### 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

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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

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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.

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