1. Solutions and Molarity
Describe the types of solutions.
Define the vocabulary words.
List and explain the factors that influence
solubility and the rate at which a solute
dissolves in a solvent.
Explain and calculate molarity.

2. Types of Solutions
 Gas in a gas – Gases mix freely and will always form a
solution unless they react (Example: air)
 Solid in a solid – Alloy: a homogeneous mixture of metals
(Example: brass is a mixture of copper and zinc)
 Liquid in a liquid
Miscible – when liquids can be mixed together to form a
solution (Example: ethylene glycol and water form
antifreeze)
Immiscible – when liquids cannot be mixed (Example: oil
and water)
“Like dissolves like” – polar dissolves polar
Polar and nonpolar (vinegar and oil) are immiscible

3.  Solid in a liquid
Energy is required to separate particles
of a solid (Endothermic)
Energy is released when solute particles
and solvent particles are attached
(Exothermic)
Goes back and forth: Dynamic
equilibrium
Differences are responsible for different
solubilities

4.  Gases in liquids – when gas is attracted
to solvent particles---release energy
Free---move toward entropy
Examples: CO2 in soda and O2 in
seawater
 Entropy (S) – disorder or randomness
Systems tend to go from a state of order
(low entropy) to a state of maximum
disorder (high entropy)

5. Solubility Curve
Solubility - quantity of solute that
will dissolve in a specific amount of
solvent at a certain temperature.
(pressure must also be specified for
gases).
-soluble vs. insoluble,
-saturated (on or above) vs.
unsaturated (below)
- solubility should NOT be confused
with the rate at which a substance
dissolves
Reading the curve:
At 30°C approximately 10g of
KClO3 will dissolve in 100g of
water

6. Refer to Solubility Curves
 At 10 °C, 135 grams of KI will dissolve
 At 50 °C, 85 grams of KNO3 will dissolve
 At 30 °C, 42 grams of NH4Cl will dissolve
 At 75 °C, 50 grams of KCl will dissolve

7. Factors Influencing the Rate at
which a Solute Dissolves in a
Solvent
 1. Agitation – stirring; a surface
phenomenon
 2. Particle size
 3. Temperature – the only factor that
affects both the rate of solution and the
solubility

8.  Solid in a liquid
Energy is required to separate particles
of a solid (Endothermic)
Energy is released when solute particles
and solvent particles are attached
(Exothermic)
Goes back and forth: Dynamic
equilibrium
Differences are responsible for different
solubilities

9.  Gases in liquids – when gas is attracted
to solvent particles---release energy
Free---move toward entropy
Examples: CO2 in soda and O2 in
seawater
 Entropy (S) – disorder or randomness
Systems tend to go from a state of order
(low entropy) to a state of maximum
disorder (high entropy)

10. Temperature Effects on
Solubility
 In general, an increase in temperature
increases the solubility of solids in liquids
 In general, an increase in temperature
decreases the solubility of gases in
liquids (gases escape)

11. Pressure Effects
 Pressure increases the solubility of gases
in liquids (nail being hammered into wood)
 Henry’s Law – At a given temperature, the
solubility of a gas in a liquid is directly
proportional to the pressure of the gas
above the liquid (Page 506)
S = solubility S1 S2
=
P = pressure P1 P2

12. Concentration of Solutions
 Dilute – contains relatively little solute
 Concentrated – contains relatively large
amount of solute

13. Vocabulary Words
 Solubility: the amount of a substance that
dissolves in a given quantity of solvent at
specified conditions of temperature and
pressure to produce a saturated solution
 Solute: dissolving particles
 Solvent: the dissolving medium in a
solution (Example- water)

14.  Saturated – a solution that holds as
much solid as it normally can at a given
temperature (If more solid is added, it
will not dissolve)
 Unsaturated – a solution which has not
yet reached the limit of solubility at a
given temperature
 Supersaturated – rare solution that
contains MORE dissolved solute than it
can normally hold at a given temperature
(Crystallization)

15.  Suspension: if the particles are
so large that they settle out
unless the mixture is constantly
stirred or agitated.
suspension
 Colloid: particles that are
intermediate in size between
those in solutions and
suspensions. AKA- colloidal
suspensions
colloid

16. Molarity
 The number of moles of solute dissolved per
liter (1000mL) of solution
 Water: 1 mL = 1 g; 1000 mL = 1 Kg
mol of solute
molarity (M) =
liters/Kg of solution
 Moles of solute = liters of solution x molarity
 Problem: What is the molarity of the solution
obtained by dissolving 90g of glucose (C6H12O6)
in 1000 grams of water?

17.  Answer
 1 mole of glucose = 180 g
 90g ÷ 180 g = .5 mole
 1000 g = 1000 mL
 .5M per 1000 mL water
 Problems: How many grams are needed
to make a molar solution of
a. 1M glucose
b. 2M glucose
c. .5M glucose

18. Answer
 1 mole of glucose = 180 grams
 a. 1M = 180 grams
 b. 2M = 2 x 180 g = 360 grams
 c. .5M = .5 x 180 g = 90 grams

19. Preparing Molar Solutions
 The solute will take up some of the available
space in the volumetric flask.
 Steps
1. The solute should be added to some of the
solvent and dissolved.
2. Then solvent is added to the 1L mark on the
volumetric flask.
 If these steps are not followed, the total volume
of the mixture is likely to exceed the desired
volume.
 A volumetric pipet measures volumes even
more accurately.

20. Making Dilutions
 Dilution reduces the moles of solute per unit
volume, however, the total moles of solute in
solution does not change.
 Moles of solute = molarity (M) x liters of solution
(V)
 Moles of solute = M1 x V1 = M2 x V2
 Problem: How many milliliters of a stock
solution of 2.00M MgSO4 would you need to
prepare 100.0 mL of 0.400 M MgSO4?

21. Answer
 0.400M x 100.0 mL ÷ 2.00M = 20.0 mL
 Thus, 20.0 mL of the initial solution must
be diluted by adding enough water to
raise the volume to 100.0 mL
 OR
0.400M is 1/5th of 2.00M
1/5th of 100mL is 20 mL

22. Percent Solutions
 If both the solute and solvent are liquids,
a convenient way to make a solution is
to measure volumes.
Example: 20 mL of pure alcohol is
diluted with water to a total volume of
100 mL – The concentration of alcohol is
20% (v/v)
 A commonly used relationship for
solutions of solids dissolved in liquids is
percent (mass/volume).

23. Problems
 1. What is the percent by volume of
ethanol (C2H6O), or ethyl alcohol, in the
final solution when 85 mL of ethanol is
diluted to a volume of 250 mL with
water?
 2. How many grams of glucose
(C6H12O6) would you need to prepare
2.0 L of 2.8% glucose (m/v) solution?

24. Answers
 1. 85/250 x 100% = 34% ethanol (v/v)
 2. In a 2.8% solution, each 100 mL of
solution contains 2.8 grams of glucose
100mL is 1/10th of a liter, so you need
28 grams per liter
2 L x 28 g = 56 grams of glucose