What does "raised the ethanol level by a factor of 80" mean?

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NewHopeR

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Supposed we have put a tablespoon of ethanol into a bowl of water, now we want to raise the ethanol level by a factor of 80, how many tablespoon of ethanol we have to put into the bowl?


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"We don't know all the answers," Clarke acknowledged. "It's possible there is a trivial explanation, but I don't think that's the case. We know that if we increase the ethanol concentration, they do not live longer. This extremely low level is the maximum that is beneficial for them."

The scientists found that when they raised the ethanol level by a factor of 80, it did not increase the life span of the worms.
 

Raymott

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Supposed we have put a tablespoon of ethanol into a bowl of water, now we want to raise the ethanol level by a factor of 80, how many tablespoon of ethanol we have to put into the bowl?


Context:

"We don't know all the answers," Clarke acknowledged. "It's possible there is a trivial explanation, but I don't think that's the case. We know that if we increase the ethanol concentration, they do not live longer. This extremely low level is the maximum that is beneficial for them."

The scientists found that when they raised the ethanol level by a factor of 80, it did not increase the life span of the worms.
It means to raise the concentration of ethanol to 80 times its current level.
I'd say you'd need another 79 tablespoons of ethanol.
(I'm not surprised that the worms' life spans didn't increase).
 

NewHopeR

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Good.
Thanks
 

BobSmith

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It means to raise the concentration of ethanol to 80 times its current level.

[not a scientist]

To be clear, you're raising the amount of ethanol by a factor of 80, not the concentration by a factor of 80. I guess it depends on what "its" refers to.
 

Raymott

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[not a scientist]

To be clear, you're raising the amount of ethanol by a factor of 80, not the concentration by a factor of 80. I guess it depends on what "its" refers to.
The way you've expressed you opinion, I can't tell whether your disagreeing with me. Who is raising the amount?
To "raise ethanol level by a factor of 80" would almost certainly mean "to raise the ethanol concentration by a factor of 80", not just the amount. The amount of ethanol used in an experiment is irrelevant if you don't keep a note of total amount of the solution into which you put it.

To be more specific than I was in my last post, you cannot just add 79 more spoonfuls of ethanol, because that would raise the ethanol concentration by slightly less than a factor of 80 (because the total amount of the solution also rises). The exact amount would depend on the relative amount of water (or solute) and ethanol. If the total solution was large, say a swimming pool size, then 79 more spoonfuls would be close enough for most purposes in achieving an 80-fold increase in concentration.

To give an example. If you put 1 litre of ethanol in 9 litres of water, you have an ethanol concentration (by volume) of 1/10 = 10%.
You might think that adding another 1 litre of ethanol would raise the concentration to 20%, but it doesn't. It gives you 2 litres ethanol and 9 litres of water, which is an ethanol concentration of (2/11) = 18.18%. To keep a 20% concentration, you must add more than another full litre of ethanol. (Exercise for the reader).

[I think my maths is right this time]
 

BobSmith

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In the original 1 tablespoon example, even at a generous 80 tablespoons (5 cups) per bowl, that's 1/81 or 1.23%. To make it an 80*1.23%, or 98.8% solution, you'd have to add 6399 tablespoons (1 original + 6399 = 6400, add in the water of 80, 6400/6480 = 98.8%).

[STRIKE]In the 10% solution example, you'd have to add 35 liters (clearly not 79). (1 original + 35 = 36, add in the water of 9, 36/45 = 80%).[/STRIKE] edit: oops, don't know what I was thinking here.. I raised it 8 times, not 80.

So, for the original question
Supposed we have put a tablespoon of ethanol into a bowl of water, now we want to raise the ethanol level by a factor of 80, how many tablespoon of ethanol we have to put into the bowl?
I think the correct answer is either 6399, or not knowable without knowing how big the bowl* is. And I'm not being facetious here; raising the ethanol level by a factor of 80 is huge, especially when talking about biology.


*And of course, the bowl has to be greater than 79 units, because at 79, the initial concentration would be 1.25%, and 80 times 1.25% is 100% and you can't achieve 100% with the stated additive method.
 
