November 25, 2020, 08:50:32 AM
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### Topic: Dumb question about finding molarity when neither solute nor solution..?  (Read 4236 times)

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#### MissDee

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##### Dumb question about finding molarity when neither solute nor solution..?
« on: November 05, 2009, 08:52:27 PM »
2Fe(NH4)2(SO4)2 x 6H2O + 3H2C2O4 + H2O2 + 3K2C2O4  2K3Fe(C2O4)3 X 3H2O + 4(NH4)HSO4 + 8H2O

I need to calculate the following:

- the number of moles in 33.0 ml of 1.00M H2C2O4
- the number of moles in 20.0 ml of 3% H2O2 (which is 1.00M)
- the number of moles in 10.0 ml of saturated K2C2O4 (which is 2.00M)

I originally thought you had to use the formula Molarity = moles solute/liters solution, but nothing is specified, just the information I posted. Do you use it anyway without regards to which is the solute and which is the solution, and just go off the molarity and mL units?

#### csrscience.com

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##### Re: Dumb question about finding molarity when neither solute nor solution..?
« Reply #1 on: November 05, 2009, 10:00:28 PM »
- the number of moles in 33.0 ml of 1.00M H2C2O4

1.00M is 1mol/1000mLs.
If you have 33ml's you are going to have the same ratio... in order to find this you'd solve for x:

1mol/1000mls = x/33mls
If this doesn't make sense.. I'll try to explain it better...

While this isn't identical, a lot of the concepts are interchangeable when it comes to converting such things.. so you may want to check this out: http://indiescience.org/basetalk/index.php?topic=24.0
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