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what volume of 0.1M KMnO4 is need to oxidise 100mg of FeC2O4 in acidic medium?

what volume of 0.1M KMnO4 is need to oxidise 100mg of FeC2O4 in acidic medium?

Grade:11

2 Answers

RON
12 Points
9 years ago
As their number of equivalents are equal i.e n1v1=n2v2
so,
We just need to find out these variables.
n1 i.e normality of KMn04= molarity x n-factor
n-factor of KMn04 in acidic medium is 5
normality = 0.1(given molarity) x 5
=0.5
so n1v1= 0.5x v1
Now to find out equivalents of Fec204 we have to to use the formula n.eq=given mass/eq.mass.
Given mass=.1g
eq.mass=mol. mass/n-factor
mol mass= roughly 144
n-factor of FeC204 in acidic medium=1
therefore. eq. mass=144
.Hence number of gram equivalents=0.1/144.
Now according to the law of equivalence .
n1v1=n2v2 (equal gram equivalents)
0.5xv1=0.1/144
v1=1/(144*5)
=1/720.
therefore our v1 comes out to be 0.00139 litres or 1.39 ml.
 
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

As their number of equivalents are equal i.e n1v1=n2v2
so,
We just need to find out these variables.
n1 i.e normality of KMn04= molarity x n-factor
n-factor of KMn04 in acidic medium is 5
normality = 0.1(given molarity) x 5
=0.5
so n1v1= 0.5x v1
Now to find out equivalents of Fec204 we have to to use the formula n.eq=given mass/eq.mass.
Given mass=.1g
eq.mass=mol. mass/n-factor
mol mass= roughly 144
n-factor of FeC204 in acidic medium=1
therefore. eq. mass=144
.Hence number of gram equivalents=0.1/144.
Now according to the law of equivalence .
n1v1=n2v2 (equal gram equivalents)
0.5xv1=0.1/144
v1=1/(144*5)
=1/720.
therefore our v1 comes out to be 0.00139 litres or 1.39 ml.

Thanks and Regards

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