Percent solutions: represent concentration and can be expressed by a) volume or b) mass

Percent by volume: = Volume of solute

x 100 Volume of solution

indicated %(v/v)Vsolution= Vsolute + Vsolvent

Percent by mass: = Mass of solute(g) x 100

Mass of solution (g)

–Indicated %(m/m)

–msolution= msolute + msolvent

1). You mix 25.0 g of salt with 225g of water. What is the %(m/m) concentration of the solution?

2) How many grams of salt are there in 1.2kg of a 6.3 % (m/m) solution?

Classwork: percent composition handout

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.

DilutionThe 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.

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

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

Colligative Properties of SolutionsOBJECTIVES:

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.

Colligative Properties -These depend only on the

number of dissolved particles -Not on what kind of particle -Two important colligative

properties of solutions are:1) Boiling point elevation2) Freezing point lowered

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.

CaCl2 will have

three particles in solution for each one particle it starts with.

Colligative Properties

Some particles in solution will IONIZE (or split), while others may not.

Boiling Point is ELEVATEDSalt water boils above 100ºCThe number of dissolved particles determines

how much, as well as the solvent itself.

Freezing Point is LOWEREDSolids 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.

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

Molality (abbreviated m)

a new unit for concentrationm = Moles of solute (mol) kilogram of solvent (kg)

Ex.1 Calculate the molality of a solution made by dissolving 45.0g of glucose, C6H12O6, in 500.0 g of water.

m = Moles of solute kilogram of solvent

CW p 144 # 1-3