Let the ED50 of a recreational drug be defined as the amount required for 50% of
ID: 1002837 • Letter: L
Question
Let the ED50 of a recreational drug be defined as the amount required for 50% of a test group to feel high or get a buzz. If the ED50 value of ethanol is 470 mg/kg body mass, what dose would a 70 kg party goer need to quickly consume in order to have a 50% chance of getting a buzz?
If a bottle of beer contains about 13.9 g of ethanol, how many bottles of beer would a 70 kg party goer need to consume in order to have a 50% chance of getting a buzz?
If the lethal dose (LD50) for ethanol is 4,700 mg/kg, how many bottles of beer would a 70 kg party goer need to chug in order to have a 50% chance of dying?
A one liter bottle of Jack Daniels contains about 356 g of ethanol. How many 1.0 liter bottles of this whiskey would a 70 kg party goer need to chug in order to have a 50% chance of dying?
Explanation / Answer
Solution:
ED50:
1) Recreational drug 470 mg/kg
for 70 kg of body the effective dose is 70 kg * 470 mg/kg = 32900 mg =32.9 gm
on 50 % --> 164500 mg is required for feeling buzz
2) For ethanol 13.9 g = 13900 mg
13900 mg / 470 mg/kg = 29.574 kg so for 70 kg body it required,
(70 kg / 29.574 kg) * 13900 mg = 32900 mg = 32.9 gm of ethanol
one bottle contains 13.9 gm of ethanol so (32.9/13.9) = 2.366 bottles of beer required for buzz
LD50:
3) For 4700 mg/kg of 70 kg is (70 kg * 4700 mg/kg) = 329000mg = 329 gm
one bottle contains 13.9 gm of ethanol so (329/13.9) = 23.669 = 24 bottles of beer is lethal for dying
4) 1.0 lt whiskey contains 356 gm of ethanol so (329/356) = 0.92 lt of whiskey is having a chance of dying.
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