Chapter 21: Colligative Properties
Section 212:Ê Freezing Point Depression, Boiling Point Elevation, and the van't Hoff Factor
Section 213:Ê Experiment  Determining Molar Mass Using Freezing Point Depression
Chapter 21 Practice Exercises and Review Quizzes
Section 211: Molality
Molality (m) is another way to express the concentration of a solution and is
found by dividing the moles of solute by the kilograms of solvent:
Sample Exercise 21A:
Calculate the molality of a
solution containing 17 grams of glucose, C_{6}H_{12}O_{6},
dissolved in 225 grams of water.
Solution:
Molality in mol/kg can be used a
conversion factor between moles of solute and kilograms of solvent:
Sample Exercise 21B:
Calculate the mass of I_{2} that must be dissolved in 115 grams of carbon disulfide to create a 0.097 m solution.
Solution:
Sample Exercise 21C:
To create a 0.105 m solution, how many grams of
cyclohexane, C_{6}H_{12}, must be used to dissolve 12.6 grams
of C_{6}H_{4}Cl_{2}?
Solution:
Section 212: Freezing Point Depression, Boiling Point Elevation, and the van't Hoff Factor
Colligative properties of
solutions, such as freezing point depression and boiling point elevation,
depend on the concentration of solute particles, but not on the identity of the
solute particles. When a solute is
dissolved in a solvent, the freezing point (t_{f}) of the solution will
be lower than the freezing point of the pure solvent, and the boiling point (t_{b})
of the solution will be higher than the boiling point of the pure solvent.
The magnitude of the freezing point
depression (Æt_{f}), or difference between the freezing points of the
pure solvent and the solution, can be calculated using the following equation
if the solute is molecular and does not ionize:
Æt_{f} = K_{f}m
The magnitude of the boiling point
elevation (Æt_{b}), or difference between the boiling points of the
solution and the pure solvent, can be calculated using the following equation
if the solute is molecular and does not ionize:
Æt_{b} = K_{b}m
The freezing point depression
constant K_{f} and the boiling point elevation constant K_{b} (not to be confused with base ionization constant) have the unit °C/m and will be different for each
solvent. The freezing point
depression can be subtracted from the freezing point of the pure solvent to
find the freezing point of the solution. The boiling point elevation can be added to the boiling point of the
pure solvent to find the boiling point of the solution:
Sample Exercise 21D:
The freezing point of water is
0.0°C and the boiling point of water is 100.0°C. Given that K_{f} = 1.86°C/m and K_{b} = 0.52°C/m for water, calculate the freezing and boiling points of a solution containing
33 grams of sucrose, C_{12}H_{22}O_{11}, dissolved in
46 grams of water.
Solution:
If the solute ionizes to some
extent during the dissolving process, the resulting concentration of solute
particles will be larger compared to a solution with the same molality in which
the solute is molecular and does not ionize. For solutions in which some or all of the solute is ionized,
the following equations can be used to calculate the freezing point depression
and boiling point elevation:
Æt_{f} = iK_{f}m Æt_{b} = iK_{b}m
The van’t Hoff factor i indicates the degree to which a solute
ionizes in water as follows:
Type of Compound

Value of i

molecular
(no ionization)

