# Very confused in chemistry?

• Glen Maverick
In summary, to calculate the number of moles of IO3- used in the titration, we first calculate the moles of KIO3 used and then determine the molarity of the solution. The mole ratio of KIO3 and IO3- is 1:1, so the number of moles of KIO3 will be equal to the number of moles of IO3-. To calculate the number of moles of IO3- in the Erlenmeyer flask, we multiply the molarity by the volume of solution in the flask. In this case, the volume is 10mL.

## Homework Statement

Calculation of the number of moles of IO3- used in the titration(i.e. in the Erlenmeyer flask):
KIO3(s) → K+(aq) + IO3-(aq): balanced eqn of dissolving KIO3 in water

1. Calculate the moles of KIO3 used: Molecular weight of KIO3 = 214.0011 g/mol
Amount of KIO3 used in grams = 0.0356g
# of moles = 0.0356g/(214.0011g/mol) (grams cancels out) = 1.6635 × 10^-5 moles
The mole ratio is 1:1:1. So # of moles of KIO3 will be equal to # of moles of IO3-

2. Calculate the Molarity
Amount of dH2O used for dissolving = 100mL = 0.1L
Molarity = (moles of solute) / (volume of solution in Liters) = 1.6635 × 10^-5 moles/0.1L
= 1.6635 × 10^-4 moles/L
I did until this. But in the question, it says that <Calculate the # of moles of IO3- in the Erlenmeyer flask> And in the Erlenmeyer flask, there was 10mL of KIO3 solution, and other solution of 54ml.

And my question is, do I have to multiply the molarity I gained by calculation by 0.064mL (because (10ml+54ml)/1000 = 0.064L=amount in Erlenmeyer flask) to gain the # of moles of IO3- in the Erlenmeyer flask, Or can I just multiply the molarity by 10mL Or am I doing wrong??

10 mL - that was the amount of solution of iodate transferred, right? You have not transferred 64 mL, or any other volume.

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I get it. I was confused when my lab TA emphasized "in the Erlenmeyer flask", so i thought she might mean something else. :)

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