Chapter 21:  Colligative Properties

 

 

Section 21-1:╩ Molality

Section 21-2:╩ Freezing Point Depression, Boiling Point Elevation, and the van't Hoff Factor

Section 21-3:╩ Experiment - Determining Molar Mass Using Freezing Point Depression

Chapter 21 Practice Exercises and Review Quizzes

 

 

 

 

 

 

Section 21-1:  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, C6H12O6, 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 I2 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, C6H12, must be used to dissolve 12.6 grams of C6H4Cl2?

 

Solution:


 

 

 

 

Section 21-2:  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 (tf) of the solution will be lower than the freezing point of the pure solvent, and the boiling point (tb) of the solution will be higher than the boiling point of the pure solvent.

 

The magnitude of the freezing point depression (Ătf), 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:

 

Ătf = Kfm

 

The magnitude of the boiling point elevation (Ătb), 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:

 

Ătb = Kbm

 

The freezing point depression constant Kf and the boiling point elevation constant Kb (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 Kf = 1.86°C/m and Kb = 0.52°C/m for water, calculate the freezing and boiling points of a solution containing 33 grams of sucrose, C12H22O11, 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:

 

Ătf = iKfm                    Ătb = iKbm

 

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 C6H12O6

0.010 m HF

0.010 m Na3PO4

0.011 m MgI2

0.015 m HCl

 

Solution:

 

Each solution has the same solvent and, therefore, will have the same value of Kf in the equation Ătf = iKfm and the same value of Kb in the equation Ătb = iKbm.  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 Ătf and Ătb at the top to largest Ătf and Ătb at the bottom as follows:

 

0.010 m C6H12O6 = 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 MgI2 = ionic, i = 3 (Mg2+ + 2 I-): i x m = 3 x 0.011 m = 0.033 m

0.010 m Na3PO4 = ionic, i = 4 (3 Na+ + PO43-):  i x m = 4 x 0.010 m = 0.040 m 

 

A solution with a larger Ătf will have a lower freezing point.  Therefore, the solutions will be ranked from lowest to highest freezing point as follows:

 

0.010 m Na3PO4 < 0.011 m MgCl2 < 0.015 m HCl < 0.010 m HF < 0.010 m C6H12O6

 

A solution with a larger Ătb will have a higher boiling point.  Therefore, the solutions will be ranked from lowest to highest boiling point as follows:

 

0.010 m C6H12O6 < 0.010 m HF < 0.015 m HCl < 0.011 m MgI2 < 0.010 m Na3PO4

 

 

Section 21-3:  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 Kf value for the solvent expressed in the unit °C•kg/mol, solve for moles of solute in the freezing point depression equation Ătf = Kf(nsolute/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 C5H4.  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 Kf 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:

 

21-1) Calculate the mass of naphthalene, C10H8, that must be dissolved in 175 grams of chloroform, CHCl3, to create a 1.2 m solution.

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21-1)

 

 

21-2) To create a 0.0855 m solution, how many grams of methanol must be used to dissolve 0.548 grams of benzoic acid, C6H5COOH?

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21-2)

 

 

21-3) The freezing point of cyclohexane, C6H12, is 6.6°C and the boiling point of cyclohexane is 80.7°C.  Given that Kf = 20.0°C/m and Kb = 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.

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21-3)

 

 

21-4) 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 CaCl2

0.014 m C12H22O11

0.014 m HC2H3O2

0.015 m KBr

0.017 m HNO3

 

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21-4) Each solution has the same solvent and, therefore, will have the same value of Kf in the equation Ătf = iKfm and the same value of Kb in the equation Ătb = iKbm.  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 Ătf and Ătb at the top to largest Ătf and Ătb at the bottom as follows:

 

0.014 m C12H22O11 = molecular (no ionization), i = 1:  i x m = 1 x 0.014 m = 0.014 m  

0.014 m HC2H3O2 = 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 HNO3 = monoprotic strong acid, i = 2 (H+ + NO3-): i x m = 2 x 0.017 m = 0.034 m

0.013 m CaCl2 = ionic, i = 3 (Ca2+ + 2 Cl-):  i x m = 3 x 0.013 m = 0.039 m 

 

A solution with a larger Ătf will have a lower freezing point.  Therefore, the solutions will be ranked from lowest to highest freezing point as follows:

 

0.013 m CaCl2 < 0.017 m HNO3 < 0.015 m KBr < 0.014 m HC2H3O2 < 0.014 m C12H22O11

 

A solution with a larger Ătb will have a higher boiling point.  Therefore, the solutions will be ranked from lowest to highest boiling point as follows:

 

0.014 m C12H22O11 < 0.014 m HC2H3O2 < 0.015 m KBr < 0.017 m HNO3 < 0.013 m CaCl2

 

 

21-5) (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 Kf 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.

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21-5) (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

C3H2Br = empirical formula

 

(b)

 

 

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