I am facing a challenge, and this is my first question in this stack exchange. I have worked out the following problem but the answer I got (number of shots) is ridiculous. Could any of the chemist here please advise me where I went wrong?
Question: Assume that you are on vacation in Florida and enjoying the company of friends at a bar. Using the information in Table 80-1 in your text, calculate the number of shots of Captain Morgan’s rum required to take your blood ethanol concentration above twice the legal limit to drive. [Note: The legal limit in Florida is 80 mg/dL]. Assume that the shots are consumed in rapid succession (the wisdom of which you were later unable to explain). Show your complete calculation, including all inputs and assumptions.
My calculation:
Let us assume that my binge-drinking has led to a blood alcohol concentration (BAC) of 162 mg/dL
.
The formula for BAC is as follows:
`BAC = A/(R x W) – (0.15 x H)` where
- A = weight of pure ethanol consumed (g)
- R = Widmark’s rho factor (0.68 and 0.55 L/kg for men and women, respectively)
- W = body weight (kg)
- 0.15 = average ethanol elimination rate in humans
- H = drinking period (hr)
Inserting the values of 162 mg/dL
for BAC, R = 0.55 L/kg, W = 50 kg, and H = 1 hr
, the equation becomes:
162 = A/(0.55 x 50) – (0.15 x 1)
162 = A/27.5 – 0.15
Therefore,
A = 27.5(162 + 0.15)
= 4459.125 g
According to table 80-1 in the book, one shot is 30 mL and contains 40% ethanol.
So one shot contains 12 mL ethanol.
Density of ethanol = 0.789 g/cm3
The volume of pure ethanol which I consumed = 4459.125/0.789
= 5651.62 mL
So, no. of shots consumed = 5651.62/12 = 471
, over the course of 1 hour
Conculsion: 471 shots of Captain Morgan Rum is absurd for any person. So I think I am missing something in my calculation. I'm not able to figure out what it is.