Read Section 5.5 before viewing the slide show.
Unit 20
Solution Terminology (5.5)
•Definition of Solution
•Terms Associated with Solutions
•Concentrations of Solutions – Molarity and % by
Mass
Definition and Types of Solutions
•A solution is a homogeneous mixture of two or more components. Solutions are
typically recognized to have two parts:
•Solvent – the component present in the largest quantity
•Solute – all of the components that are not the solvent
•Solutions can be made from combinations of any of the three states of matter:
State of
Solvent
State of
Solute
Example
gas
gas
air
liquid
gas
dissolved gases in water, such as oxygen
liquid
liquid
alcohol in water
liquid
solid
sugar in water
solid
gas
hydrogen in a metal
solid
liquid
mercury in silver
solid
solid
silver in gold
Terms Associated with Solutions
•The quantity of solute in a solution may vary over a wide range. Terms have come
into use to describe qualitatively the levels of solute dissolved.
•Dilute solution – does not contain very much solute per quantity of solution
•Concentrated solution – has a large amount of solute per quantity of solution
•Saturated solution – contains all of the solute the given amount of solvent can hold at the
temperature and pressure being considered
•Unsaturated solution – contains less solute than the solvent could hold at the temperature and
pressure being considered
•Supersaturated solution – solvent contains more solvent than it can at the saturated level. These
solutions are unstable and the solute will come out of solution if disturbed.
•Pictorially
Quantitative Solution Expressions
•As useful as it is to have qualitative descriptors of solution concentrations, numeric
descriptors are much more helpful in describing chemical processes.
•Three primary expressions we will consider:
Molarity 
# mol solute
# L solution
% by volume 
% by mass 
(abbreviated as M )
mass solute in g
100%
volume of solution in mL
mass solute
100%
mass solution
•Each has the form of mass or mol of solute divided by mass or volume of solution
•Each can serve as a conversion factor for conducting calculations
Molarity
•Consider molarity first:
Molarity 
# mol solute
# L solution
(abbreviated as M )
•Examples:
•Suppose 15.0-g of NaCl is dissolved in 250 mL of solution. Find the molarity.
•First need to find the number of moles of solute, NaCl. That is given by:
# mol NaCl 
15.0 g NaCl 1 mol NaCl

 0.256 mol NaCl
1
58.5 g NaCl
•The molarity then is given by:
Molarity 
# mol solute 0.256 mol NaCl

 1.03 M NaCl
# L solution 0.250 L solution
•How many mol of NH3 are contained in 75.0 mL of 0.450 M NH3?
•Try the conversion factor approach where 0.450 M NH3 can be thought of as 0.450 mol NH3 per 1 L solution.
# mol NH 3 
0.075 L solution 0.450 mol NH 3

 0.0338 mol NH 3
1
1 L solution
% by Volume
•Next consider % by volume:
% by volume 
volume of solute
100%
volume of solution
Examples:
•Suppose 15.0-mL of ethyl alcohol is dissolved in 250 mL of solution. Find the % by volume.
•This becomes a simple substitution:
% by volume 
volume solute
15.0 mL ethyl alcohol
100% 
100%  6.0% NaCl
volume of solution
250 mL solution
•How many mL of ethyl alcohol are contained in 75.0 mL of 15.5% by volume ethyl
alcohol?
• Try the conversion factor approach where 15.5% by volume of ethyl alcohol can be thought of as 15.5 mL of
ethyl alcohol per 100 mL solution – that is what a percent amounts to.
# g NH 3 
75 mL solution 15.5 mL ethyl alcohol

 11.6 mL ethyl alcohol
1
100 mL solution
% by Mass
•And finally consider % by volume:
% by mass 
mass solute
100%
mass of solution
Examples:
•Suppose 15.0-g of NaCl is dissolved in 250 g of water. Find the % by mass.
•Notice the denominator is the mass of solution – it has both the solute and solvent masses included so the two
masses must be added together for a total solution mass of 15.0 g + 250 g = 265 g of solution.
% by mass 
mass solute
15.0 g NaCl
100% 
 100%  5.66% by mass NaCl
mass of solution
265 g solution
•How many g of NH3 are contained in 75.0 g of a solution that is 15.5% by mass NH3?
• Try the conversion factor approach where 15.5% by mass NH3 can be thought of as 15.5 g NH3 per 100 g
solution – that is what a percent amounts to.
# g NH 3 
75 g solution 15.5 g NH 3

 11.6 g NH 3
1
100 g solution