I would like to ask something quite simple but I can't get my head around it. If I have 2 cm^3 of iodine, HCL and water together with 4cm^3 of propanone. What is the concentration of iodine in mol dm^-3?
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Finding number of moles in a reaction watch
- Thread Starter
- 06-03-2013 16:16
- 06-03-2013 16:38
Firstly, what you need to tackle is your units. In chemistry it tends to be in decimeters for volume.
First change 2cm^3 to dm^3
1cm^3 = 0.001dm^3
2cm^3 = 0.002dm^3
Plugging into our equation gives us:
n = c x 0.002dm^3
You have your volume there but tbh you don't have a lot of information to go on without a molar ratio.
Did you mean you have 2cm^3 of Iodine on it's own, or 2cm^3 of Iodine, HCl and water all together in solution?
The latter is not possible as you don't know what the composition of those substances in that 2cm^3 is. You need more information either in the form of individual volumes for the Iodine, HCl and water or moles, etc.
I don't want to confuse you but if it's the former, it could be roughly worked out by this method and using your trusty periodic table:
1 mole of Iodine has a mass of 127.00grams (this is the molar mass) so if you're calculating for 2cm^3 of iodine
then you can do the calculation of n = Mass/Molar Mass
2cm^3 or 0.002dm^3 is roughly equivalent to 2 grams
so moles = mass/molar mass:
2/127 = 0.015 moles of Iodine
Then rearrange n = c x v, to make c the subject
n/v = c
0.015/0.002 = 7.5 mol/dm^3 or (mol dm ^-3)
But yeah, you don't have all the necessary information there. Is this a problem from a teacher or exam question? Might be helpful to see it.Last edited by Dr_H; 06-03-2013 at 16:40.