1. CONCENTRATION OF
SOLUTIONS

2. Concentration = amount of solute per
quantity of solvent
Mass/volume % = Mass of solute (g) x 100%
Volume of solution (mL)
CONCENTRATION AS A MASS/VOLUME PERCENT
Usually for solids dissolved in liquids

3. SAMPLE PROBLEM:
2.00mL of distilled water is added to 4.00g of a
powdered drug. The final volume is 3.00mL. What is the
concentration of the drug in g/100mL of solution? What
is the percent (m/v) of the solution?
CONCENTRATION AS A MASS/VOLUME PERCENT

4. = (4.00g) x 100%
(3.00mL)
CONCENTRATION AS A MASS/VOLUME PERCENT
SAMPLE PROBLEM:
2.00mL of distilled water is added to 4.00g of a powdered drug. The final volume
is 3.00mL. What is the concentration of the drug in g/100mL of solution? What is
the percent (m/v) of the solution?
Mass/volume % = Mass of solute x 100%
Volume of solution
Therefore the mass/volume percent is 133%
= 133%

5. 1.7% = Mass of solute x 100%
(2000mL)
SAMPLE PROBLEM:
Many people use a solution of Na3PO4 to clean walls before
putting up wallpaper. The recommended concentration is 1.7%
(m/v) . What mass ofNa3PO4 is needed to make 2.0L of solution?
CONCENTRATION AS A MASS/VOLUME PERCENT
Mass/volume % = Mass of solute x 100%
Volume of solution
Mass of solute = 34g
Therefore the mass required is 34g

6. CONCENTRATION AS A MASS/MASS PERCENT
Concentration = amount of solute per
quantity of solvent
Mass/Mass % = Mass of solute (g) x 100%
Mass of solution (g)
Usually for solids dissolved in liquids

7. Mass/Mass % = Mass of solute x 100%
Mass of solution
SAMPLE PROBLEM:
Calcium chloride, CaCl2, can be used instead of road salt to melt the
ice on roads during the winter. To determine how much calcium
chloride had been used on a nearby road, a student took a sample of
slush to analyze. The sample had a mass of 23.47g. When the
solution was evapourated, the residue had a mass of 4.58g (assume
that no other solutes were present).
a) What was the mass/mass percent of calcium chloride in the slush?
b) How many grams of calcium chloride were present in 100g solution?
CONCENTRATION AS A MASS/MASS PERCENT

8. Mass/Mass % = Mass of solute x 100%
Mass of solution
= (4.58g) x 100%
(23.47g)
= 19.5%
Therefore the mass/mass percent of calcium chloride is
19.5%
SAMPLE PROBLEM:
Calcium chloride, CaCl2, can be used instead of road salt to melt the ice on roads during the winter. To
determine how much calcium chloride had been used on a nearby road, a student took a sample of
slush to analyze. The sample had a mass of 23.47g. When the solution was evapourated, the residue had
a mass of 4.58g (assume that no other solutes were present).
a) What was the mass/mass percent of calcium chloride in the slush?
CONCENTRATION AS A MASS/MASS PERCENT

9. If your mass/mass % of CaCl2 is 19.5% and you
have 100g, then you have 19.5g of CaCl2
Ex: 19.5% x 100g = 19.5g
SAMPLE PROBLEM:
Calcium chloride, CaCl2, can be used instead of road salt to melt the ice on roads during the winter. To
determine how much calcium chloride had been used on a nearby road, a student took a sample of
slush to analyze. The sample had a mass of 23.47g. When the solution was evapourated, the residue had
a mass of 4.58g (assume that no other solutes were present).
b) How many grams of calcium chloride were present in 100g solution?
CONCENTRATION AS A MASS/MASS PERCENT

10. Concentration = amount of solute per
quantity of solvent
Volume/Volume % = Volume of solute (mL) x 100%
Volume of solution (mL)
CONCENTRATION AS A VOLUME/VOLUME PERCENT
Usually for liquids dissolved in liquids

11. CONCENTRATION AS A VOLUME/VOLUME PERCENT
SAMPLE PROBLEM:
Rubbing alcohol is sold as a 70% (v/v) solution of isopropyl alcohol in
water. What volume of isopropyl alcohol is used to make 500mL of
rubbing alcohol?
Volume/Volume % = volume of solute x 100%
volume of solution
(70%) = volume of solute x 100%
(500mL)
= 350mL
Therefore the volume is 350mL

12. CONCENTRATION AS PARTS PER MILLION (ppm)
ppm = Mass of solute (g) x 106
Mass of solution (g)
Usually mass/mass relationships BUT does not refer to
the number of particles
Mass of solute (g) = x
Mass of solution (g) 106
g of solution
OR

13. CONCENTRATION AS PARTS PER BILLION (ppB)
ppb = Mass of solute (g) x 109
Mass of solution (g)
To the power of 9 instead…
Mass of solute (g) = x
Mass of solution (g) 109
g of solution
OR

14. CONCENTRATION AS PARTS PER BILLION (ppB)
SAMPLE PROBLEM:
A fungus that grows on peanuts produces a deadly toxin.
When ingested in large amounts, this toxin destroys the liver
and can cause cancer. Any shipment of peanuts that contains
more than 25ppb of this fungus is rejected. A company
receives 20 tonnes of peanuts to make peanut butter. What is
the maximum mass (in g) of fungus that is allowed?

15. CONCENTRATION AS PARTS PER BILLION (ppB)
SAMPLE PROBLEM:
A fungus that grows on peanuts produces a deadly toxin. When ingested in large amounts,
this toxin destroys the liver and can cause cancer. Any shipment of peanuts that contains
more than 25ppb of this fungus is rejected. A company receives 20 t of peanuts to make
peanut butter. What is the maximum mass (in g) of fungus that is allowed?
ppb = Mass of fungus (g) x 109
Mass of peanuts (g)
Convert 20 t to grams:
20t x 1000kg/t x 1000g/kg = 20 x 106
g
25ppb = Mass of fungus x 109
20 x 106
g
25ppb = Mass of fungus x 103
20
500 = Mass of fungus
103
0.5g = Therefore the maximum mass of the
fungus allowed is 0.5g

16. MOLAR CONCENTRATION

17. MOLAR CONCENTRATION
The number of moles of solute in 1L of solution
(this is the standard measure of concentration in chemistry!!!)
Molar concentration (mol/L) = moles of solute
Volume of solution (L)
C = n
V
Simplified…

18. MOLAR CONCENTRATION
SAMPLE PROBLEM:
A saline solution contains 0.90g of NaCl dissolved in 100mL
of solution. What is the molar concentration of the solution?
C = n
V
Given:
m = 0.90g
n = m/M = 0.90g / 58.44g/mol = 0.0154mol
V = 100mL = 0.1L
C = (0.0154mol)
(0.1L)
= 0.154mol/L
Therefore the molar concentration of the saline solution is 0.15 mol/L

19. MOLAR CONCENTRATION
SAMPLE PROBLEM:
At 20°C, a saturated solution of CaSO4 has a concentration of
0.0153mol/L. A student takes 65mL of this solution and
evaporates it. What mass (in g) is left in the evapourating
dish?
C = n
V
Given:
C = 0.0153mol/L
V = 65mL = 0.065L
n = ? 0.0153mol/L = n
(0.065L)
= 0.000994mol
NOT DONE YET! Convert to grams!
MCaSO4
= 136.15g/mol
mCaSO4
= n x M
= 0.000994mol x 136.15g/mol
= 0.135g
Therefore 0.14g of
CaSO4 are left in the
evapourating dish