1. Here is a helping hand with some of the new material on Molarity:
Remember:
Not all substances that we use in chemical reactions are in the same PHASE of matter
Solids (s)
Liquids (l)
Gases (g)
And then to complicate matter (that’s a joke) –
Liquids that are a uniform mix of solute and solvent are aqueous solutions (aq)…they are made from ions.
It is easier for many reactions to take place in a liquid because they can move more easily in liquids…. That’s another chapter in itself!

2. Molarity = Number of moles of solute liters of solution
What is the molarity of 40.0 grams of calcium carbonate dissolved in 1.0 L of water?
What if it is dissolved in 2.0 L of water?
What if 20.0 grams of the rock salt are dissolved in 1.0 L of water?
…10.0 grams in 500mL?
…40.0 grams in 250 mL?

3. A saline (salt) contains 0. 90 g of NaCl in 100 mL of solution
A saline (salt) contains 0.90 g of NaCl in 100 mL of solution. Calculate M:
.90 g ( 1 mole)
58.5 g
= moles NaCl
.0154 moles
.100 L
= .154 M

4. A salt solution has a volume of 250 mL and contains 0. 70 mole of NaCl
A salt solution has a volume of 250 mL and contains 0.70 mole of NaCl. What is the molarity?
0.70 moles of NaCl =
.250 L
2.8M

5. Now, how many grams is this?
How many moles of silver nitrate are needed to make 350 mL of a 2.0 M AgNO3 solution?
X moles
.350L
= 2.0M
So, x 2.0 = .70 moles
Now, how many grams is this?
There are 170 g in a mole of silver nitrate…
.70 moles x 170 g = 119 = 120 g Ag NO3
1 mole

6. 400.0 mL of a .40 M solution of BaCl2
First let’s find the amount of moles of barium chloride that one must gather from a container of the salt.
Now convert that amount of moles to grams: .

7. e. A solution that is made from dissolving 17
e. A solution that is made from dissolving g of NaNO3 to make mL…
Would have a molarity calculated as follows:
Molarity is moles
liter …
So we must find moles,
Then plug that in and,

8. f. 35.5 g of solid Na2SO4 can make how many liters of .25 M solution?
X mole = M
? Liters
Cross multiply and divide =
.moles x 1 / (.25) = 1 liter
Well, how many moles is 35.5 g of sodium sulfate?
Call that X moles
How many Liters would .25 moles dissolve in to make a .25 M solution? .

9. How many grams of Silver will be produced from 85. 0 mL of 1
How many grams of Silver will be produced from 85.0 mL of 1.50 M silver nitrate and copper metal?
For this problem you will have to begin with a balanced equation:
__AgNO3 + __Cu 
Now you need to add the phase symbols and coefficients for the mole to mole ratio.
X moles = 1.50 M
.085 L,

10. Now we need to determine the mole to mole ratio of the given and find:
The balanced equation shows us that for every 2 moles of silver nitrate reacted with 1 mole of Cu that 2 moles of Ag will be produced along with 1 mole of copper nitrate.
(It is important to note that since the Cu metal is ‘in excess’ that all the silver nitrate gets used up and the full 2 moles of product are produced!)
2 Ag NO3 (aq) Cu (s)  2 Ag (s) Cu(NO3)2 (aq)

11. h. Calculate the # of Barium Sulfate grams when 125mL of 1
h. Calculate the # of Barium Sulfate grams when 125mL of 1.20 M Barium nitrate reacts with excess sulfuric acid.
Write the equation:
__Ba(NO3)2 +__H2SO4 Now balance it.
Find the moles of the given:
x moles = 1.20 M ,

12. Now determine the balanced molar ratio for all the other reactants and products.
1 Ba(NO3)2 + 1H2SO4  1BaSO4 + 2HNO3
Find the Barium Sulfate grams: