Molarity

• Molarity is defined as the amount of moles of a compound dissolved in an amount of solvent (usually water).

• It can be solved with the equation:

Molarity (M) = moles soluteliters of solution

Molarity – What is it?• A measure of how concentrated a solution is.• Think about making Kool-Aid.• What makes Kool-Aid more

delicious?• The more sugar you add, the

sweeter the solution gets.

• Adding more moles of sugar, without changing the volume makes the Molarity “Concentration” Increase

Molarity Practice• 1) 80.5 grams of sodium chloride is

dissolved in 3.5 Liters. Determine the Molarity.

• 2) 125 grams of Calcium Fluoride is dissolved in 250 mL of water. Solve for the Molarity.

12.3

What mass of KI is required to make 500. mL ofa 2.80 M KI solution?

2.80 M = x mol / 0.500 L = 1.40 mol 1.40 mol

166 g KI 1 mol KI

= 232 g KI

Dilution of Solutions• Dilution is the procedure for preparing a

less concentrated solution from a more concentrated solution.

• When you want to dilute a solution, what happens to the number of moles present in the solution?– Do they increase?– Decrease?– Stay the same?

• Since moles are constant before and after dilution, we can use the following formula for dilution calculations.

• M1V1=M2V2M1 = Starting Molarity , V1 = Initial Volume of solution

M2 = Final Molarity, V2 = Final Volume of solution

Solution Stoichiometry

• We use the same ideas when we converted from grams of one compound to grams of another, except we are in mL of solution instead of grams.

• Use the Molarity equation: M = mol/L

• Use stoichiometry to convert moles of one compound to moles of the other compound.

X mol X grams

Y mol X mol

X grams

Y grams Y mol

Solution Stoich Practice• How many mL of 2.00 M Hydrobromic acid,

when reacted with aluminum hydroxide, will be required to produce 60.0 grams of water in the acid base reaction?

• ___HBr +___Al(OH)3 ___H2O + ___ AlBr3

1 mol H2O 18.02 g H2O

3 mol HBr3 mol H2O

60.0 g H2O

= 3.33 mol HBr

13 3 1

2.00 M HBr = 3.33 mol HBr

X ml HBr = 1.67 mL HBr