calculate the number of shots of Captain Morgan’s rum 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:

Table reference

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.
As Zhe mentioned, you have inconsistent units in your calculation, so your numerical results are not physically meaningful.

Take for instance this expression:

162 = A/(0.55 x 50) – (0.15 x 1)

Here, you have different units for each term; from the values you supplied, the units are

162 mg/dL = A/(0.55 L/kg x 50 kg) - 0.15 g/(L•h) x 1 h

[Note that the units for the elimination rate weren't stated in the question, but a quick search of the literature gave studies measuring it at about 0.01-0.02 g/(dL•h), so I infer the 0.15 value refers to g/L]
If we use uniform units throughout, we get

1.62 g/L = A/27.5 L - 0.15 g/L

from where you get a more reasonable value for A:

A = (1.62 + 0.15) * 27.5 = 48.675 g

Each shot contains 

V = 30 mL * 0.40 / 0.789 g/mL = 15.2 g

of ethanol, so you need to drink 48.675 / 15.2 = 3.2 shots of rhum over the course of an hour.

I'd suggest, however, that drinking shots in "rapid succession" implies a negligible drinking time (∆t ~ 0 h). In that case, A = 44.5 g and you need about 3 shots (2.93).