1. Molarity
2. Molarity (M) this is the most common expression of concentration
M = molarity = moles of solute = mol
liters of solution L
Units are mol L-1 or mol/L (moles per liter)
A solution that is 1.0 molar (1.0M ) contains 1.0 mol of solute per liter of solution
M = molarity (mol L-1)
n = number of moles (mol)
V = volume (L) n
M V
M = n/V (mol L-1)
n = M  V (mol)
V = n /M (L)

2. Example 1:
Calculate the molarity of a solution made by dissolving 11.5g of solid NaOH in enough water to make 1.50 L of solution.
Step 1: Find the # of moles of NaOH
n = m = g
MM g mol-1
= mol
Step2: Find the molarity
M olarity = moles of solute = 0.288mol
liters of solution 1.50 L
= M NaOH

3. Example 2:
Calculate the molarity of a solution made by dissolving 1.56g of gaseous HCl in enough water to make 26.8 mL of solution.
Step 1: Find the # of moles of HCl
n = m = g
MM g mol-1
= mol HCl = 4.27 10-2 mol HCl
Step2: Convert milliliters to Liters 26.8mL/1000L = L
= 2.68  10-2 L
Step3: Find the molarity
M olarity = moles of solute = 4.27 10-2 mol
liters of solution  10-2 L
= 1.59 M HCl

4. Example 3: Give the concentrations of all the ions in each of the following solutions:
a M Co(NO3)2 b. 1 M FeCl3
a. Co(NO3)2 (s)  Co2+ (aq) + 2NO3- (aq)
1 mol Co(NO3)2 (s)  1 mol Co2+ (aq) + 2 mol NO3- (aq)
Therefore a solution that is 0.5 mol Co(NO3)2 contains 0.50 M Co2+ and 1 M NO3-
b. FeCl3 (s)  Fe3+ (aq) + 3Cl- (aq)
1 mol FeCl3 (s)  1 mol Fe3+ (aq) + 3 mol Cl- (aq)
Therefore a solution that is 1 M FeCl3 contains 1 M Fe3+ and 3 M Cl-

5. Example 4: How many moles of Ag+ ions are present in 25mL of a 0
Example 4: How many moles of Ag+ ions are present in 25mL of a 0.75 M AgNO3 solution?
Molarity of solution = 0.75 M
Volume of solution = 25mL
moles of Ag+ ions = ?
A 0.75 M AgNO3 solution contains 0.75 M Ag+ ions and 0.75 M NO3- ions
Convert mL to L 25mL / 1000L = 2.5  10-2 L
Molarity = n or n = Molarity  Volume V
(2.5  10-2 L)  (0.75 M Ag+)
= 1.9  10-2 mol Ag+

6. Calculating mass from Molarity
A chemist is analyzing the alcohol content of a wine from Napa. To do this she needs to use 1.00L of an M K2Cr2O7 solution. (molar mass of K2Cr2O7 = g mol-1)
How much solid K2Cr2O7 must be weighted out to make this solution?
Molarity of solution = M
Volume of solution = 1.00 L
grams of K2Cr2O7 = ?
Step 1: Calculate the moles of K2Cr2O7 present.
Molarity = n or n = Molarity  Volume V
= (1.00 L)  (0.200 M K2Cr2O7) = mol K2Cr2O7

7. Step 2: Convert moles into grams
n = mass or mass = Molar mass  n
molar mass
= (0.200 mol K2Cr2O7)  (294.2g K2Cr3O7) = 58.8g K2Cr2O7

8. Making a Standard Solution
A standard solution is a solution whose concentration is accurately known.
This done by weighing out a sample of solute, transferring it completely to a volumetric flask, and adding enough solvent to bring the volume up to the mark on the neck of the flask.

9. Formulae for solution problems
n = number of moles (mol)
m = mass (g)
MM = molar mass (g mol-1)
M = Molarity (mol L-1)
n = number of moles (mol)
V = volume (L)

10. Dilution
To save space in out Prep room we buy solutions in concentrated form, i.e. 18M HCl (18 mol L-1). We call these stock solutions.
The process of adding more solvent to a solution is called dilution.
A typical dilution involves determining how much water must be added to an amount of stock solution to achieve a solution of the desired concentration.
The amount of solute after dilution = moles of solute before dilution
Remains constant
M = moles of solute volume (L)
Decreases
Increases (water added)

11. Example: We want to prepare 500. mL of 1
Example: We want to prepare 500. mL of 1.00 M acetic acid, CH3COOH, from a 17.5 M stock solution of acetic acid. What volume of the stock solution is required?
(b) water is added to dilute it
(c) final diluted solution has a volume of 0.5L and a molarity of 1 M acetic acid
(a) What volume of 17.5 M acetic acid do you have to remove?

12. Example: We want to prepare 500. mL of 1
Example: We want to prepare 500. mL of 1.00 M acetic acid, CH3COOH, from a 17.5 M stock solution of acetic acid. What volume of the stock solution is required?
Step1: Find the number of moles of acetic acid needed in final solution
n = molarity of dilute solution  volume of dilute solution
n = 1.00 molL-1  0.5 L
n = mol CH3COOH
Step2: Find the volume of 17.5 M acetic acid that contain mol of CH3COOH. We will call this unknown volume V.
V= moles of solute = mol CH3COOH Molarity mol L-1
V = L or 28.6 mL

13. (a) 28.6 mL of 17.5 M acetic acid solution is transferred to a volumetric flask that already some water.
(b) water is added to the flask up to the 500 mL mark.
(c) The final solution is 1.00 M acetic acid

14. M1  V1 = moles of solute = M2  V2
Because the moles of solute remain the same before and after dilution, we can write
Final Conditions
Initial Conditions
M1  V1 = moles of solute = M2  V2
Volume before dilution
Volume after dilution
Molarity before dilution
Molarity after dilution
17.5 M  L = moles of solute = 1.0 M  0.5 L
17.5 M  L = moles of solute = mol
1.0 M  0.5 L = moles of solute = mol
M1  V1 = moles of solute = M2  V2

15. Use this equation to solve this problem: M1  V1 = M2  V2
What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H2SO4 solution?
M1 = 16 molL-1 M2 = 0.10 molL-1
V1 = ? V2 = 1.5 L
= 9.4  10-3 L
= 9.4 mL