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    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?

    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.
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Updated: March 6, 2013


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