Transcript Slide 1

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 solute
liters 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.
What mass of KI is required to make 500. mL of
a 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
12.3
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=M2V2
M1 = 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 grams
X mol
X grams
Y mol
X mol
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?
1
3 2O + ___
1 AlBr3
• ___HBr
+___Al(OH)
3
3  ___H
60.0 g H2O 1 mol H2O 3 mol HBr
= 3.33 mol HBr
18.02 g H2O 3 mol H2O
3.33 mol HBr
2.00 M HBr =
X ml HBr
= 1.67 mL HBr