Interesting math problem - Convert Grams to ML

[illusion88]

according to https://fscimage.fishersci.com/msds/89308.htm">this site</a> the LD50 for grain alcohol (when ingested oraly from a rat) is LD50 = 7060mg/kg or 7.06 grams.

For those who don't know the LD50 is the point at which the substance kills half of the subjects tested on.

So I wanted to know how much alochol would kill me. So I take my weigh in KG (71KG) and multiple it by 7.06 to figure out how much alochol I would need to have a 50% chace of dieing right?

7.06 * 71 = 501.26 G, now how do I convert that into liters? Grams is a mass and liters is a volume, I would need to find the displacement right? How muych does a ml of liquid weigh? Anyone?

 

[dighn]

u need the density

 

[EpsiIon]

The answer depends on your drink. For the sake of simplicity, let's say that the part of your liquor that is not ethanol is entirely water. Again for simplicity, we'll consider any mixer (like soda) to be water. That said, divide the proof of your drink by two to get the % ethanol. If you were to drink straight rum (usually 80 proof), you would be drinking 40% ethanol.

Define some variables:
x = % ethanol of your chosen liquor (40% = .4).
y = % of your drink that is your chosen liquor (10% = .1)

So, for example, let's say you're drinking a simple rum and coke. I don't know how you like to mix yours, but I would consider 1 part rum to 3 parts coke pretty strong. In this case:

x = .4 (rum is 40% alcohol)
y = .25 (1/4 of your drink is rum).

The density of ethanol is 0.789 g/ml.
The density of water is 1 g/ml.

So the density of what you're drinking is given by the following:

(percent of your drink that is liquor * ( (percent of your liquor that is ethanol * density of ethanol) + (percent of your liquor that is water * density of water) )) + (percent of your drink that is not liquor * density of water).

Density = y * ( (x*.789) + (1-x) ) + (1-y) g/ml.

Or in the case of my rum and coke:

Density = .25 * ( (.4 * .789) + .6 ) + .75 = .9789 g/ml.

Or if you were just drinking rum (y = 1):

Density = (.4*.789) + .6 = .9156 g/ml.



Ok, what we need is this:

z = amount (g) of drink to give you a 50% chance of dying.

(x * (z * y)) = 501.26 g

Or, staying with the rum idea (y = 1):

z = 501.26 g / x g

Still staying with the rum idea:

z = 501.26 g / .4 = 1253.15 g of rum.



This is where we use the density of rum calculated earlier:

1253.15 g * (1 ml / .9156 g) = 1368.67 ml of rum to have 50% chance of dying.

1.3 liters. That's a LOT of alcohol.

By the way: these numbers could be wrong. I take NO responsibility for what you do with this (mis)information.


EDIT:
Just because I was curious:

I weigh 75kg => I need 529.5 g of alcohol to have a 50% chance of dying.

I probably make a rum and coke with 15% rum. In this case:

x = .4
y = .15

Density of my rum and coke = y * ( (x*.789) + (1-x) ) + (1-y) g/ml = .98734 g/ml

z = 529.5 / (.4 *.15) = 8825 g of my rum and coke

8825 g * (1 ml / .98734 g) = 8938 ml of my rum and coke.

There's no way I'm fitting 9L of anything into my body. Guess I don't have to worry about dying.

 

[TheStu]

Why oh why would I want to convert from grams to MegaLiters for? milliLiters maybe... but now MegaLiters

 

[biostud]

alcohol is 0.785 g/ml so that would be 638 ml pure alcohol. for a liquer with 40% vol that would be ~1.6l of liquer.