1. Percent by volume: = Volume of solute x 100 Volume of solution
Percent solutions: represent concentration and can be expressed by a) volume or b) mass
Percent by volume: = Volume of solute x Volume of solution
indicated %(v/v)
Vsolution= Vsolute + Vsolvent

2. Percent by mass: = Mass of solute(g) x 100 Mass of solution (g)
Indicated %(m/m)
msolution= msolute + msolvent

3. 1). You mix 25. 0 g of salt with 225g of water
1). You mix g of salt with 225g of water. What is the %(m/m) concentration of the solution?

4. Classwork: percent composition handout
2) How many grams of salt are there in 1.2kg of a 6.3 % (m/m) solution?
Classwork: percent composition handout

5. Dilution
Adding water to a solution will reduce the number of moles of solute per unit volume
but the overall number of moles remains the same!
Think of taking an aspirin with a small glass of water vs. a large glass of water
You still have one aspirin in your body, regardless of the amount of water you drank, but a larger amount of water makes it more diluted.

6. Dilution
The number of moles of solute in solution doesn’t change if you add more solvent!
The # moles before = the # moles after
Formula for dilution: M1 x V1 = M2 x V2
M1 and V1 are the starting concentration and volume; M2 and V2 are the final concentration and volume.
Stock solutions are pre-made solutions to known Molarity.

7. 1. You need to prepare 250. mL of a 0. 5M KCl solution
1. You need to prepare 250. mL of a 0.5M KCl solution. What volume of a 2.0 M KCl solution do you need?
M1 x V1 = M2 x V2 M1=2.0M V1= ? M2= 0.5M V2= 250.mL

8. M1 x V1 = M2 x V2 M1=3.0M V1= 50.0mL M2= ? V2= 250.mL
2. You add 200 mL of water to 50.0 mL of a 3.0M NaCl solution. What is the new concentration of the solution?
M1 x V1 = M2 x V2 M1=3.0M V1= 50.0mL M2= ? V2= 250.mL
Classwork: Dilutions handout

9. Colligative Properties of Solutions
OBJECTIVES:
Identify three colligative properties of solutions.
Explain why the vapor pressure, freezing point, and boiling point of a solution differ from those properties of the pure solvent.
Solve problems related to the molality and mole fraction of a solution.
Describe how freezing point depression and boiling point elevation are related to molality.

10. Colligative Properties
-These depend only on the number of dissolved particles
-Not on what kind of particle
-Two important colligative properties of solutions are:
Boiling point elevation
Freezing point lowered

11. Colligative Properties
Some particles in solution will IONIZE (or split), while others may not.
Colligative Properties
CaCl2 will have three particles in solution for each one particle it starts with.
Glucose will only have one particle in solution for each one particle it starts with.
NaCl will have two particles in solution for each one particle it starts with.

12. Boiling Point is ELEVATED
Salt water boils above 100ºC
The number of dissolved particles determines how much, as well as the solvent itself.
Freezing Point is LOWERED
Solids form when molecules make an orderly pattern called “crystals”
The solute molecules break up the orderly pattern.
Makes the freezing point lower.
Salt water freezes below 0ºC
How much lower depends on the amount of solute dissolved.

13. The addition of a solute would allow a LONGER temperature range, since freezing point is lowered and boiling point is elevated.

14. Molality (abbreviated m)
a new unit for concentration
m = Moles of solute (mol) kilogram of solvent (kg)

15. Ex. 1 Calculate the molality of a solution made by dissolving 45
Ex.1 Calculate the molality of a solution made by dissolving 45.0g of glucose, C6H12O6, in g of water.
m = Moles of solute kilogram of solvent

16. CW p 144 # 1-3