COLLIGATIVE PROPERTIES OF SOLUTIONS - PART II

Lecture Video Link:

H.W.: 4a - 4c, 5a, & 5b


BOILING POINT ELEVATION and FREEZING POINT ELEVATION:

Boiling Point Elevation is the property of solutions that describes the _______________________________ between the boiling point of a PURE SOLVENT and the temperature at which a solution begins to boil.

Freezing Point Depression is the property of solutions that describes the _______________________________ between the freezing point of a PURE SOLVENT and the temperature at which a solution begins to freeze.

The _______________________________ the concentration of the solute, the _______________________________ the solution deviates from the boiling temperature and freezing temperature of the PURE SOLVENT.

ELECTROLYTE: A substance (when dissolved in water) that is capable of carrying an _______________________________.


COLLIGATIVE PROPERTIES: The _______________________________ of solute particles - NOT the _________________________________ of the solute particle, determines how the solute will affect the boiling and freezing points of water.

(sodium chloride placed in water) NaCl(s) ® Na+1(aq) + Cl-1(aq) . . . ( 2 ions )

(calcium chloride placed in water) CaCl2 (s) ® Ca+2(aq) + 2 Cl-2(aq) . . . (3 ions)

(glucose placed in water) C6H12O6 ® C6H12O6. . . (1 molecule)


MOLALITY: is used to calculate the boiling point and freezing point changes.

Molality is the number of moles of solute per _______________________________ of solvent, and is abbreviated by the lowercase letter, m.


EXAMPLE: How would you prepare a 0.50 m solution of sucrose, C12H22O11, using 500.0 grams of water?

First, find the number of grams of sucrose to measure by first finding the number of moles of sucrose needed to make a 0.50 m solution.

Change 500.0 grams of water to kilograms:

Multiply the number of kilograms by the molality, the number of moles is the result:

Converting moles to grams using the molar mass of sucrose gives the number of grams of sucrose:

C12H22O11 = [(12 x 12.0g)+(22 x 1.0g)+(11 x 16.0g) = 342 grams sucrose

Answer: Mix 86 grams of sucrose in 500 grams of water to achieve a 0.5 m sucrose solution.


PRACTICE PROBLEM:

1. How many moles of sucrose, C11H22O10, are there in a 2.33 m solution made with 1450 milliliters of water?

 

 

 

2. How many grams of magnesium carbonate are there in a 0.25 m solution made with 850.0 grams of water?

 

 

 


HOMEWORK QUESTIONS:

4a. How many grams of glucose, C6H12O6, are there in a 0.77 m glucose solution made with 450.0 grams of water?

 

 

 

 

4b. How many grams of NaCl would you need to prepare a 0.005 m saline solution (NaCl) using 2.0 kg of water?

 

 

 

 

4c. How many grams of BaCl2 would you need to prepare a 0.40 m BaCl2 solution using 100.0 grams of water?

 

 

 


CALCULATIONS INVOLVING COLLIGATIVE PROPERTIES:

Now you can apply your knowledge of molality to solve problems that ask you to find the boiling point elevation or freezing point depression of a solution.

The equation that relates to the concepts discussed earlier to the _______________________________ in boiling temperatures between a solution and the pure solvent is:

In this equation is the change (increase)in the boiling point of the pure solvent and the temperature at which the solution begins to boil; kb is the constant that relates to the molality of the solute, m.

Most of the solutions you work with are water solutions. The value of kb for water is:


The equation that relates the difference in the freezing point of a pure solvent and the initial freezing point of the solvent in a solution is much like the previous equations.

Here represents the difference between the freezing point of the pure solvent and the temperature at which the solution begins to freeze; kf is the constant that relates to the molality of the solute, m. The value of kf for water is:


EXAMPLE: At what temperature will a solution that is composed of 0.73 moles of glucose in 650.0 mL of water begin to boil?

First, the problem asks you to find , the difference between the boiling point of pure water, 1000C, and the temperature at which the solution begins to boil.

You have been given the value for kb earlier. Calculate the molality, m, from the information given in the problem.

Calculate the mass of the water:

Find the value of m:

Now you can substitute the known values in the equation to find .

Since the increase in boiling temperature is 0.570C and the boiling point of pure water is 100.000C, the temperature at which the solution will begin to boil is 100.570C.


PRACTICE PROBLEM:

1. At what temperature will a sucrose solution, C11H22O10, boil if it contains 175 grams of sucrose in 750 mL of water?

 

 

 

 

2. At what temperature will an ethanol solution, CH3OH, freeze if it contains 55.0 grams of ethanol in 1200 mL of water?

 

 

 


HOMEWORK QUESTIONS:

5a. At what temperature will a sucrose solution boil if it contains 1.55 moles of sucrose in 600.0 mL of water?

 

 

 

 

5b. At what temperature will a solution of ethylene glycol (the major ingredient in antifreeze) freeze if it contains 120 grams of ethylene glycol in 500.0 mL of water? (The molar mass of ethylene glycol is 62.1 grams/mole.)

 

 

 

 


Lecture Video Link:

H.W.: 6a - 6c


An ionic substance that is dissociated - _______________________________ the freezing point or _______________________________ the boiling point of water more than a molecular substance of the same concentration.

Recall when we talked about when ionic substances dissociating into ions in an aqueous state.

The number of ions dissolved change the number of moles of solute:

CaCl2 ® 1 Ca+2 + 2 Cl-1

1 mole of CaCl2 ® 3 moles of ions

Therefore, when you add one mole of CaCl2 to water (you now have 3 moles of solute) . . . So the equation is now:

Therefore, 1000C + 1.530C = 101.530C.


When you add one mole of NaCl to water (you now have 2 moles of solute). Remember that the freezing point depression will LOWER the freezing point of a solution.

NaCl ® 1 Na+1 + 1 Cl-1

1 mole of NaCl ® 2 moles of ions

Therefore, 00C - 3.760C = -3.760C


EXAMPLE: What is the lowest freezing temperature for a saltwater solution? The solubility of sodium chloride is 280 grams per 1000 grams of water at 00C.

1000 grams = 1 kilogram

NaCl = 23g + 35.4 g = 58.4 g/mol

First, calculate the molality:

Next, determine that the effective molality is twice the calculated molality (since the 1 mole of sodium chloride dissociates into 2 moles of ions)

NaCl ® 1 Na+1 + 1 Cl-1

1 mole of NaCl ® 2 moles of ions

4.8 m x 2 = 9.6 m

The equation for finding freezing temperatures of solution is:

Substituting the values for this problem, you get the following:

00C - 17.90C = -17.90C


PRACTICE PROBLEM:

1. At what temperature will a magnesium nitrate solution boil if it contains 320 grams of magnesium nitrate in 2547 mL of water?

 

 

 

 

2. At what temperature will a potassium phosphate solution freeze if it contains 144 grams of potassium phosphate in 487 grams of water?

 

 

 


HOMEWORK PROBLEMS:

6a. At what temperature will a sodium chloride solution boil if 120 grams of salt is dissolved in 1000 grams of water? (answer: 102.10C)

 

 

 

 

6b. At what temperature will a solution begin to freeze if 200.0 grams of calcium chloride is dissolved in 850.0 mL of water?

 

 

 

 

6c. At what temperature will a solution begin to boil if 155 grams of magnesium chloride is dissolved in 2300 grams of water?

 

 

 

 


REVIEW WORKSHEET, PART II


link to the Extra Practice ; link to the video explation