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

AI Thread Summary
Mary's chemistry lab mixture initially weighed 20 pounds and was 90% water. After a weekend, it became 50% water, prompting calculations about its new weight. The discussion revealed that the non-water component remained unchanged despite evaporation. Various calculations suggested that the mixture now weighs 4 pounds, as the water content decreased significantly. Participants debated the accuracy of an external website's answer, with some asserting that the problem could be misleading if interpreted through different scientific principles, such as stoichiometry versus weight. Ultimately, the consensus leaned towards the mixture weighing 4 pounds, emphasizing the importance of assuming mass percentages for accurate calculations.
Shackleford
Messages
1,649
Reaction score
2
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.
 
Physics news on Phys.org
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. (-:
 
Last edited:
Can you show your equations?
 
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?
 
Last edited:
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.
 
Last edited:
I remember these sorts of questions showing up in like 11th Grade Algebra or so, haha.
 
  • #11
Here's your mistake:
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?
 
  • #12
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.
 
  • #13
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.
 
  • #14
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.
 
  • #15
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. :approve:
 

Similar threads

Back
Top