1. Chapter 18 - Solutions

2. 18.1 Properties of Solutions
Solution Formation
Solvent
This is the liquid that is doing the dissolving
Solute
This is what is being dissolved
Form a homogenous mixture

3. Saturated vs. Unsaturated Solutions

4. Solubility
Two liquids that dissolve in each other are said to be miscible
Immiscible liquids are insoluble in each other

5. Immiscible vs. Miscible Solutions

6. Factors Affecting Solubility
Solubility increases with the increase in temperature (Easier to dissolve something when the temperature is increased)
Solubility increases with an increase in surface area (ex. CRUSHING)
Few exceptions that occur in the reverse

7. 18.2 Concentrations of Solutions
Molarity is the concentration of moles per liters
Molarity (M) = moles of solute / liters of solution
Dilute solution contains a low concentration of solute
Concentrated solution contains a high concentration of solute

8. Practice Problems:
Calculate the molarity of a solution which contains 0.40 mol of C6H12O6 dissolved in 1.6 L of a solution.
What is the molarity of a solution containing 325 g of NaCl dissolved in 750. mL of solution?(1000 ml = 1L)

9. Do Now:
In complete sentences, describe how you would prepare 500 mL of a 1.4M solution of sodium chloride

10. Making Dilutions Formula for making a dilution C1 V1 = C2 V2
“Stock solution” is the same as the original solution.

11. Example
A stock solution of HCl has a concentration of 12M. How much of the stock solution would be required to make 325 mL of a 6M solution?

12. Molality
Molality is another way to represent the concentration of a solution.
It is represented by a lower case m.
Molality = moles of solute/kg of solvent
Example: A 4.9m solution of NaCl is dissolved in 1000 grams of water. How many grams of NaCl is this?

13. Determine the molality of a NaCl solution in which 17
Determine the molality of a NaCl solution in which 17.3 moles of solute are dissolved in 1400 grams.

14. In your notebook practice:
A solution containing 250 grams of BaCl2 is dissolved in 2000 g of water creating a solution that has a total volume of 2450 mL.
Given: mL = 1L g = 1 kg
Determine the molarity from the information above.
 Determine the molality from the information above.

15. 18.3 Colligative Properties of Solutions
Properties that depend on the number of particles dissolved in a given mass of solvent. These properties can change depending on the solute that is dissolved.
Boiling Point Elevation
Freezing Point Depression

16. Boiling-Point Elevation
∆Tb = Kb x m x i
m is the molality
i is the number of ions in solution (covalent: i = 1)
Kb is the molal boiling point elevation constant
It is dependent on the solvent

17. Freezing Point Depression
∆Tf = Kf x m x i
m is the molality
i is the number of ions in solution (covalent: i = 1)
Kf is the molal freezing point constant
It is also dependent on the solvent

18. Example:
Determine the boiling point and freezing point of a solution in which 600 grams of MgCl2 is dissolved in 2400 grams of water. (Kb = and Kf = 1.86)