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    I have been told by my chem teacher that wr will be using a 1% iodine in potassium iodide solution for a vitamin c redox titration. I dont know how to convert the 1% to molarity and need some help please! I will be using 75 ml of this stock solution but need help calculating the mol/liter. Also the experiment I found online said to use a 0.025 mol/L iodine solution and I am wondering what % this solution is?
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    Molarity

    Have you seen this formula?:
    M (moles) = C (conc) * V (volume)
    You know the Concentration of the Iodine (1%) and you know the Volume (75ml).
    Volume needs to be in decimeters. To convert from milliliters to decimeters you divide by 1000. Therefore to work out the Moles you would simply calculate: 0.01 * 0.075 = 7.5x10^-4 mol/dm^3

    0.025*100% = 2.5% is 0.025mol as a percentage of 1 litre.
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    (Original post by Avacado)
    Molarity

    Have you seen this formula?:
    M (moles) = C (conc) * V (volume)
    You know the Concentration of the Iodine (1%) and you know the Volume (75ml).
    Volume needs to be in decimeters. To convert from milliliters to decimeters you divide by 1000. Therefore to work out the Moles you would simply calculate: 0.01 * 0.075 = 7.5x10^-4 mol/dm^3

    0.025*100% = 2.5% is 0.025mol as a percentage of 1 litre.
    Thanks very much! I was trying to make it more complicated than I needed to!
 
 
 
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