Grade 11Physical Chemistry

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

keerthi priya

11 Years agoGrade 11

2 Answers

RON

Approved Tutor Answer11 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

5 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