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Raymott

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In the original 1 tablespoon example, even at a generous 80 tablespoons (5 cups) per bowl, that's 1/81 or 1.23%. To make it an 80*1.23%, or 98.8% solution, you'd have to add 6399 tablespoons (1 original + 6399 = 6400, add in the water of 80, 6400/6480 = 98.8%).

In the 10% solution example, you'd have to add 35 liters (clearly not 79). (1 original + 35 = 36, add in the water of 9, 36/45 = 80%).
Your figures could be right. But in the 10% solution example I gave, the aim wasn't to raise it to 80%, nor was the solution 79 litres.
What about the real 10% solution example I gave. Is that correct? The answer is not 35 litres.

What do you think of my statement that "If the total solution was large, say a swimming pool size, then 79 more spoonfuls would be close enough for most purposes in achieving an 80-fold increase in concentration"? That is, I was making the point that the "bowl" had to be sufficiently large. Isn't increasing the amount of something from 1 part ethanol per billion parts water to 80 parts per billion an increase in the concentration by a factor of 80?

I don't know how many spoonfuls a swimming pool holds, (in Aus. we don't use either spoonfuls or bowlfuls in the lab) but the principle remains that if you have a large enough solution to start with and you add one small unit of a substance, you can increase the concentration of that substance by a factor of 80 by adding roughly 79 more small units - given that the original volume is huge compared to the units you are adding.

I'll give another example, and you can tell me if this is wrong. If you have a 1000 litre tank of water, and you put 1ml (millilitre) of ethanol it in, you have 1 part ethanol to a million part of water, and and ethanol concentration 0.99999... * 10^-6% by volume. If you add 79 extra mls of ethanol, you have 80 mls ethanol in a total solution of 1,000,080 mls = 7.9994 * 10-^5%.This is a 79.994/0.99999... fold increase (slightly less than 80, hence my statement that you need to add slightly more than 79 spoonfuls).
I could be confused. Is this right?
 

emsr2d2

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Please excuse me while my head explodes.
 

BobSmith

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I'll give my best answers, though I am not a teacher.

Your figures could be right. But in the 10% solution example I gave, the aim wasn't to raise it to 80%, nor was the solution 79 litres.
What about the real 10% solution example I gave. Is that correct? The answer is not 35 litres. I got lost in 80% versus 80 times. My bad. 35 liters will give you an 80% solution, but that's different than the OP's 80 times.

What do you think of my statement that "If the total solution was large, say a swimming pool size, then 79 more spoonfuls would be close enough for most purposes in achieving an 80-fold increase in concentration"? That is, I was making the point that the "bowl" had to be sufficiently large. Isn't increasing the amount of something from 1 part ethanol per billion parts water to 80 parts per billion an increase in the concentration by a factor of 80? Of course it is. I am/was trying to answer the OP. To suggest putting in 79 units to offset the 1 is, IMO, not the same intent as raising the concentration by 80 times.

I don't know how many spoonfuls a swimming pool holds, (in Aus. we don't use either spoonfuls or bowlfuls in the lab or swimming pools, I assume) but the principle remains that if you have a large enough solution to start with and you add one small unit of a substance, you can increase the concentration of that substance by a factor of 80 by adding roughly 79 more small units - given that the original volume is huge compared to the units you are adding. I think you already covered this in the previous paragraph.

I'll give another example, and you can tell me if this is wrong. If you have a 1000 litre tank of water, and you put 1ml (millilitre) of ethanol it in, you have 1 part ethanol to a million part of water, and and ethanol concentration 0.99999... * 10^-6% by volume. If you add 79 extra mls of ethanol, you have 80 mls ethanol in a total solution of 1,000,080 mls = 7.9994 * 10-^5%.This is a 79.994/0.99999... fold increase (slightly less than 80, hence my statement that you need to add slightly more than 79 spoonfuls).
I could be confused. Is this right? I think you know that it is right. It could be me, but I'm sensing some defensiveness. If I am, then I apologize if my early reply came across snarky. That wasn't my intent. If I am wrong, then I apologize for misreading you here.
 
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