A nurse is preparing to administer promethazine 12.5 mg IM. Available is promethazine 25 mg/mL. How many mL should the nurse administer? (Round the answer to the nearest tenth.
Use a leading zero if it applies.
Do not use a trailing zero.). mL.
The Correct Answer and Explanation is:
To determine how many milliliters (mL) of promethazine 25 mg/mL solution are needed to administer a dose of 12.5 mg, follow these steps:
1. Understand the Problem:
You need to administer 12.5 mg of promethazine, and you have a solution where each milliliter contains 25 mg of promethazine.
2. Use the Formula:
To find the volume required, use the formula:
[
\text{Volume (mL)} = \frac{\text{Desired Dose (mg)}}{\text{Concentration (mg/mL)}}
]
In this case:
- Desired Dose = 12.5 mg
- Concentration = 25 mg/mL
3. Substitute the Values into the Formula:
[
\text{Volume (mL)} = \frac{12.5 \, \text{mg}}{25 \, \text{mg/mL}}
]
4. Perform the Calculation:
[
\text{Volume (mL)} = \frac{12.5}{25} = 0.5 \, \text{mL}
]
5. Round the Answer:
The result is 0.5 mL. Since the answer is already at one decimal place, no further rounding is required.
Explanation:
When administering medications, precise dosing is crucial to ensure patient safety and therapeutic efficacy. In this scenario, the nurse needs to calculate how much of a stock solution to use to provide the correct dose of medication. Promethazine is available in a concentration of 25 mg/mL, which means each milliliter of this solution contains 25 mg of the drug.
To determine the amount of solution needed, divide the prescribed dose (12.5 mg) by the concentration of the solution (25 mg/mL). This calculation ensures that the nurse draws up the correct volume from the vial to administer the exact dose required.
In practice, nurses must be meticulous with these calculations and ensure that the medication is administered as prescribed. Using the correct volume not only ensures that the patient receives the appropriate amount of medication but also helps in minimizing potential side effects or adverse reactions.