T16: Mixed types from T12-T18

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:

286.16 mg of a drug is parenterally administered to a patient weighing 163 Kg. If the volume of distribution of the drug is 0.66 L/Kg, and the AUC was measured to be 9.71 mg × h/L, then determine the elimination rate constant of the drug.


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 0.274 h-1 kg

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

    Rearranging the equation: `K=(dose)/(AUC × 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.

    `∴ K=(286.16 \quad mg)/((9.71 \quad mg × h)/L × (0.66 \quad L)/(Kg) × 163 \quad Kg)`

    `∴ K=0.274 \quad h^{-1}`. Ans.

    Clarification: The AUC shown in this question (9.71) has been rounded from its more precise value (9.7080291970803), the dose shown in this question (286.16) has been rounded from its more precise value (286.1628). That is why the displayed answer may look slightly distorted.


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