1

monoprotic
weak acid

1 < i < 2

monoprotic
strong acid

2

ionic

i = # ions in formula unit

Sample Exercise 21E:
Rank the following aqueous
solutions in order from lowest to highest freezing point and from lowest to
highest boiling point without performing detailed calculations:
0.010 m C_{6}H_{12}O_{6}
0.010 m HF
0.010 m Na_{3}PO_{4}
0.011 m MgI_{2}
0.015 m HCl
Solution:
Each solution has the same solvent
and, therefore, will have the same value of K_{f} in the equation Æt_{f} = iK_{f}m and the same value of K_{b} in the equation Æt_{b} = iK_{b}m. As such, a solution
with a larger product of (i x m) will have a larger freezing point
depression and a larger boiling point elevation. The solutions can be ranked from smallest Æt_{f} and
Æt_{b} at the top to largest Æt_{f }and Æt_{b} at the
bottom as follows:
0.010 m C_{6}H_{12}O_{6} = molecular (no ionization), i =
1: i x m = 1 x 0.010 m = 0.010 m
0.010 m HF = monoprotic weak acid, 1 < i < 2: 0.010 m < i x m < 0.020 m
0.015 m HCl = monoprotic strong acid, i = 2 (H^{+} + Cl^{}): i x m = 2 x 0.015 m = 0.030 m
0.011 m MgI_{2} = ionic, i = 3 (Mg^{2+} + 2 I^{}): i x m = 3 x 0.011 m = 0.033 m
0.010 m Na_{3}PO_{4} = ionic, i = 4 (3 Na^{+} + PO_{4}^{3}): i x m = 4 x 0.010 m = 0.040 m
A solution with a larger Æt_{f }will
have a lower freezing point. Therefore, the solutions will be ranked from lowest to highest freezing
point as follows:
0.010 m Na_{3}PO_{4} < 0.011 m MgCl_{2} < 0.015 m HCl < 0.010 m HF < 0.010 m C_{6}H_{12}O_{6}
A solution with a larger Æt_{b} will have a higher boiling point. Therefore, the solutions will be ranked from lowest to highest boiling
point as follows:
0.010 m C_{6}H_{12}O_{6} < 0.010 m HF < 0.015 m HCl < 0.011 m MgI_{2} < 0.010 m Na_{3}PO_{4}
Section 213: Experiment  Determining Molar Mass Using Freezing Point Depression
The molar mass of a molecular
solute that does not ionize (i = 1)
can be determined using freezing point depression as follows:
1. Measure the mass of the solute.
2. Measure the mass and freezing
point of the pure solvent.
3. Dissolve the solute and measure
the freezing point of the resulting solution.
4. Given the K_{f} value
for the solvent expressed in the unit °C•kg/mol, solve for moles of solute in
the freezing point depression equation Æt_{f} = K_{f}(n_{solute}/kg
solvent).
5. Divide the grams of solute by
the moles of solute to obtain the molar mass of the solute.
If we have determined the empirical
formula of the solute, we can use the molar mass we have calculated to
determine the molecular formula of the solute as follows:
Sample Exercise 21F:
A molecular solute that does not
ionize has the empirical formula C_{5}H_{4}. A solution containing 1.48 grams of the
solute dissolved in 55.7 grams of benzene was found to freeze at 4.47°C. If K_{f} for benzene is 5.12°C/m and the freezing point of pure benzene
is 5.53°C, determine the molar mass and molecular formula of the solute.
Solution:
Chapter 21 Practice Exercises and Review Quizzes:
211) Calculate the mass of naphthalene, C_{10}H_{8}, that must be dissolved in 175 grams of chloroform, CHCl_{3}, to create a 1.2 m solution.
Click for Solution
212) To create a 0.0855 m solution, how many grams of methanol
must be used to dissolve 0.548 grams of benzoic acid, C_{6}H_{5}COOH?
Click for Solution
212)
213) The freezing point of
cyclohexane, C_{6}H_{12}, is 6.6°C and the boiling point of
cyclohexane is 80.7°C. Given that
K_{f} = 20.0°C/m and K_{b} = 2.79°C/m for cyclohexane, calculate
the molality, the freezing point, and the boiling point of a solution
containing 18 grams of carbon tetraiodide dissolved in 88 grams of cyclohexane.
Click for Solution
213)
214) Rank the following aqueous
solutions in order from lowest to highest freezing point and from lowest to highest
boiling point without performing detailed calculations:
0.013 m CaCl_{2}
0.014 m C_{12}H_{22}O_{11}
0.014 m HC_{2}H_{3}O_{2}
0.015 m KBr
0.017 m HNO_{3}
Click for Solution
214) Each solution has the same
solvent and, therefore, will have the same value of K_{f} in the equation
Æt_{f} = iK_{f}m and the same value of K_{b} in
the equation Æt_{b} = iK_{b}m. As such, a solution with a larger product of (i x m) will have a larger
freezing point depression and a larger boiling point elevation. The solutions can be ranked from
smallest Æt_{f} and Æt_{b} at the top to largest Æt_{f }and
Æt_{b} at the bottom as follows:
0.014 m C_{12}H_{22}O_{11} = molecular (no ionization), i =
1: i x m = 1 x 0.014 m = 0.014 m
0.014 m HC_{2}H_{3}O_{2} = monoprotic weak acid, 1 < i <
2: 0.014 m < i x m < 0.028 m
0.015 m KBr = ionic, i = 2 (K^{+} + Br^{}): i x m = 2 x 0.015 m = 0.030 m
0.017 m HNO_{3} = monoprotic strong
acid, i = 2 (H^{+} + NO_{3}^{}): i x m = 2 x 0.017 m = 0.034 m
0.013 m CaCl_{2} = ionic, i = 3 (Ca^{2+} + 2 Cl^{}): i x m = 3 x 0.013 m = 0.039 m
A solution with a larger Æt_{f }will
have a lower freezing point. Therefore, the solutions will be ranked from lowest to highest freezing
point as follows:
0.013 m CaCl_{2} < 0.017 m HNO_{3} < 0.015 m KBr <
0.014 m HC_{2}H_{3}O_{2} < 0.014 m C_{12}H_{22}O_{11}
A solution with a larger Æt_{b} will have a higher boiling point. Therefore, the solutions will be ranked from lowest to highest boiling
point as follows:
0.014 m C_{12}H_{22}O_{11} < 0.014 m HC_{2}H_{3}O_{2} < 0.015 m KBr < 0.017 m HNO_{3} < 0.013 m CaCl_{2}
215) (a) A molecular solute that
does not ionize was found to be 30.6% carbon and 1.7% hydrogen by mass, with
the remainder being bromine. Determine the empirical formula of the solute.
(b) A
solution containing 3.60 grams of the solute dissolved in 33.4 grams of carbon
tetrachloride was found to freeze at 36.5°C. If K_{f} for carbon tetrachloride is 29.8°C/m and the freezing point of pure carbon
tetrachloride is 22.9°C, determine the molar mass and molecular formula of the
solute.
Click for Solution
215) (a) 100%  30.6% C –
1.7% H = 67.7% Br by mass
Assume one hundred grams of solute:
2.548 mol
C: 1.69 mol H: 0.8473 mol Br (divide each by 0.8473)
= 3 mol C:
2 mol H: 1 mol Br
C_{3}H_{2}Br
= empirical formula
(b)