Chemistry Reference
In-Depth Information
Practice Problem 15.5
Calculate the mass of 0.650% solution that
can be prepared by diluting 25.0 g of 3.00% stock solution.
EXAMPLE 15.6
A nurse needs 525 g of 2.00% sterile saline solution (NaCl). He has two ster-
ile stock solutions of NaCl: 5.000% and 0.5000% by mass, but no sterile water.
How can he make the required solution?
Solution
2.00 g NaCl
100.0 g solution
525 g solution
a
b
10.5 g NaCl required
Let
x
the mass of the 5.000% solution.
Then
525
x
the
mass of the 0.5000% solution.
x
(5.000)
100.0
(525
x
) (0.5000)
100.0
10.5
0.05000
x
2.62
5
0.005000
x
10.5
0.0450
x
7.87
5
x
175 g of 5.000% solution
525
x
350 g of 0.5000% solution
Snapshot Review
❒
Percent by mass is the number of grams of solute in exactly 100 g of
solution. Note that unlike molarity (Chapter 11), the denominator of
this ratio is a mass, not a volume.
ChemSkill Builder 15.2
❒
Be sure to distinguish between mass of solvent and mass of solution.
A. Calculate the percent by mass of glucose in a solution containing 11.1 g
of glucose in 100.0 g of water.
B. State how to prepare 125.0 g of a 2.500% dextrose solution in water.
15.4
Molality
The concept of molarity was defined and used in Chapter 11. Molarity is based
on the volume of the solution, which may change with temperature. That is, if
the solution is heated, its volume increases and its molarity decreases, even if
the components of the solution do not change. Because chemists sometimes
need to measure concentrations in experiments in which different temperatures
