T11: Mixed problems from T7 - T11.
50. Pharmacokinetics 50.1) PK Basic parameters 50.1.2) Clearance
50.1.2.1) Clearance basic concepts Normal 1
Total tried: Correct: Wrong:
A drug with the elimination rate constant of 0.223 h-1 is administered as an IV bolus to a patient. The weight of the patient is 173 Kg. If the volume of distribution of the drug is 0.34 L/Kg, then calculate the clearance of the drug in this patient's body.
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 13.117 L/h kg
-
We know that, clearance (Cl) is the product of elimination rate constant (K) and volume of distribution (V).
` Cl = K × V `.
However, the total volume of distribution will be the product of volume per Kg body weight and the weight of the patient. Thus,
` V = 0.34 \quad L/(Kg) × 173 \quad Kg `.
Plugging the values in the first equation we get:
` Cl = 0.223 \quad h^{-1} × 0.34 \quad L/(Kg) × 173 \quad Kg = 13.12 \quad L/h ` Ans