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
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
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Finding moles of an ion present in a solution watch
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Last edited by pizzacammed; 29-02-2016 at 21:48.
- 29-02-2016 21:44
- 02-03-2016 10:54