1. Molar Concentrations

2. Molarity is the number of moles of solute that can dissolve in 1 L of solution.
Molar concentration (mol/L) =
Amount of solute (mol)
Volume of solution (L) n C V

3. Step 1: Find the number of moles of calcium chloride using n=m/M n=
What is the concentration, in mol/L, of a solution formed by dissolving 28.0g of calcium chloride in enough water to make 225 mL of solution?
Step 1: Find the number of moles of calcium chloride using n=m/M n=
= = mol CaCl2 m M
28.0
110.98

4. Step 2: Use c=n/V to calculate the molar concentration c= =
= 1.12 mol/L n V
Notice the volume
has been converted from mL
to L
0.252
0.225 L

5. Solve using 1 step using factor label

6. Step 1: Find the number of moles using n= cV n= (6.00 mol/L)(0.425L)
How many grams of sodium nitrate would be needed to make 425 mL of 6.00 mol/L solution?
Step 1: Find the number of moles using n= cV
n= (6.00 mol/L)(0.425L)
= 2.55 mol of sodium nitrate
Step 2: Calculate how many grams is needed using m = nM
m = (2.55 mol) (85.00 g/mol)
= 217 g of NaNO3

7. Solve in 1 step using factor label

8. What final solution volume would be required to prepare
A mol/L solution of AgNO3(aq) if 1.20 g of the solid salt will be used?
MMAgNO3 = g/mol
C = mol/L
mAgNO3 = 1.20 g
V = ?
nAgNO3 = m/MM
= 1.20 g
169.9g/mol
= mol
V = n/C = mol/0.100 mol/L = L

9. Factor label method

10. Conversion of % to mol/L
Calculate the concentration, in mol/L, of a 96% solution of sulphuric acid which has a density of 1.84 g/mL
Step 1. Calculate the number of moles of sulphuric acid.
n= m/M =
= 0.98 mol of H2SO4
96g
98.0 g/mol

11. Step 2: Find the volume of the solution using the mass and the density.
Recall: V= m/D =
= 54.3 mL
= L of solution m
100 g
1.84 g/mL D V

12. Step 3: Calculate the concentration using c= n/V c = n/V = = 18 mol/L
The concentration is 18 mol/L
0.98 mol of H2SO4
0.0543L

13. Factor label method

14. Examples C = ? mNaCl = 14.0 g Vsolution = 250 mL
Eg1) A Solution of sodium chloride in water contains
14.0 g of NaCl dissolved in 250 mL of solution. What is the
molarity of the solution?
What do we need to
find?
Given?
MMNaCl = g/mol
mNaCl = 14.0 g
Vsolution = 250 mL
C = ?
C = n/V
n = m/MM
= 14.0 g/58.44 g/mol = mol NaCl
C = n/V
= mol/0.250 L = mol/L

15. Factor label method

16. Examples
Eg 2) What mass of sucrose, C12H22O11, is dissolved in
50.0 mL of a mol•L-1 solution of sucrose in water?
Given, MMsucrose = g/mol
C = mol/L
V = 50.0 mL = L !
nsucrose = C x V = mol/L x L = mol sucrose
msucrose = n x MM = mol x g/mol = 6.85 g of
sucrose

17. Factor label method