PHYSICAL PROPERTIES
OF SOLUTIONS
Everyday Solutions
Solutions
• A solution is a homogenous mixture of 2 or more substances
• The solute is(are) the substance(s) present in the smaller amount(s)
• The solvent is the substance present in the larger amount
Table 12.1 Types of Solutions
Component 1
Component 2
State of Resulting
Solution
Examples
Gas
Gas
Gas
Air
Gas
Liquid
Liquid
Soda water (CO2 in water)
Gas
Solid
Solid
H2 gas in palladium
Liquid
Liquid
Liquid
Ethanol in water
Solid
Liquid
Liquid
NaCl in water
Solid
Solid
Solid
Brass (Cu/Zn), solder (Sn/Pb)
•
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Saturation
• A saturated solution contains the maximum
amount of a solute that will dissolve in a given
solvent at a specific temperature.
• An unsaturated solution contains less solute
than the solvent has the capacity to dissolve at
a specific temperature.
• A supersaturated solution contains more
solute than is present in a saturated solution at
a specific temperature.
Sodium acetate crystals rapidly form when a seed
crystal is added to a supersaturated solution of
sodium acetate.
Solutions
• A concentrated solution has a relatively
large quantity of dissolved solute(s)
• A dilute solution has only a small
quantity of solute
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Formation of a Solution
• Three types of interactions in the solution process:
• solvent-solvent interaction
• solute-solute interaction
• solvent-solute interaction
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∆𝐻soln = ∆𝐻1 + ∆𝐻2 + ∆𝐻3
Molecular view of the formation of solution
“Like Dissolves Like”
• Two substances with similar intermolecular
forces are likely to be soluble in each other.
• That is:
•
non-polar molecules are soluble in non-polar
solvents
•
•
polar molecules are soluble in polar solvents
•
•
Example: CCl4 in C6H6
C2H5OH in H2O
ionic compounds are more soluble in polar
solvents
•
NaCl in H2O or NH3(l)
Predicting solubility
• Br2 in CCl4 and in water
• KCl in benzene and in ethanol
• Formaldehyde (CH2O) in CO2 and in water
Concentration Units
• The concentration of a solution is the amount of solute present in a given quantity
of solvent or solution.
Percent by Mass
mass of solute
% by mass =
x 100%
mass of solute + mass of solvent
=
mass of solute
x 100%
mass of solution
Practice Problem
A sample of 6.44 g of naphthalene (C10H8) is dissolved in 80.1 g of benzene (C6H6).
Calculate the percent by mass of naphthalene in this solution.
6.44 g N
6.44 g N +. 8.1 g B
x
100
= 7.44%
Concentration Units
Percent by Volume
volume of solute
% by volume =
x 100%
volume of solute + volume of solvent
volume of solute
=
x 100%
volumeof solution
Concentration Units
Percent by Mass/Volume
mass of solute (g)
% by mass =
x 100%
volumeof solution (mL)
Practice Problem
A solution was prepared by dissolving 1210 mg of K3Fe(CN)6 (329.2 g/mol) in
sufficient water to give 775 mL solution. Calculate the weight/volume percentage of
K3Fe(CN)6.
% mass/vol =
1.21 g
775 mL
= 0.156%
x
100 mL soln./100 mL soln.
Concentration Units
Mole Fraction (X)
moles of A
XA =
sum of moles of all components
Practice Problem
A vapor mixture of 30.0 g of methanol (MM = 32.04 g/mol) and 45.0 g of ethanol
(MM = 46.07 g/mol) is prepared. Calculate the mole fraction of methanol and
ethanol in the vapor.
No. of moles methanol =
30.0 g
= 0.936 mol
32.04 g/mol
No. of moles ethanol
=
45.0 g
x = 0.936
= o.489
1.913
= 0.977 mol
46.07 g/mol
Total no. of moles = 0.936 + 0.977 = 1.913 mol
x = 0.977
1.913
=. 0.511
Molarity (M)
o Or molar concentration is the concentration of a substance in solution,
expressed as the number moles of solute per liter of solution
moles of solute
M=
liters of solution
(unit: mol/L)
e.g. 1 mole of HCl in 1 liter of solution 1 M HCl
Practice Problems
Calculate the molar concentration of ethanol in an aqueous solution that contains
2.30 g of C2H5OH (46.07 g/mol) in 3.50 L of solution.
2.30 g ethanol
= 0.499 mol
46.07 g/mol ethanol
0.499 mol
3.50 L soln.
=
0.0143 mol/L or. 0.0143 M
Practice Problems
What mass of KI is required to prepare 500. mL of a 2.80 M KI solutio
500. mL x
1L
1000 mL
x
2.80 mol KI
1 L soln
x
166 g KI
1 mol KI
= 232 g KI
Molality (m)
o Or molal concentration is the number of moles of solute per kilogram of solvent
moles of solute
m=
kilogram of solvent
Practice Problem
What is the molality of a solution containing 7.78 g of urea [(NH2)2CO] in 203 g of
water?
Molar mass of urea = 60.07 g/mol
Mass of solvent = 203 g =. 0.203 kg
No. of moles urea =
7.78 g
=
0.1295 mol
60.07 g/mol
molality =
0.1295 mol
0.203 kg
= 0.638 mol/kg
or 0.638 m
Example
An ethanol water solution is prepared by dissolving 10.00 mL of ethanol, CH3CH2OH
(ρ = 0.789 g/mL, MM = 46.07 g/mol) in a sufficient volume of water to produce 100.0
mL of a solution with a density of 0.982 g/mL.
What is the concentration of ethanol in this solution expressed as (a) volume
percent; (b) mass percent; (c) mass/volume percent; (d) mole fraction; (e) molarity;
(f) molality?
Dilution
When you dilute a solution, the amount of solute is still the same even if the
solution volume changed due to the added solvent.
moles of solute before dilution = moles of solute after dilution
Dilution
We can prepare a known volume of a dilute solution from a stock solution
with known concentration.
C1V1 = C2V2
Example
What volume of 5 M NaOH is needed to create a 100 mL solution of 1 M
NaOH?
c1 V1 = c2 V2
V1 = (1 mol/L) (0.100 L)
5 mol/L
= 0.020 L (20 mL)
0
