Name:Solutions Notes Part 2

A.Colligative Properties of Solutions - Characteristics of a solution that do NOT depend on size or type of solute particles, but only upon the concentration of particles. The physical properties of solutions are different from those of the pure solvent. Colligative properties depend on the number of ionized particles dissolved in a given mass of solvent:

NaCl  Na+ + Cl- 2 particles

Na2SO4 2Na+ + SO42- 3 particles

(NH4)3PO4 3NH41+ + PO43- 4 particles

What do you notice about the solutes that contained polyatomic ions?

Organic and molecular molecules do not ionize so they stay as one particle.

Practice Problems: How many particles will the following solutions become?

1.magnesium chloride2. calcium bromide3. aluminum sulfate

4. iron (III) nitrate5. ethanol (C2H5OH)6. dinitrogen pentoxide

B. Types of Colligative Properties

1. Boling Point Elevation (BPE) – some solutes increase the point at which a solution boils

Example: salt added to water increases the boiling point of the water

This occurs because in water there are fewer particles to “energize” or boil, once you add salt the number of particles increases and more energy must be added to break the intermolecular forces that hold the molecules together.

Another explanation: Solutions have a lower vapor pressure than the pure solvent. Because the vapor pressure is decreased, more kinetic energy is required to raise the vapor pressure to atmospheric pressure; therefore the boiling point of a solution is higher than the boiling point of the pure solvent.

2. Freezing Point Depression (FPD) – some solutes decrease the point at which a solution freezes

Example: salt added to water decreases the freezing point of the water

Occurs because in water there are fewer particles to freeze, once you add salt the number of particles increases and more energy must be given off (lost) until the solution freezes. (Intermolecular forces are allowed to become stronger as more energy is lost that is why energy is lost during freezing.)

Another explanation: When a solute is present, the orderly pattern that the pure solvent normally takes when freezing is disrupted, so more kinetic energy must be lost from the solution for it to solidify (freeze). This lowers the freezing point of the solution.

Factors that affect the amount which the boiling point or freezing point changes are:

1. Molality of the solution

2. Whether the solution is ionic or molecular (# of particles)

3. The boiling point and freezing point constants (given or on page 438)

Colligative Property Constants

Solvent / Boiling Point (°C) / Kb (° C/m) / Freezing Point ( °C) / Kf (°C/m)
Water / 100 / 0.5l2 / 0 / l.86
Benzene / 80.l / 2.53 / 5.48 / 5.12
Acetic Acid / 118.l / 3.07 / 16.6 / 3.90
Nitrobenzene / 210.88 / 5.24 / 5.70 / 7.00
Phenol / 182 / 3.56 / 43 / 7.40

C. Formulas for solving boiling point elevation or freezing point depression problems:

1. BPE = (molality)(Kb)(# particles of solute)

2. FPD = (molality)(Kf)(# particles of solute)Both BPE and FPD are always positive.

Molality = moles of soluteORMolality = grams solute/molar mass of solute

kg of solvent kg of solvent

D. Formulas to find freezing point or the boiling point are:

1. BP solution = BP pure solvent +BPE

2. FP solution = FP pure solvent - FPD

Practice Problems:

1. What is the boiling point of a solution in which 75.00 grams of sodium sulfate are dissolved in 800.0 grams of water? Kb = 5.12 °C/m

1st solve for BPE: BPE = grams solute/molar mass of solute (Kb) (# particles)

kg of solvent

BPE = 75.00 g/142 g/mol (5.12°C/m) (3) = 10.1°C

0.8000 kg

2nd solve for the boiling point of the solution: BP solution = BP pure solvent +BPE

BP solution = 100°Ct + 10.1°C = 110.1°C

2. What is the freezing point of a solution in which 35.0 grams of magnesium bromide are dissolved in 400.0 grams of acetic acid? Kf = 3.90°C/m

  1. A benzene solution, C6H6, boils at 87.3°C and freezes at -1.12°C. Find the BPE and FPD.

4. When 50.0 grams of a molecular solute are dissolved in 600.0 grams of nitrobenzene, the boiling point of the solution is 214°C. Find the molar mass of the solute. (BP of nitrobenzene = 210.88°C; Kb = 5.24°C/m)

5. When 37.5 grams of a solute known to ionize into 4 particles is dissolved in 750.0 grams of benzene, the freezing point of the solution is -1.16°C. What is the molar mass of the solute? (FP of benzene = 5.48°C; Kf = 5.12°C/m)

VII. Chemical Reactions and how they are Affected by Various Concentrations

If 1 liter of a 3M HCl solution was allowed to react with excess magnesium and then one liter of a 6M HCl solution was allowed to react with excess magnesium, in which case, the 3M or the 6M, did more magnesium react?

2HCl + Mg  H2 + MgCl23M HCl = 3 moles HCl6M HCl = 6 moles HCl

1 liter solution 1 liter solution

Obviously we can see that in the 6M solution 6 moles of HCl were present to react with the magnesium, and in the 3M solution only 3 moles of HCl were present to react with the magnesium.

?grams of H2  6 mol HCl X 1 mol H2 X 2.02 g H2 = 6.06 grams H2

2 mol HCl 1 mol H2

?grams of H2  3 mol HCl X 1 mol H2 X 2.02 g H2 = 3.03 grams H2

2 mol HCl 1 mol H2

The reaction that used 1 liter of 6M HCl produced more hydrogen and magnesium chloride then the reaction that used 1 liter of 3M HCl solution. We only showed the work for the amount of hydrogen produced, but we could do the same for the magnesium chloride.

Example: H2SO4 + 2Na  Na2SO4 + H2 What mass of sodium sulfate results if 0.75 liters of 4M H2SO4 reacts with excess sodium according to the above equation?

Name: Solutions Worksheet #3 (Colligative Properties)

1. Solutions Homework from Book: Page 420: 13, 20, 21, Page 448: 18

2. What is the freezing point of a solution that contains 0.50 mole of sucrose dissolved in l.0 kg of water?

3. What is the boiling point of a solution that consists of l7l g of sucrose in 2000 grams of water?

4. What is the freezing point of a solution that contains 4.50 grams of C9Hl0O2 dissolved in 50.0 g of benzene?

5. Forty grams of an organic compound dissolved in l.00 kg of water have a boiling point of l00.52 C. What is the molar mass of the

solute?

6. 35.0 grams of an organic compound dissolved in 500. g of water have a freezing point of -3.72 C. What is the molar mass of the solute?

7. An organic compound whose molar mass is l40 g/mole is dissolved in l000 grams of nitrobenzene. The freezing point of the solution is 4.75 C. How many grams of the compound were dissolved in nitrobenzene?

8. What is the freezing point of 67.3 grams of sodium chloride dissolved in 755 grams of acetic acid?

9. What is the boiling point of l4.0 grams of glucose, C6H12O6, dissolved in 250 g of phenol?

10. How many grams of aluminum sulfate need to be dissolved in l500. grams of acetic acid in order to raise the boiling point by 4.3 C?

11. What is the freezing point of l7.3 grams of Cl0H8Cl4 in 250. grams of benzene?

12. What is the molar mass of a non-ionizing solute if 42.6 g of it dissolved in l89 grams of water raise the boiling point to l00.680 C?