Provided that a physician has ordered 325 mg of atropine, intramuscularly and the atropine was available as 0.50 g/mL of solution, the volume of atropine needed to be given in millimeters is 0.65 mL.
From the given information:
The mass of the drug (atropine) ordered for = 325 mg
The density of the available solution of atropine = 0.50 g/mL
The objective is to determine the volume of the atropine that will be given by the physician.
Using the relation for [tex]\mathbf{Density = \dfrac{mass}{volume}}[/tex]
[tex]\mathbf{0.50 \ g/mL= \dfrac{325 \ mg}{volume}}[/tex]
[tex]\mathbf{volume= \dfrac{325 \ mg}{0.50 \ g/mL}}[/tex]
Now, the next step is to make the unit to be similar.
Since the volume is to be determined in milliliters;
We know that:
1 mg = 0.001 grams
∴
325 mg to grams will be = (325× 0.001 grams)
= 0.325 grams
Now;
[tex]\mathbf{volume= \dfrac{0.325 \ g}{0.50 \ g/mL}}[/tex]
[tex]\mathbf{volume= 0.65 \ mL}[/tex]
In conclusion, the volume that will be needed by the physician is 0.65 mL.
Learn more about atropine here:
https://brainly.com/question/17005997?referrer=searchResults
