Chapter 2, Lesson 1: Properties of Solutions
In this lesson you must be familiar with the following terminologies and computations …. (please refer to your
textbook pp. 48 – 62)
Homogeneous Mixtures
Solute
Solvent
Soluble / Insoluble
Miscible / Immiscible
Aqueous solution
Electrolyte / Nonelectrolyte
Gaseous solution
Liquid solution
Solid solution
Concentration …. Dilute / Concentrated
Parts per million
ppm
=
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡
𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
× 106
Example: If 100 L of a gas mixture over a metropolitan area contains 0.0060 L of CO,
what is the concentration in ppm of CO?
Solution:
I Given:
volume of component CO = 0.0060 L
total volume of solution = 100 L
II Required:
ppm of CO = ?
III Computation:
ppm CO =
IV Answer:
0.0060 𝐿
100 𝐿
× 106 = 60 ppm
ppm CO = 60 ppm
Mass or volume percent
Mass % of component
=
Volume % of component
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑖𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
=
× 100%
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑖𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
× 100%
Proof = Mass % of component × 2
or
Proof = Volume % of component × 2
Example: 15 mL of alcohol is mixed with 85 mL of water. Find a) volume % of the solution and
b) proof of the solution.
Solution:
I Given:
volume of alcohol = 15 mL
Volume of water = 85 mL
II Required:
a) volume %
b) proof of the solution
III Computation:
Total volume of solution = 15 mL alcohol + 85 mL water = 100 mL
Volume % of component
IV Answers:
Proof = 15% × 2 = 30
a) volume % = 15%
=
15 𝑚𝐿
100 𝑚𝐿
× 100
b) Proof = 30
= 15%
Mole fraction
Mole fraction
=
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡
𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Molarity (M)
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
=
Molarity
𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Molality (m)
Molality
=
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑘𝑖𝑙𝑜𝑔𝑟𝑎𝑚 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Example: A solution is prepared by dissolving 4.00 g of NaOH in 100.00 g of water. The volume
of the resulting solution is 102.00 mL. Calculate the following:
a) Mole fraction of NaOH b) molarity of the solution
c) molality of the solution d) mass percent of NaOH
Solution:
I Given:
weight NaOH = 4.00 g
Weight water = 100.00 g
Volume of solution = 102.00 mL = 0.102 L
II Required:
a) Mole fraction of NaOH b) molarity of the solution
c) molality of the solution d) mass percent of NaOH
III Computation:
a) Volume of solution in liters = 102.00 mL ÷ 1000 = 0.102 L
Weight of solution = weight NaOH + weight water = 4.00 g + 100.00 g = 104.00 g
Weight of solution in kilograms = 104.00 g ÷ 1000 = 0.104 kg
atomic weight Na (from periodic table) = 23 g
atomic weight O (from periodic table) = 16 g
atomic weight H (from periodic table) = 1 g
weight per mole NaOH = 23 + 16 + 1 = 40 g / mole
weight per mole H2O = (1 × 2) + 16 = 18 g / mole
1 𝑚𝑜𝑙𝑒
number of moles NaOH in 4.00 g NaOH = 4.00g ×
40 𝑔
number of moles H2O in 100.00 g H2O = 100.00g ×
Mole fraction NaOH
Molarity
=
Molality
=
=
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓𝑁𝑎𝑂𝐻
𝑡𝑜𝑡𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓𝑠𝑜𝑙𝑢𝑡𝑒 (𝑁𝑎𝑂𝐻)
𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓𝑠𝑜𝑙𝑢𝑡𝑒 (𝑁𝑎𝑂𝐻)
𝑘𝑖𝑙𝑜𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Mass % of component
=
=
=
=
0.10
0.102
0.10
0.104
1 𝑚𝑜𝑙𝑒
=
18 𝑔
=
=
0.10 mol
5.56 mol
0.10
0.10 + 5.56
=
=
0.018
0.98 mol / L or 0.98 M
=
1.0 mol/kg
𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
104 𝑔
5.66
=
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 (𝑁𝑎𝑂𝐻)𝑖𝑛 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
4.00 𝑔
0.10
or
1m
× 100%
× 100% = 3.85%
Answers: a) Mole fraction of NaOH = 0.018 b) molarity of the solution = 0.98 M
c) molality of the solution = 1 m d) mass percent of NaOH = = 3.85%
Dilution
Stock solution
Titration
Number of moles before dilution = number of moles after dilution
M1V1
= M2V2
Where M1
= molarity of initial solution
V1
= volume of initial solution
M2
= molarity of final solution
V2
= volume of final solution
Example: How would you prepare a 0.1 M HCL solution from 10 mL of a 0.5 M HCl
stock solution.
Solution:
I Given:
M1
= 0.5 M
V1
= 10 mL
M2
= 0.1M
II Required:
V2
=?
III Computation:
M1V1 = M2V2
or
M2V2 = M1V1
V2
= (M1V1 ) ÷ M2 = (0.5 M × 10 mL) ÷ 0.1 M = 5 ÷ 0.1 = 50 mL
****