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Topic: Really hard problem {solutions and dilutions}  (Read 1620 times)

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Offline lilianag

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Really hard problem {solutions and dilutions}
« on: November 13, 2012, 08:21:09 PM »
How do i go about doing this....
Suppose you are given exactly 1 liter of a 1.00 x 10^-3 M solution of toluene C6H5CH3 dissolvedin benzene C6H6. Exactly 1 ml of this solution is taken and diluted to exactly 1 liter with benzene. Call the resulting solution #2. Next take exactly 1 ml of solution #2 and dilute it to exactly 1 liter with benzene. Call the resulting solution #3. Continue in this manner until you get to solution #8. What is the concentration of solution #8? Calculate the number of molecules of toluene in 1 liter of solution #8. Interpret your result...
« Last Edit: November 13, 2012, 10:21:48 PM by Arkcon »

Offline discodermolide

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Re: Really hard problem
« Reply #1 on: November 13, 2012, 09:31:43 PM »
1 Liter = 1000 mL, you know the concentration per liter, therefore how much toluene (in moles) is present in 1 mL?
Keep going until you reach solution no 8.
You know the concentration in moles of solution 8 therefore you know the number of molecules.

Or you can start by working out how many molecules are in 1 mL of solution 1 and so on until you reach solution 8, therefore you know the concentration.
There is probably a mathematical formula for this but I don't know it!
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Offline lilianag

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Re: Really hard problem
« Reply #2 on: November 13, 2012, 10:13:51 PM »
what is the concentration?

Offline discodermolide

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Re: Really hard problem {solutions and dilutions}
« Reply #3 on: November 13, 2012, 10:35:37 PM »
What do you mean by that? I don't quite understand.
The concentration is mass per volume.
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