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    The question is 'How many moles of MnO4- are present in 1000cm^3 of the standard 0.01 M KMnO4 solution'

    So i recently did a titration experiment in class and I got 12.1cm^3 of 0.01 M KMnO4.

    So i had a go at the question and did moles = c*v/1000

    1000cm^3=L

    moles MnO4-: 1L x 0.01M= 0.01

    moles MnO4-: 0.0121 L x 0.01 M= 0.000121 moles.

    Is this right?? I'm not too sure if it is

    since it's talking about releasing anions would it be

    1000cm^3 = 1 L

    0.01 M = 0.01 mole/1000

    0.01 mole/1L * 1 L = 0.01 M KMnO4

    1 mole of KMnO4 releases 1 mole of MnO4- anions


    instead?
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    A 0.01M solution has 0.01 Mole of the solute in each litre. This means that there is 0.01 Mole of KMnO4, K, Mn, or MnO4.

    If you have 12cm^3 then you will have \frac {12}{1000}\times 0.01=0.00012M \quad\text{of}\quad KMnO_4 which suggests you are correct.

    Whilst this is true "1 mole of KMnO4 releases 1 mole of MnO4- anions". I don't see what you are driving at?
 
 
 
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