Put the ratio in standard form by dividing each side of the ratio by the sample
---- ---------- = 1:8
The dilution of this specimen is 1:8 and if this information were to be used in
calculating the concentration of the specimen, the concentration of the diluted sample
would be multiplied by 8 in order to determine the concentration of the original,
Dilutions and dilution factors have no unit of report, because the units divide
b. Example 2. A 5 mol/L standard solution has been diluted 1:5, what is the
concentration of the dilute solution?
Solution. Read the problem carefully, and determine the unknown quantity.
The unknown quantity is the concentration of the dilute solution. In order to solve this
type of problem you must express the dilution as a fraction. If you have a fraction of a
number, as is the case with this problem, you are in effect multiplying the fraction times
the original number.
5 mol/L X -- = 1 mol/L
c. Example 3. A glucose determination was performed on a serum specimen
that was diluted 1:10. The concentration of the diluted specimen was determined to be
73 mg/dL. What is the concentration of the undiluted serum?
Solution. Read the problem carefully and determine the unknown quantity. In
this particular instance you need to determine the concentration of a serum specimen,
based upon a glucose determination that was performed on a diluted sample.
The dilution factor is the reciprocal of the dilution. This is the factor by which
the quantitative results are multiplied to determine the concentration of the undiluted
(73 mg/dL)(10) = 730 mg/dL