1. Molar Concentrations

2. Molarity is the number of moles of solute that can dissolve in 1 L of solution.
Solute= a solid that will be dissolved
Solvent= the liquid doing the dissolving
Molar concentration (mol/L) =
A 1.00 molar (1.00 M) solution contains 1.00 mol solute in every 1 liter of solution.
Units of molarity are: mol/L = M
Amount of solute (mol)
Volume of solution (L) n C V

3. Steps involved in preparing solutions from pure solids
Calculate the amount of solid required
Weigh out the solid
Place in an appropriate volumetric flask
Fill flask about half full with water and mix/shake.
Fill to the mark with water and invert to mix/shake.
You should be able to describe this process (including calculating the mass of solid to use) for any solution I specify.

4. Preparing a 1.0 Molar Solution
One liter of a 1.00 M NaCl solution
need 1.00 mol of NaCl
weigh out 58.5 g NaCl (1.00 mole) and
add water to make 1.00 liter (total volume) of solution.
Copyright © by Pearson Education, Inc.
Publishing as Benjamin Cummings

5. Steps involved in preparing solutions from pure solids

6. Step 1: Find the number of moles of calcium chloride using n=m/M n=
Ex#1: 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 g
g/ mol

7. 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 divide by 1000
0.252mol
0.225 L

8. Step 1: Find the number of moles using n= cV n= (6.00 mol/L)(0.425L)
Ex#2: 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

9. nAgNO3 = m/MM C = 0.100 mol/L mAgNO3 = 1.20 g V = ? Ex #3:
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 x 1000 = 70.6 mL

10. Ex #4: 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. Example #5 C = ? mNaCl = 14.0 g Vsolution = 250 mL
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

14. Example#6
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