Allopurinol 115 mg/m² is prescribed to a client who weighs 55 kg and is 148 cm tall. Allopurinol is supplied in vials containing 25 mg/mL after reconstitution.
How much solution should a nurse draw for one dose?
4.6 mL
5.9 mL
66.9 mL
6.49 mL
12.4 mL
The Correct Answer is D
To find the amount of solution to draw for one dose, you need to calculate the following:
- The body surface area (BSA) of the client in square meters (m²) using the formula: BSA (m²) = √(Height (cm) x Weight (kg) / 3600)
- The dose of allopurinol in milligrams (mg) using the formula: Dose (mg) = BSA (m²) x Prescribed dose
(mg/m²)
- The volume of solution in milliliters (mL) using the formula: Volume (mL) = Dose (mg) / Concentration (mg/mL)
First, use the formula for BSA to find the client's body surface area in square meters:
BSA (m²) = √(Height (cm) x Weight (kg) / 3600)
Since the client's height is 148 cm and weight is 55 kg, plug in these values into the formula:
BSA (m²) = √(148 cm x 55 kg / 3600)
Simplify and solve for the BSA:
BSA (m²) = 1.41 m²
Next, use the formula for dose to find the amount of allopurinol in milligrams:
Dose (mg) = BSA (m²) x Prescribed dose (mg/m²)
Since the client's BSA is 1.41 m² and the prescribed dose is 115 mg/m², plug in these values into the formula:
Dose (mg) = 1.41 m² x 115 mg/m²
Simplify and solve for the dose:
Dose (mg) = 162.15 mg
Then, use the formula for volume to find the amount of solution in milliliters:
Volume (mL) = Dose (mg) / Concentration (mg/mL)
Since the dose is 162.15 mg and the concentration is 25 mg/mL, plug in these values into the formula:
Volume (mL) = 162.15 mg / 25 mg/mL
Simplify and solve for the volume:
Volume (mL) = 6.486 mL
Therefore, the nurse should draw **6.486 mL** of solution for one dose of allopurinol.
Nursing Test Bank
Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is B
Explanation
Dextrose 5% in water (D5W) is an IV fluid that contains **5 grams of dextrose** per 100 mL of water². To calculate how many grams of dextrose are in 500 mL of D5W, you can use a simple proportion:
5 g / 100 mL = x g / 500 mL
Cross-multiply and solve for x:
x = (5 g * 500 mL) / 100 mL
x = 25 g
Therefore, there are **25 grams of dextrose** in 500 mL of D5W.
Correct Answer is ["B","C","D"]
Explanation
The three components of communication are the sender, the receiver, and the message. The sender is the person who initiates the communication by sending a message. The receiver is the person who receives the message. The message is the information that is being communicated.
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