Water is the solvent in mostof the solutions that we encounter. Unless otherwise
indicated, we assume that water is the solvent. When the solvent is other than water, we
state this explicitly.
EXAMPLE 3-17 Molarity
Calculate the molarity (M) of a solution that contains 3.65 grams of HCl in 2.00 liters of
solution.
Plan
We are given the number of grams of HCl in 2.00 liters of solution. We apply the definition
of molarity, remembering to convert grams of HCl to moles of HCl.
Solution
0.0500 mol HCl/L soln
The concentration of the HCl solution is 0.0500 molar, and the solution is called 0.0500 M
hydrochloric acid. One liter of the solution contains 0.0500 mol of HCl.
You should now work Exercise 60.
EXAMPLE 3-18 Mass of Solute
Calculate the mass of Ba(OH) 2 required to prepare 2.50 L of a 0.0600 Msolution of barium
hydroxide.
Plan
The volume of the solution, 2.50 L, is multiplied by the concentration, 0.0600 mol Ba(OH) 2 /L,
to give the number of moles of Ba(OH) 2. The number of moles of Ba(OH) 2 is then multiplied
by the mass of Ba(OH) 2 in one mole, 171.3 g Ba(OH) 2 /mol Ba(OH) 2 , to give the mass of
Ba(OH) 2 in the solution.
Solution
__? g Ba(OH) 2 2.50 L soln
25.7 g Ba(OH) 2
You should now work Exercise 62.
The solutions of acids and bases that are sold commercially are too concentrated for
most laboratory uses. We often dilute these solutions before we use them. We must know
the molar concentration of a stock solution before it is diluted. This can be calculated
from the specific gravity and the percentage data given on the label of the bottle.
EXAMPLE 3-19 Molarity
A sample of commercial sulfuric acid is 96.4% H 2 SO 4 by mass, and its specific gravity is 1.84.
Calculate the molarity of this sulfuric acid solution.
171.3 g Ba(OH) 2
1 mol Ba(OH) 2
0.0600 mol Ba(OH) 2
1 L soln
1 mol HCl
36.5 g HCl
3.65 g HCl
2.00 L soln
__? mol HCl
L soln
We place 3.65 g HCl over 2.00 L of
solution, and then convert g HCl to
mol HCl.
106 CHAPTER 3: Chemical Equations and Reaction Stoichiometry