Answer :
You would need to give 0.65 mL.
Since the solution concentration is given in g/mL
We have to convert the mass of atropine to g
1000 mg make 1 g
325 mg = 325/1000 * 1 = 0.325 g
Now;
0.50 g is contained in 1 mL
0.325 g should be contained in 0.325 g * 1 mL/0.50 g
= 0.65 mL
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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.
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