Question

how do i convert 77 mg to moles and what is the answer

(26 points) Kinetics of an Antiviral Drug. In 2016, Nature reported the discovery of a small molecule (GS-5734) that exhibited antiviral activity against the Ebola virus in rhesus monkeys (Nature, 2016, 531,381-385). That drug, now called remdesivir, is drawing renewed attention as a possible treatment for COVID-19. In the original study, the drug was administered by intravenous injection, 10 mg of GS-5734 per kg body weight of the rhesus monkey. A. (6 points) What's the typical body weight of a rhesus monkey? Cite your source for the above answer: https://en.m.wikipedia.org/wiki/Rhesus machique So, in this study, how much GS-5734 (in mg) would they have injected into that monkey? Show your work. long per 10mg x 1kg 7.749 = 77 mg TKY B. (4 points) The study established a short half-life (t12 0.39 hr) for the elimination of GS-5734 from the blood stream. Assuming first-order kinetics, determine the rate constant for elimination of GS-5734 from the blood plasma in rhesus monkeys. Show your work and be sure to include the correct units in your answer. t/ = to 0.39 hr = 77 mg 2K 2K 0 XSK= 77mg K= 0.78 0.78hr mg/hr C. (14 points) One subject in the study received an initial GS-5734 dose of 68 mg, 1. Assume the minimum effective amount of GS-5734 was known to be 41 mg. How long would it take for the GS-5734 in the subject's blood plasma to drop from 68 mg to 41 mg? Show your work

Answer 1

A.

Weight of rhesus monkey = 7.7 Kg

hence mass of Gs-5734 drug needed

= 7.7 (Kg) $\times$ $\frac{10 mg}{1 Kg}$

= 77 mg.

molar mass of the drug remdesivir = 602.6 g/mol

mass of the drug = 77 mg

= 77$\times$10-3  g

the moles of the drug = ( mass / molar mass)

= (77$\times$10-3 /602.6)

= 1.3 $\times$ 10-4  

B.

Rate constant for first order reaction

K = $\frac{ln2}{0.39}$

K = $\frac{0.693}{t_{1/2}}$

given , t1/2 = 0.39 hr

then

K = $\frac{0.693}{0.39}$ hr-1

or, K = 1.78 hr-1

so, rate constant is 1.78 hr-1

C.

Integrated rate law of first order reaction is

K = $\frac{1}{t}$ ln $\frac{a_{0}}{a_{t}}$

given, a0  = 68 mg ; at = 41 mg , Rate constant (K) = 1.78 hr-1

then ,

1.78 =  $\frac{1}{t}$ ln

68(mg) 41(mg)

or, 1.78 = $\frac{1}{t}$ 0.51

or, t = (0.51/1.78) hr

or , t = 0.284 hr

or, t = 0.284 (hr)  $\times$

60(min) 1(hr)

or , t = 17 min

so, it will take 17 min to drop from 68 mg to 41 mg in blood plasma.