Calculating Molarity for a Diluted Ethanol Solution - Quick Dilution Question

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SUMMARY

The discussion centers on calculating the molarity of a diluted ethanol solution, specifically 1.72x10^-4 moles of ethanol in a total volume of 100 mL after dilution. The correct molarity is determined by dividing the moles of ethanol by the total volume in liters, which is 0.1 L. The calculation confirms that the molarity of the solution is 1.72x10^-3 mol/L. The participants clarify that the original 10 mL of ethanol mixed with 90 mL of water results in a diluted solution, and they emphasize the importance of accurate volume measurement in lab procedures.

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Homework Statement


The mols of ethanol is 1.72x10^-4. I am trying to find its volume. However, I am confused on what the molarity of this solution is. (Molarity here is in mol/L)

(the 10mL sample contained 1.72x10^-4 mols of ethanol, but it was diluted in 90 mL water)

The lab said:
"Dilute the 10.0 mL ethanol sample as such: (10.0 mL ethanol in 90.0 mL of distilled water, for a total of 100.0 mL)."

The Attempt at a Solution



The mols of ethanol is 1.72x10^-4. Do I divide this figure by .01 liters? Or do I do it by .1? I think it's one or the other but either way I need this calculation perfect or the rest of the lab is moot!
 
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If you want to find the molarity, you divide the number of moles by the volume (here: .1 liter).

I don't see which volume you want to find.

10ml ethanol plus 90ml distilled water won't give exactly 100ml, but that's probably a negligible detail.
 
Oops, sorry! My original post was a bit unclear.

The original dilution scheme was 10 mL of the ethanol in 90mL water [I think my teacher isn't really concerned with the negligible detail(s)]

1.72x10^-4 mols of ethanol is the mols of ethanol oxidized by dichromate ion in my lab. I am trying to find the volume of those mols. The procedure included preparing a solution by adding 10 mL ethanol into 90 mL of water. We shook it up and added into a separate flask 1 mL of that solution and 21 mL of 0.01M dichromate in 5M H2SO4.

Here's the reaction if it is relevant:
2 Cr2O72− + 16 H+ + 3 C2H5OH → 4 Cr3+ + 11 H2O + 3 CH3COOH

Hopefully that clears things up! For such a simple dilution question I can't believe I'm having trouble with it, of all things to be troubled with in this lab.
 
how silly. I think I just had to convert the grams to mL using ethanol's density - (0.789g/mL).

Resulting figure would be of the 10 mL ethanol originally put in, 0.01 mL was oxidized.

7.92x10^-3 g ethanol X (1 mL/0.789g) = 0.0100mL

(7.92x10^-3 g of ethanol = 1.72x10^-4 mols of ethanol )
 
Ritzycat said:
Resulting figure would be of the 10 mL ethanol originally put in, 0.01 mL was oxidized.

You took 1 mL out of 100 mL of the diluted solution, so 1/100th of the original sample. If 0.01 mL is 1/100th of 10 mL, then you got it right.
 

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