles. In this case
= iMRT
>> Example 3
What is the van't Hoff factor of a 0.010 M solution
of CaCl2 if the osmotic pressure at 25°C is 0.70 atm?
Solution:
| 0.70 |
|
| 0.010 0.08206 298 |
|
= |
i |
2.86 = i
(For calcium chloride, the predicted van't Hoff factor is 3.
Ion pairing causes the true value to be less than the calculated
one.)
>> Calculating Molar Mass from Osmotic
Pressure
>> Example 4
What is the molar mass of a nonelectrolyte if 1.50 g in 250.0
mL of solution has an osmotic pressure of 0.521 atm at 15 °C?
Solution:
molar mass = g/mol
g = 1.50 g
= MRT
= 0.521 atm
R = 0.0821 L•atm/mol•K
T = 15 °C + 273.15 = 288 K
| ML |
= |
mol |
= |
|
|
0.2500 L |
= |
0.00551 mol |
>> Example 5
What is the molar mass of a nonelectrolyte if 141 mg in 1.00
L of solution has an osmotic pressure of 0.0184 atm at 30 °C?
Solution:
molar mass = g/mol
= MRT
0.0184 atm = M (0.0821 L•atm/mol•K)(303
K)
0.000740 mol/L = M
| mol |
= |
|
|
1.00 L |
= |
0.000740 mol |
| grams |
= |
141 mg |
|
|
= |
0.141 g |
>> back
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| B. Boiling Point Elevation,
Equation 5.6 |
 T = Kbm or
T = iKbm
>> Calculating Change in Boiling Point
The change in boiling point is T in the preceding
formula. The key to doing these problems is the calculation of molality
and choosing the right constant (Kb) from the table (Table
5.2 in the text). Always choose the constant based on the solvent
and be sure to choose the boiling point constant for boiling point
problems and freezing point constant for freezing point problems.
>> Example 6
What is the change in boiling point if 3.69 g of K2SO4
was added to 100.0 mL of water?
Solution:
Method 1:
T = Kbm
Kb = boiling point constant for water =
0.52 °C/m
| mol K2SO4 |
= |
3.69 g |
|
|
= |
0.0212 mol |
| kg water |
= |
100.0 mL |
|
|
|
|
= |
0.100 kg |
Remember that you need the molality of particles and when K2SO4
dissolves in water it dissociates into two K+ ions
and one SO42– ion. Therefore, i
= 3 and molality of particles = im = 3(0.212 m) = 0.636 m
| T |
= |
|
|
0.636 m |
= |
0.33 °C |
Method 2:
Determine Kb, i, and molality
of K2SO4 the same way as method 1; then
use the equation
| T |
= |
iKbm |
= |
3 |
|
|
|
0.212 m |
= |
0.33 °C |
For both examples, the answer has two significant figures because
the Kb only has two.
>> Calculating Actual Boiling Point
Problems involving actual boiling point are different because they
ask for the boiling point of the solution rather than for
the change in boiling point of the solvent. The process is initially
the same. You must still determine the change in boiling point,
but there is an extra step. The extra step is to add the change
in boiling point, T, to the boiling point of the
solvent. It will help you to remember to add if you remember the
name of this colligative property is boiling point elevation.
>> Example 7
What is the boiling point of a solution made by mixing 13.63
g ethanol and 540.00 g water?
Solution:
The sneaky part of this question has to do with both ethanol
and water being listed as solvents. Remember that the solvent
is the substance in greater quantity. In this case, it is obviously
water.
solute = 13.63 g ethanol (C2H5OH)
solvent = 540.00 g water (H2O)
T = Kbm
The value of Kb is the value of
the solvent = 0.52 °C/m
The value of m is the value of particles of solute. Since
ethanol is a molecular compound, i = 1. The molality
of ethanol is the same as the molality of particles.
| mol ethanol |
= |
13.63 g C2H5OH |
|
|
= |
0.2959 mol |
| kg water |
= |
540.00 g |
|
|
= |
0.54000 kg |
| T |
= |
Kbm |
= |
|
|
0.5479 m |
= |
0.28 °C |
Don't forget (and it is the most common mistake) that this does
not answer the question!
| boiling point (bp) solution |
= |
bp solvent |
+ |
T |
= |
100 °C |
+ |
0.28 °C |
= |
100.28 °C |
Recall that the 100°C was defined as the boiling
point of water; therefore it is an exact number with infinite
significant figures. The significant figure rule for addition
is to use the value with the fewest decimal places.
>> Example 8
What is the boiling point of 0.360 m Ca(ClO4)2(aq)?
Solution:
T = Kbm
Kb is the boiling point constant for the
solvent, which is water (from "(aq)") = 0.52
°C/m.
m is the molality of the particles. Since Ca(ClO4)2
is an ionic compound, it dissociates into ions in solution. The
ions are one Ca2+ and two ClO4–,
so i = 3 and the molality of particles = im = 3(0.360) = 1.08 m.
| T |
= |
|
|
1.08 m |
= |
0.56 °C |
bp solution = bp solvent + T
= 100 °C + 0.56 °C = 100.56 °C
>> Example 9
What is the boiling point of a solution made by mixing 72.5 g
Br2 with 660.0 g CCl4?
Solution:
T = Kbm
solute = Br2
solvent = CCl4
The solvent determines the Kb. CCl4
has a Kb = 5.02 °C/m and a boiling
point of 76.8 °C. (Found in Table 5.2 in the text.)
Since Br2 is a molecular compound it will not dissociate
when dissolved. (The nonaqueous solvent is another tip-off that
it will not dissociate.) Therefore i = 1 and the molality
of particles = molality of Br2.
| mol Br2 |
= |
72.5 g |
|
|
= |
0.454 mol |
| kg CCl4 |
= |
660.00 g |
|
|
= |
0.66000 kg |
| m |
= |
| 0.454 mol Br2 |
|
| 0.66000 kg CCl4 |
|
= |
0.688 m |
| T |
= |
Kbm |
= |
|
|
0.688 m |
= |
3.45 °C |
(three significant figures this time) |
bp solution = bp solvent + T
= 76.8 °C + 3.45 °C = 80.3 °C
Remember to round for significant figure to the fewest decimal
places.
| C. Freezing Point Depression,
Equation 5.8 |
T = Kfm
or T
= iKfm
Use these equations as you would Equation 5.6 with the three following
exceptions.
- Use for freezing point (melting point) problems.
- The value of Kf will be different from
the value of Kb.
- When determining the value of the freezing point, subtract
T from the freezing point of the pure solvent.
>> Example 10
What is the freezing point of 0.544 m KMnO4(aq)?
Solution:
T = Kfm
Kf is the value for the solvent, which is
water in this example. Kf = 1.86 °C/m
for water (the value is different from that with boiling point
problems) and the freezing point of pure water is exactly
0 °C.
The solute, KMnO4, is an ionic compound (metal combined
with nonmetal). You might not recognize the anion, but the cation,
K+, should be obvious. Since most ionic compounds consist
of only two ions, it is best to assume that the rest of the formula
is the anion. That anion, MnO4–, must have
a charge of –1, so that the compound is electrically neutral.
This makes the van't Hoff factor for this compound = 2 and the
molality of particles = 2(0.544) = 1.09 m (significant figures!).
| T |
= |
Kfm |
= |
|
|
1.09 m |
= |
2.02 °C |
The other difference in this type of problem is that it is freezing
point depression. The solute lowers the freezing point,
so you subtract T from the value of pure solvent.
| freezing point (fp) solution |
= |
fp solvent – T |
| |
= |
0 °C – 2.02 °C |
| |
= |
–2.02 °C |
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