The calculated value of 26.5 g/L is now used as a factor to determine the

volume of solution that can be prepared. However, the reciprocal of the value must be

used as a factor. If there are 26.5 grams per liter then for every liter there are 26.5

grams of solute. This relationship is true of all factors we have used.

L

106.0 g X ------ = 4.00 L

26.5 g

Milliequivalent per liter problems are based on the same principle as normality. If

we examine the definition of normality, we will find that a normal solution contains one

gram equivalent weight per liter of solution. Equivalently, we can say that a normal

solution is one equivalent per liter of solution (Eq/L). A milliequivalent per liter solution

would, therefore, contain one milliequivalent weight per liter of solution.

a. A milliequivalent = 0.001 equivalent and 1 equivalent = 1000 milliequivalents.

b. A milliequivalent has the numerical value of one equivalent weight expressed

in mg/mEq.

c. To convert Eq/L to mEq/L and vice versa multiply the concentration by the

conversion factor 1000 mEq/1 Eq as appropriate.

a. **Example 1**. How much NaCl is needed to prepare 500 mL of a 25.0 mEq/L

NaCl solution?

Solution. Read the problem carefully and determine the desired quantity.

Milligrams of NaCl.

Determine the milliequivalent weight of the compound.

NaCl

Na

23.0

C

+ 35.5

58.5 mg/mmol

MD0837

4-8

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