### 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

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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

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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

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