The most direct approach to conclude solving the problem would be to simply
express the given information as weight per volume and to evaluate the numerical data
-------- = 20.0 g/dL or equivalently, 20.0
Ratio and proportion could also be used to determine the concentration of the
solution obtaining the same results.
-------- = ----
(50.0 g)(1 dL) = (2.50 dL)(x g)
(50.0 g)(1 dL)
x g = ------------= 20.0 g
Substituting for x in the original expression yields the desired results.
------ = -------- = 20.0 g/dL
Some salts come in several forms, such as the anhydrous form (no water) and in
the form of one or more hydrates. In a hydrate, a number of water molecules are
attached to each molecule of salt. The water that is attached to the salt contributes to
the molecular weight. Thus, if we were to weigh out equal amounts of a desired
chemical and one of its hydrates, the hydrate would not yield as much of the desired
chemical, per unit weight, as the anhydrous form, since some of the weight is
attributable to the water molecules. In some cases, the prescribed form of a salt may
not be available for the preparation of a solution. One must be able to determine how
much of the available form is equivalent to the quantity of the form prescribed. To do
this, we must first determine the amount of the prescribed form that is needed. Then,
using the molecular weights of both substances involved, use ratio and proportion to
determine the amount of available form needed.
a. Example 1. A procedure requires that 100 mL of a 10.0 solution be
prepared. Only CuSO4.H2O is available. How much of the hydrate is needed to
prepare this solution?