Not really sure how to go about solving this \"challenge\" question. Any help wo

Question

Not really sure how to go about solving this "challenge" question. Any help would be greatly appreciated, thank you!

10. The rate of alcohol metabolism is a zeroth order process with a rate constant of around 0.015 by mass per hour. (Recall that percentage is a form of concentration.) Jane Doe is a typical 70 kg female with 58 percent free water. John Doe is a typical 83 kg male with 65 percent free water. A drink -30 mL (1 oz) of spirits (40 alcohol), 150 mL ofwine (8.0 or 350 mL of beer (3.5%) each contains 12 mL of alcohol. The alcohol distributes itself through the aqueous regions of the body. a) How many drinks do Jane and John need to achieve an instantaneous blood alcohol level of 0.12% by mass (Answer: Jane: 5.1 drinks) b) How many drinks per hour can Jane and John have and maintain a blood alcohol of 0.00 percent one hour after they stop drinking (Answer: Jane: 0.64 drinks) c Jane and John each have three drinks per hour from 2115 to 0115 hours. What is the blood alcohol content at 0115? at 0800 the next morning? (Answer: Jane: 0.22 at 0115; 0.12% at 0800) d Continuing from c, when are Jane and John legally allowed to drive home? Assume the legal limit is 0.05 percent. (Answer: Jane: 12:30 pm the next day)

Explanation / Answer

10) the rate of alcohol metabolism is of zeroth order

so the rate law expression will be

[A] = [A0] - Kt

K = rate constant = 0.015 % by mass / hour

[A0] = initial concentration

[A]= concentration at time t

a) the alcohol content in 1 drink = 12mL , density of alcohol =0.78 g / mL so mass of alcohol in each drink = density X volume = 12 X 0.78 = 9.36 grams

The Mass of water in Jane = 0.58 X total body weight = 0.58 X 70 = 40.6 Kg

The % of alcohol in blood needed = 0.12%

0.12% of 40.6Kg = 0.04872Kg = 48.72 Kg

number drinks required = total mass required / mass in each drink = 48.72 / 9.36 = 5.2 drinks for Jane

Similalry for jone

The mass of water = 0.65 X 83 = 53.95 Kg

0.12% of 53.95Kg = 0.0647 Kg = 64.7 grams

so number drinks required = 64.7 / 9.36 = 6.92 drinks for Jone

b) The final concentration required = 0 = [At]

Intial concentration = ?

Time = 1Hour

[A0] = [A] + Kt = Kt = 0.015 % by mass

For Jane

0.015% = 0.015 X 40.6 / 100 Kg = 0.00609Kg = 6.09grams

Number of drinks required = 6.09 / 9.36 = 0.65 drinks for Jane

For Jone

0.015% = 0.015 X 53.95 / 100 Kg = 0.00809Kg = 8.09grams

Number of drinks required = 8.09 / 9.36 = 0.86 drinks

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