# Mary's Puzzling Lab Experiment: Calculating the New Weight of a Mixture

• Shackleford
In summary, Mary had a mixture of chemicals in a chemistry lab that was 90% water and weighed 20 pounds. After returning from a weekend break, she found that the mixture was now 50% water. Assuming the non-water part was not affected by evaporation, the mixture now weighs 4 pounds. The website that posed this riddle may have incorrect answers.
Shackleford
Mary was working in a chemistry lab with a mixture of chemicals that was 90% water and weighed 20 pounds. After returning to the lab from a weekend break, she calculated the mixture was now 50% water. How much does the mixture now weigh? For purposes of this puzzle, assume the non-water part of the mixture was not affected by evaporation.

Isn't there a brain teaser subforum here?

Anyway, x-other substance. y-water, x,y in pounds.
0.9(x+y)=20
10y/9=20
y=18
18*0.5=9=y

First the mixture had 200/9 pounds after that it lost 11 pounds, so it weighs 200/9-11 or so I think, I myself just woke up an hour ago. (-:

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In my sleep deprived state, I now have four pounds. lol.

10/9 x = 20 pounds
x = 18 pounds

10/1 y = 20 pounds
y = 2 pounds

It says the non-water part was unaffected so that would mean the 2 pounds of non-water is still there, correct?

Now, the mixture is 50% water or is a 1:1 ratio. Two pounds to two pounds? The mixture now weighs four pounds?

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At least I know your'e honest.

Ok, I'll try to find a way.
Mary was working in a chemistry lab with a mixture of chemicals that was 90% water and weighed 20 pounds. After returning to the lab from a weekend break, she calculated the mixture was now 50% water. How much does the mixture now weigh? For purposes of this puzzle, assume the non-water part of the mixture was not affected by evaporation.
At first the water 20/0.9 pounds per percent
40 percent of the water evaporated meaning is gone, i.e 0.4*0.9 times 20/0.9 means
8 pounds were evaporated which means the water now weighs 12 pounds.
As I said the whole compound weighs 20/0.9 so if it loses 8 pounds it still should weigh more than 12 pounds, or am I way off here.

loop quantum gravity said:
At least I know your'e honest.

Ok, I'll try to find a way.

At first the water 20/0.9 pounds per percent
40 percent of the water evaporated meaning is gone, i.e 0.4*0.9 times 20/0.9 means
8 pounds were evaporated which means the water now weighs 12 pounds.
As I said the whole compound weighs 20/0.9 so if it loses 8 pounds it still should weigh more than 12 pounds, or am I way off here.

That's what I thought, too, but it just didn't seem logical to me. See if my previous post makes sense.

Assuming your going by weight and not by stoichiometry, I got 4 lbs.

That website is wrong.

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I remember these sorts of questions showing up in like 11th Grade Algebra or so, haha.

loop quantum gravity said:
40 percent of the water evaporated meaning is gone

The answer is indeed 4 lbs.

Topher925 said:
Assuming your going by weight and not by stoichiometry, I got 4 lbs.
What is going by stoichiometry, and how would that give you a different answer?

If you go my moles of molecules instead of weight you will get a different answer do to the chemical and water having different densities.

Topher925 said:
If you go my moles of molecules instead of weight you will get a different answer do to the chemical and water having different densities.

Yeah. I just made the assumption by weight. That stupid website probably has so many incorrect answers to the riddles.

Topher925 said:
If you go my moles of molecules instead of weight you will get a different answer do to the chemical and water having different densities.
If you assume the percentages are in fact mole fractions (or volume fractions), then there is no way to solve the problem with the given data. You will additionally need the molecular weight (or density). So, in order to solve it, you have to assume the numbers are mass percentages.

Gokul43201 said:
If you assume the percentages are in fact mole fractions (or volume fractions), then there is no way to solve the problem with the given data. You will additionally need the molecular weight (or density). So, in order to solve it, you have to assume the numbers are mass percentages.

Or you could just claim that the riddle is a trick question and not bother answering it.

## 1. How do I calculate the new weight of a mixture in Mary's lab experiment?

To calculate the new weight of a mixture in Mary's lab experiment, you will need to know the individual weights of each component in the mixture and the percentage of each component in the original mixture. You can then use the formula: New Weight = Original Weight x (100% - % Component 1) / 100%.

## 2. Can this experiment be done with any type of mixture?

Yes, this experiment can be done with any type of mixture as long as you have the necessary information about the individual weights of each component and the percentage of each component in the original mixture.

## 3. What if the mixture contains more than two components?

If the mixture contains more than two components, the formula to calculate the new weight will be slightly different. You will need to subtract the percentage of each component from 100% and then multiply the remaining percentages together. This will give you the percentage of the final mixture, which you can then use in the formula: New Weight = Original Weight x (100% - % Final Mixture) / 100%.

## 4. Is there a certain order in which the components should be mixed in this experiment?

No, there is no specific order in which the components should be mixed in this experiment. As long as you have the necessary information about the individual weights and percentages of each component, you can mix them in any order you prefer.

## 5. How accurate is this method for calculating the new weight of a mixture?

This method is fairly accurate as long as the necessary information is provided and the calculations are done correctly. However, it may not account for any potential changes in the weight of the components due to chemical reactions or other factors.

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