CLINICAL LAB SCIENEC

CHAPTER 9: LABORATORY MATHEMATICS 219
A
Problem:
Solution:
Prepare 500 mL of 0.85% saline.
1. A 0.85% solution contains 0.85 g of the solute in every 100 mL of solution.
2. Therefore, to prepare 500 mL, 5 × 0.85 g, or 4.25 g, of sodium chloride (NaCl)
must be used.
3. To prepare the solution:
a. Weigh out 4.25 g of NaCl.
b. Fill a 500 mL volumetric flask approximately half full with water.
c. Add 4.25 g of NaCl and swirl gently to dissolve.
d. Add water to the flask's fill line.
B
Problem:
Solution:
Prepare 500 mL of 10% bleach solution.
1. A 10% solution of bleach contains 10 mL bleach (hypochlorite) per 100 mL
of solution.
2. Therefore, 500 mL of solution would contain 50 mL of bleach (5 × 10 mL).
3. To prepare the solution:
a. Place 450 mL of water into a flask or bottle.
b. Add 50 mL of bleach.
c. Carefully mix; label the container.
FIGURE 9-4 (A) Percent solution as a weight-to-volume measurement; (B) percent solution as a volume-tovolume
measurement.
Source: Delmar/Cengage Learning.
Problem:
Solution:
Explanation:
Prepare one liter of a 2 M solution of NaOH.
The gram molecular weight of NaOH is 40. A 2 M solution of NaOH contains 40 X 2 g/L.
One liter of a 2 M solution contains 80 grams NaOH in one liter of solution
As described previously, 1 gmw is 1 mole of a substance. For sodium hydrox-ide,
1 mole equals 40 grams as derived by adding the molecular weights for the three
atoms: 1 mole of sodium (Na) is 23, 1 mole of oxygen (O) is 16 grams, and
1 mole of hydrogen (H) is 1.
Solution: 80 g NaOH/40 (gmw) = 2.0 mol/L or 2 molar solution or 2 M
FIGURE 9-5 Preparation of a 2 M solution of NaOH (sodium hydroxide).
Source: Delmar/Cengage Learning.
Molal Solutions
Molality is sometimes confused with molarity. Molality is always used as weight
per weight, where a relationship is established between the weight of the solute
and the weight of the solvent. Therefore, molality is unlike molarity as it is always
expressed as moles per kilogram of solvent, or mol/kg.
Normal Solutions
Normality differs from molarity and molality by using the term equivalent
weight, which is the gram molecular weight divided by its valence, or the number
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.