Answer:
After 9.0 hours
Explanation:
Using the decay equation, we can write the amount of drug propanol after time t as:
[tex]m(t)=m_0 e^{-\lambda t}[/tex] (1)
where
m(t) is the mass left at time t
[tex]m_0[/tex] is the initial mass of the substance
[tex]\lambda[/tex] is the decay constant
t is the time
The decay constant is related to the half-life of the substance as follows:
[tex]\lambda=\frac{ln2}{t_{1/2}}[/tex]
where [tex]t_{1/2}[/tex] is the half-life.
Here we have
[tex]t_{1/2}=3.9 h[/tex]
So the decay constant is
[tex]\lambda=\frac{ln 2}{3.9}=0.178 h^{-1}[/tex]
We want to find the time t after which the dose is 80% of the initial dose is eliminated, so the time t after which 20% of drug is left, so
[tex]\frac{m(t)}{m_0}=0.20[/tex]
Substituting eq(1) and solving for t, we find:
[tex]t=-\frac{ln(\frac{m(t)}{m_0})}{\lambda}=-\frac{ln(0.20)}{0.178}=9.0 h[/tex]
