T08: Clearance from K, dose and concentration

50. Pharmacokinetics 50.1) PK Basic parameters 50.1.2) Clearance Clearance basic concepts Normal 1

Total tried:       Correct:       Wrong:

34.66 mg of a drug was administered to a patient as an IV push, which resulted the plasma concentration of 0.66 mg/L. The elimination rate constant of the drug is 0.185 h-1. Determine the clearance of the drug in this patient.

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 9.714 L/h kg

    We know that, clearance (Cl) is the product of elimination rate constant (K) and volume of distribution (V).

    ` Cl = K × V `.

    In this case, we hav the dose of the drug administered (`X_{0}`) and the plasma concentration (`C_{0}`). We know that:
    `C_{0} = X_{0}/V ⇒ V = X_{0}/C_{0}`.

    Therefore, clearance (Cl) will be:
    `Cl = 0.185 \quad h^{-1} \quad × (34.66 \quad mg)/(0.66 \quad (mg)/L) = 9.71 \quad L/h`. Ans

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