To prepare a 14.1 kg dog (assume volume of 14.1 L) for surgery, 0.0191 moles of
ID: 916913 • Letter: T
Question
To prepare a 14.1 kg dog (assume volume of 14.1 L) for surgery, 0.0191 moles of an anesthetic was administered intravenously. The reaction in which the anesthetic is metabolized displayed a first and second half life of 162.3 minutes. After 50.32 minutes of surgery the drug began to lose its effect, but more surgery was required. How many moles of the anesthetic must be re-administered to restore the original level of the anesthetic?
To prepare a 14.1 kg dog (assume volume of 14.1 L) for surgery, 0.0191 moles of an anesthetic was administered intravenously. The reaction in which the anesthetic is metabolized displayed a first and second half life of 162.3 minutes. After 50.32 minutes of surgery the drug began to lose its effect, but more surgery was required. How many moles of the anesthetic must be re-administered to restore the original level of the anesthetic? Submit Answer Tries 0/99Explanation / Answer
Given,
half life of drug (t 1/2) = 162.3 minutes
K = ln 2 / t 1/2 = 0.693 / 162.3 = 4.27 x 10^-3
We know that for a first order reaction,
C(t) = C0 exp (-Kt)
where,
C(t) = conc. of the drug at time t
C0 = initial conc. of drug
K = rate constant
t = time
=> C (at t = 50.32 min.) = 0.0191 x exp (-0.00427 x 50.32) = 0.0154 moles
=> 0.0154 moles of drug were remaining after 50.32 minutes
=> Moles of drug needed = 0.0191 - 0.0154 = 0.0037 moles
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