T20: Mixed types from T01-T20
50. Pharmacokinetics 50.1) PK Basic parameters 50.1.3) Area Under Curve (AUC)
50.1.3.1) AUC basic concepts Normal 1
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The elimination rate constant of a drug is 0.044 h-1. If its volume of distribution in a patient is 115.34 L, then determine the clearance of the drug.
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lb equals 5.075 L/h kg
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We know that, clearance (Cl) is the product of elimination rate constant (K) and volume of distribution (V).
` Cl = K × V `.
Plugging the values in this question we get:
` Cl = 0.044 \quad h^{-1} × 115.34 \quad L = 5.07 \quad L/h ` Ans