1. Bellwork Dec 4 – graded Take out a sheet of paper and answer…
What would be the Molarity if 3.5 moles of sodium chloride were dissolved in 1.75 Liters of water?
2. What would the Molarity be if 0.75 Moles of glucose dissolved in 750 ml of water? (Something tricky is happening here)
grams of sodium chloride is dissolved in 3.5 Liters. Determine the Molarity.
grams of Calcium Fluoride is dissolved in 250 mL of water. Solve for the Molarity. 2M 1M
0.39 M
6.4M
Crash course chemistry: Solutions

2. A solution is a homogenous mixture of 2 or more substances.
The solute is(are) the substance(s) present in the smaller amount(s). What is being dissolved.
The solvent is the substance present in the larger amount. What is doing the dissolving.
12.1

3. Concentration
As we make solutions, we care about how much solute is actually dissolved inside
Concentration is a measure of this!
We say that something is concentrated if there is a lot of it dissolved, or we say that something is dilute if there is not a lot dissolved.

4. total liters of solution
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
total liters of solution

6. 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

7. Practice:
1) What would be the Molarity if 3.5 moles of sodium chloride were dissolved in 1.75 Liters of water?
2) What would the Molarity be if 0.75 Moles of glucose dissolved in 750 ml of water? (Something tricky is happening here)

8. What to look out for!
Molarity is defined as moles per LITER. So every calculation, you need to make sure you convert your volume to Liters.
Additionally, Molarity is in Moles. If your starting amount is in grams, be sure to convert to moles.

9. Molarity Practice Round 2
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.

10. Changing the question a bit…
What mass of KI is required to make 500. mL of
a 2.80 M KI solution?
2.80 moles x 1L x 500mL = 1.4 moles
L mL
1.4 moles KI x g KI = 232 g KI
1 mole KI
12.3

11. (Mass solute + Mass solvent)
% by Mass
Another way of referring to concentration
Basically % composition of a solution
Key info: The density of water is handy! Every 1mL = 1gram of H2O
% Mass =
Mass of solute
(Mass solute + Mass solvent)
X 100

12. % Mass Practice
If 31 grams of ammonium bromide are dissolved to make a 287 gram solution, what is the % by mass of the solute?
If a 64 gram solution is 5.2% sodium acetate, how many grams of solute are inside?
11%
3.3g

13. % Mass Practice Round 2
1.940 grams of potassium nitrate are dissolved in 32.0 mL of water. What is the % Mass?
0.243 moles of sodium chloride are dissolved into 250mL of water. What is the % Mass?
5.7%
5.4%

14. Dilution of Solutions using Molarity
Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution by adding more solvent.
When you add more solvent, what happens to the number of moles already present in the solution?
Do they increase?
Decrease?
Stay the same?
Stay the same. 14

15. 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 (total) Volume of solution

16. Example Type 1
If you take 75 mL of a 1.7 M CaO solution and add in an additional 75 mL of water, what is the new concentration?
M1V1 = M2V2
(1.7M ) (75mL) = (x) (150mL)
M2= 0.85 M
Note that V2 is the TOTAL volume of the solution, not just how much water was added.

17. Example Type 2 How much water will need to be added?
You need 800.mL of a 2.00 M H2SO4 solution, but only have 6.00 M stock solution. How many mL of your stock solution do you need to use for the dilution?
M1V1 = M2V2
6.0M x V1 = 2.0M x 800.mL
V1= 267mL
So we need to start with 267mL of the concentrated acid…
How much water will need to be added?
We want a total of 800.mL so…
800. – 267 = 533mL of water!

18. How would you prepare 60.0 mL of 0.2 M
HNO3 from a stock solution of 4.00 M HNO3?
MiVi = MfVf
Mi = 4.00
Mf = 0.200
Vf = 60.0 mL
Vi = ? mL
Vi =
MfVf Mi =
0.200 x 60.0
4.00
= 3 mL
3 mL of acid
+ 57 mL of water
= 60 mL of solution
4.5

19. Solution Stoichiometry
When you mix solutions, there can be chemical reactions. Knowing the Molarity, or concentration, of these solutions is critical when determining theoretical calculations, percent yield, and limiting reactant.

20. Need to Balance the Coefficients:
Write the equation for this problem: Hydrochloric acid reacts with Calcium Hydroxide to produce…
_HCl + _Ca(OH)2 
Need to Balance the Coefficients:
_CaCl2 +_H2O

21. 2 HCl + 1 Ca(OH)2  1 CaCl2 +2 H2O
If 255 mL of a 1.75 M HCl solution was used, determine how many grams of calcium chloride would be produced:

22. _H3PO4 + _Zn  _Zn3(PO4)2 + _H2
If 55.0 grams of zinc are going to completely react, how many Liters of 0.75 M Phosphoric acid solution are required?