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Page Title:
Advantages and Disadvantages of Percent Solutions |
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  b. Example 2. How many milligrams of Na2CO3 are needed to prepare 100 mL
of a 10.0 mg/dL sodium standard?
Solution. Read the problem carefully and determine the unknown quantity.
Milligrams of Na2CO3.
Calculate the amount of the sodium needed.
1 dL
10.0 mg
100 mL X --------X --------= 10.0 mg
100 mL
1 dL
Calculate the gram molecular weight of each substance.
Na 23.0 X 2 = 46.0
Na 23.0 X 2 = 46.0
C
12.0 X 1 = 12.0
46.0 g/mol
O
16.0 X 3 = + 48.0
106.0 g/mol
Use ratio and proportion to determine the amount of available form needed.
46.0 g/mol Na
10.0 mg Na
------------------------ = --------------
106.0 g/mol Na2CO3
x mg Na2CO3
(106.0 g/mol NaCl) (10.0 mg Na)
x mg Na2CO3 = --------------------------------
46.0 g/mol Na
x mg Na2CO3 = 23.0 mg
2-33. ADVANTAGES AND DISADVANTAGES OF PERCENT SOLUTIONS
The main advantage of percent solutions is the ease in calculations. There are,
however, two distinct disadvantages associated with percent solutions: (1) the
technician does not know the number of molecules or ions that are present in the
solution, and (2) the technician has no idea of the reacting strength of the solution.
MD0837
2-18
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