

Molarity
Presentation
•
Chemistry
•
10th Grade
•
Hard
Joseph Anderson
FREE Resource
5 Slides • 7 Questions
1
2 Step Molarity
Today we will be combining the knowledge we gained during the mole calculations and the molarity

2
Open Ended
How many moles are present in 300 grams of CaCO3 (gram formula mass = 100 g/mol)
Remember:
moles =
mass =
gfm =
moles = (mass/gfm)
3
Multiple Choice
What is the molarity of a 1.5 L solution containing 3 moles of CaCO3?
What is the molarity in this question?
3 moles
1.5 L
unknown
4
Multiple Choice
What is the molarity of a 1.5 L solution containing 3 moles of CaCO3?
What is the liters of solution in this problem?
3 moles
1.5 L
unknown
5
Multiple Choice
What is the molarity of a 1.5 L solution containing 3 moles of CaCO3?
What is the moles of solute in the problem?
3 moles
1.5 L
unknown
6
Open Ended
What is the molarity of a 1.5 L solution containing 3 moles of CaCO3?
Molarity = M
moles of solution = 3
Liters of solution = 1.5
Remember:
Molarity = moles of solute/liters of solution
7
Why 2 steps?
Some questions will ask for the molarity, but instead of giving you the number of moles, it will provide you with the mass and the gram formula mass.
Step 1: Find the # of moles
# of moles = given mass
gram formula mass
Step 2: Find the molarity of the solution
Molarity = moles of solute
Liters of solution
8
Sample Problem 1:
What is the molarity of a 2.0 L solution containing 145 grams of NaCl (gram formula mass = 58 g/mol)
9
Sample Problem 2:
What is the molarity of a 6 L solution containing 303 grams of KNO3 (gram formula mass = 101 g/mol)?
mass =
gfm =
moles =
10
Open Ended
Step 1: Find the number of moles.
mass = 303
gfm = 101
moles = mol
Remember: moles = mass/gfm
11
Sample Problem 2:
What is the molarity of a 6 L solution containing 303 grams of KNO3 (gram formula mass = 101 g/mol)?
moles = 3
Liters of solution =
Molarity =
12
Open Ended
Step 2: Find the molarity
moles = 3 mol
liters of solution = 6L
Molarity = M
Remember:
Molarity = moles of solute/Liters of solution
2 Step Molarity
Today we will be combining the knowledge we gained during the mole calculations and the molarity

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