T20: Mixed types from T12-T20

50. Pharmacokinetics 50.1) PK Basic parameters 50.1.3) Area Under Curve (AUC) AUC basic concepts Normal 1

Total tried:       Correct:       Wrong:

322.04 mg of a drug (elimination rate constant of 0.019 h-1) is administered as an IV bolus to a patient. The weight of the patient is 194 Kg. If the volume of distribution of the drug is 1 L/Kg, then calculate the AUC.

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 87.368 mg × h/L kg

    We know that:
    `AUC=(dose)/(cl earance)=(dose)/(K × V)`.

    However, the total volume of distribution (V) will be equal to the volume of distribution per Kg body weight times the weight of the patient in Kg. Therefore, `V=1 \quad L/(Kg) × 194 \quad Kg = 194 \quad L`.

    `∴ AUC=(322.04 \quad mg)/(0.019 h^{-1} × 194 \quad L)`.

    `∴ AUC=87.37 \quad (mg × h)/L`. Ans.

Notice: Undefined index: TOTALTRY in /home/jshzub/web/rxcal.org/public_html/practice/practice/includes/body.php on line 675