1. Concentration of Solutions
Concentration of a Solution- amount of solute present in a given quantity of solvent or solution.
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Concentrated- solution that contains large amounts of solute
Dilute- solution contains relatively small amounts

2. Concentration of solutions (cont.)
Most useful way to describe the concentration is, Molarity (M).
Molarity can also be called Molar Concentration.
Ex M = 6-molar solution
Molarity is the number of moles of solute per liter of solution.

3. Concentration
The measure of how much solute is dissolve in a specific amount of solvent or solution.
Concentrated large amount of solute high Molarity
Diluted small amount of solute low Molarity

4. Molarity = Molarity ( M ) moles solute liters of solution
The number of moles of solute dissolved in per liter of a solution.
Molarity = moles of solute = mol
liters of solution L
Volume must be converted to Liters (L)
Molarity ( M ) =
moles solute
liters of solution

5. Example: Calculate Molarity
Calculate the molarity of 0.35 moles of sodium sulfate (Na2SO4) dissolved in 1.5 L solution.
Find your equation:
Substitute known values:
M= mol L
0.35 mol Na2SO4
1.5 L solution M=
M= 0.23 M Na2SO4

6. Class Practice
What is the molarity of 65g of KCl if 90 mL of solution are formed?
1) Convert grams to mol
= .87 mol
2) Convert mL to L (90 mL/1000) = 0.09 L
3) Calculate molality = mol/liters = .87mol//.09 L
= 9.67 M
65 g KCl
1 mol KCl
74.6 g KCl
REMEMBER:
M is the same as writing mol L

7. Calculating Molarity
What is the molarity of an aqueous solution containing 40.0 g of glucose (C6H12O6) in a 1500 ml solution?
If given 340 g of CuCl2, what would be the molarity of the solution if it was dissolved in 6425 cm3 of water? Note: (1 cm3 = 1 ml)

8. Practice Problems (to turn in)
1) .25 mol of NaCl is dissolved in mL of water, calculate the molarity.
2) 525 mL of water dissolves 1.5 mol of FeO2, calculate the molarity.
3) What is the molarity of the following solutions?
a) g MgCl2 in 500 mL of solution
b) 2.48 g CaF2 in 375 mL of solution
4) 15.8 g of KCl is dissolved in 225 mL of water. Calculate the molarity.
5) Calculate the volume of 0.30M  KCl solution that contains .05 mol of KCl

9. Molarity and dilutions
Molarity is important for calculating changes in the concentration as a solution is diluted
In dilutions solute stays the same as solvent is added
Diluting a solution reduces the number of solute per unit of volume, but the total number of moles in a solution stays the same.
Molarity of solute
Molarity of solute after dilution
before dilution =

10. Calculating Dilutions
The final solution will have a larger volume but lower concentration, and the moles of te solute present will be the same after as before the dilution.
The relationship can be expressed with the following equation:
M1V1=M2V2

11. Example 1: Calculating Dilutions
Suppose a .2L solution of 15-molar sulfuric acid (H2SO4) was diluted to a volume of 1.0L. What would the resulting concentration be?
1st step identify the variables:
M1= 15M H2SO4 = 15 mol H2SO4 /L
V1= 0.2L
M2= ?
V2=1.0 L

12. Example 1: Calculating Dilutions
Suppose a .2L solution of 15-molar sulfuric acid (H2SO4) was diluted to a volume of 1.0L. What would the resulting concentration be?
2nd step Rearrange 3rd step substitute
the formula: and solve :
M1V1=M2V2
M2= 15 mol H2SO4 /L x 0.2L L
M2= M1V1 V2
M2= 3M H2SO4

13. Example 2: Calculating Dilutions
If I have a container of a concentrated 6.0M sodium hydroxide (NaOH), and I want to prepare 450mL of 0.10M Sodium Hydroxide, How much of the concentrated solution will I need to use?
1st step identify the variables:
M1= 6M NaOH = 6 mol NaOH/L
V1= ?
M2= 0.10M NaOH
V2=450mL= 0.450L

14. Example 2: Calculating Dilutions
If I have a container of a concentrated 6.0M sodium hydroxide (NaOH), and I want to prepare 450mL of 1.0M Sodium Hydroxide, How much of the concentrated solution will I need to use
2nd step Rearrange 3rd step substitute
the formula: and solve :
M1V1=M2V2
V1= 1.0M NaOH x 0.450L
6M NaOH
V1= M2V2 M1
V1= 0.075L or 75ml