|Molarity is a way of specifing the amount of solute in one
solvent. Molarity is also known as the concentration
a solution. The concentration of a solution is the ratio of
a given quantity of solvent. We can use whatever units we wish but the
commonly used units are moles of solute per litre of solution.
special name for this ratio is the molar concentration or molarity
which is abbreviated 'M'.
mol solute =
|Suppose we had a bottle with the label "0.5 M KBr".
This means that it contains a solution of potassium bromide with a
concentration of 0.10
mol of potassium bromide per litre of solution (or per 1000 mL of
It does not tell us the amount of solution in the bottle.
|Solutions are considered to be homogenous
substances once the
solute has completely dissolved.
|A student requires 0.250 moles of NaCl for an
only thing available to them is a bottle with a solution labeled "0.400
NaCl." What volume of the solution should be used? Give the
|Use the information on the bottle. There is 0.400
moles of solute
per litre. We need 0.250 moles.
0.400 moles = 0.400 moles = 0.250 moles
x = 625 mL. Thus 625 mL of 0.400 M NaCl contains 0.250 mol of
|A glucose solution with a molar concentration of 0.200 M
is available. What volume of this solution must be measured to
obtain 0.001 mol of glucose?
|Preparation of Solutions|
|Another very common problem involving molar concentration
is the calculation
of the number of grams of solute needed to make a given volume of
having a specific molarity. The best way to see this is by
|Problem: How can 500 mL of 0.150 M Na2CO3
solution be prepared?
|Solution: This is stated in a way that normally arises in
the lab? What we really want to know is how many grams
of Na2CO3 that are going to be in 500 mL of 0.150
M Na2CO3 solution.
Although the label reads in M, the balance reads in grams. Before we can calculate the number of grams we need to know the number of moles.
0.150 M = 0.150 mol = 0.150 mol
x = 0.075 mol of solute. Therefore we need to weigh out 0.075 mol of Na2CO3.
The number of grams of Na2CO3 are therefore:
g = n * mm
|To answer the question: Weigh out 7.95 grams
of sodium carbonate. Add it to 100 mL of water in a 500 mL
volumetric flask. Swirl to dissolve. Top the
flask up with 400 mL more water up to the
500 mL meniscus mark.
|Practise problem: How can we prepare 250 mL of 0.200
Go to the Molarity and Solution Creation Worksheet
Go to the Stoichiometry of Solution Molarity Worksheet