T20: Mixed types from T01-T20
50. Pharmacokinetics 50.1) PK Basic parameters 50.1.3) Area Under Curve (AUC)
50.1.3.1) AUC basic concepts Normal 1
Total tried: Correct: Wrong:
If a patient weighing 103 Kg is given 39.861 mg of drug as an IV bolus, then determine the concentration of the drug in the blood. For this drug, the volume of distribution is 0.9 L/Kg.
Click on the button below to see the answer and explanations
Notice: Undefined variable: input_value1 in /home/jshzub/web/rxcal.org/public_html/practice/practice/includes/body.php on line 220
lb equals 0.43 mg kg
-
Total volume of distribution (V) for this patient will be:
`0.9 \quad L/(Kg) × 103 \quad Kg = 92.7 \quad L`
Drug quantity (`X_{0}`) =39.861 mg
Concentration (`C_{0}`) = ?
We know that,
`C_{0}=X_{0}/V`
`therefore C_{0} = (39.861 \quad mg)/ (92.7 \quad L) = 0.43 \quad (mg)/L`. Ans.