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The order states: D5W 500 mL plus KCl 20 mEq at 42 mL/h IV. KCl is supplied in a 20-mL ampule containing 40 mEq. How many milliliters of KCl will be added to the 500 mL of D5W?

Answer :

To determine the amount of KCl to be added, we start by calculating the concentration of KCl in the ampule. The ampule contains [tex]$40$[/tex] mEq of KCl in a volume of [tex]$20$[/tex] mL, so the concentration is given by

[tex]$$
\text{Concentration} = \frac{40\text{ mEq}}{20\text{ mL}} = 2\text{ mEq/mL}.
$$[/tex]

Next, the order requires [tex]$20$[/tex] mEq of KCl. To find the volume of KCl solution needed, we use the concentration:

[tex]$$
\text{Volume} = \frac{20\text{ mEq}}{2\text{ mEq/mL}} = 10\text{ mL}.
$$[/tex]

Thus, [tex]$10$[/tex] mL of the KCl solution should be added to the [tex]$500$[/tex] mL of D5W.

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