Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 15 - Solutions. A solution is a homogeneous mixture of two or more substances in a single physical state. Examples: air, brass, salt water, carbon.

Similar presentations


Presentation on theme: "Chapter 15 - Solutions. A solution is a homogeneous mixture of two or more substances in a single physical state. Examples: air, brass, salt water, carbon."— Presentation transcript:

1 Chapter 15 - Solutions

2 A solution is a homogeneous mixture of two or more substances in a single physical state. Examples: air, brass, salt water, carbon dioxide in water

3 Miscible - pairs of liquids that will mix Immiscible - pairs of liquids that will not mix Aqueous solutions - solutions in which water is the solvent

4 Concentration - the amount of solute in a given amount of solvent Molarity (M) = mol of solute/L of solution Demonstrate sample problem 1 on page 507 Assignment: C-15 page 507 practice problems 1 & 2

5 Molality (m) = mol of solute/Kg of solvent 1 L of water = 1 kg of water 1 mL of water = 1 g Demonstrate sample problem page 508 Assignment: C-15 page 508 practice problems 3 & 4

6 Describe how you would make each of the following. 150 mL of.27 M NaOH solution Start by finding how many moles of NaOH this will require and then change that to grams. Mix this with water to dissolve the NaOH and then continue to add water until the total volume of the solution is 150 mL.

7 Approx. 150 mL of.27 m NaOH solution This will require the same number of grams of NaOH as above, only now we will just add this to 150 mL of water. The final volume will not be exactly 150 mL, however the solution will be.27 m.

8 Mole fraction = mol of solute/total mol of solution If 2 moles of water is mixed with 1 mole of sugar and 1 mole of alcohol. The mole fraction of sugar is.25 and the mole fraction of water is.5. Demonstrate sample problem 3 on page 510. Assignment: C-15 page 510 practice problems 5 & 6

9 Unsaturated - less solute than the solution can hold under existing conditions. Saturated - as much solute as the solution can possibly hold at the current temperature and pressure Supersaturated - more solute that the solution can hold (some will precipitate out)

10 The Dissolving Process When something dissolves, bonds between solute molecules are broken (which require energy) and new bonds with water are formed (which give off energy). If breaking the solute bonds takes more energy than bonding with water molecules gives back then the overall dissolving process will require energy (endothermic) and the heat of solution (∆H) will be positive.

11 If the solute - water bonds provide more energy than it takes to break the bonds between solute molecules then the dissolving process will give off heat (exothermic) and the heat of solution (∆H) will be negative.

12 Solubility - The three factors that affect solubility are: the nature of the solvent and solute (like dissolves like) temperature - The solubility of most gases decrease with temperature. The solubility of most solids increases with temperature. pressure - The solubility of gases increases with increasing pressure. See solubility graphs on pages 516 and 517.

13

14 The factors that affect the rate of dissolving are: The surface area of the solute. Making a powder increases the surface area. Stirring. Stirring brings less concentrated solvent in contact with the solute. Temperature. Increasing temperature speeds up molecule movement.

15 Colligative Properties - properties that are dependent on the number of solute particles but independent of their chemical identity.

16 Four colligative properties vapor pressure reduction - As the mole fraction of the solute increases, the vapor pressure of the solvent decreases.

17 boiling point elevation - This is related to vapor pressure reduction. A liquid boils when its vapor pressure is equal to the atmospheric pressure. Since the overall vapor pressure is lowered, it takes a higher temperature to increase the vapor pressure enough to equal the atmospheric pressure. ∆T b = K b m(see page 522 for K b )

18 Demonstrate sample problem 4 on page 523 Assignment: C-15 page 523 practice problems 7 & 8

19 freezing point depression - ∆T f = K f m (see page 526) d)Osmotic Pressure - Osmotic pressure is due to the fact that water tends to move through a semipermeable membrane from the side of lower solute concentration to the side of higher concentration.


Download ppt "Chapter 15 - Solutions. A solution is a homogeneous mixture of two or more substances in a single physical state. Examples: air, brass, salt water, carbon."

Similar presentations


Ads by Google