Question

The purity of \[{H_2}{O_2}\] in a given sample \[85\% \] . Calculate the weight of impure sample of \[{H_2}{O_2}\] which requires \[10\,ml\]of \[\dfrac{M}{5}KMn{O_4}\] solution in a titration in acidic medium.
A. \[2g\]
B. \[0.2g\]
C. \[0.17g\]
D. \[0.15g\]

Answer 1

Hint: The chemical formula given i.e. \[{H_2}{O_2}\] in the question is Hydrogen peroxide. We should know about the properties of hydrogen peroxide. It is a strong oxidizer as well as a reducing agent. It can behave as oxidizing as well as reducing agent in both acidic and alkaline medium.

Complete answer:
In this question, first we need to calculate the weight of hydrogen peroxide, \[{H_2}{O_2}\].
According to the given question, the equivalent of hydrogen peroxide is equal to the equivalent of Potassium permanganate, \[KMn{O_4}\] . We know that in acidic medium, \[KMn{O_4}\] undergoes reduction and gets reduced to \[Mn{O_4}^ - \]. The value of n- factor is \[5\]
So, equivalent of \[Mn{O_4}^ - \]= equivalent of \[{H_2}{O_2}\]
Equivalent = \[normality\, \times \,volume\]
Also, normality = \[Molarity\, \times \,n\,factor\]
It is given that molarity = \[\dfrac{1}{5}\]
Equivalent of \[KMn{O_4}\]= \[\dfrac{1}{5} \times 5 \times \dfrac{{10}}{{1000}}\]….(I)
Also, we know that equivalent = \[no.\,of\,moles\,\, \times \,\,n\,factor\]
The n- factor of \[{H_2}{O_2}\]= \[2\]
 Equivalent of \[{H_2}{O_2}\] = \[no.\,of\,moles\,\, \times \,\,n\,factor\]
Equivalent of \[{H_2}{O_2}\] = \[no.\,of\,moles\,of\,{H_2}{O_2}\,\, \times \,\,2\]….(II)
Now, Equivalent of \[Mn{O_4}^ - \]= equivalent of \[{H_2}{O_2}\]….(III)
Put the value of (I), (II) in (III)
We get,
\[\dfrac{1}{5} \times 5 \times \dfrac{{10}}{{1000}}\]= \[no.\,of\,moles\,of\,{H_2}{O_2}\,\, \times \,\,2\]
No. of moles of \[{H_2}{O_2}\] = \[\dfrac{1}{{200}}\]….(IV)
Now, we also know that Number of moles = \[\dfrac{{Given\,\,weight}}{{Molar\,mass}}\]
No. of moles of \[{H_2}{O_2}\]= \[\dfrac{{Given\,\,weight}}{{34}}\]
Put No. of moles in equation (IV)
We get,
\[\dfrac{{Given\,\,weight}}{{34}}\, = \dfrac{1}{{200}}\]
Given weight of \[{H_2}{O_2}\]= \[\dfrac{{34}}{{200}}g\]
We can write it as
Mass of \[100\% \]\[{H_2}{O_2}\] required = \[\dfrac{{34}}{{200}}g\]
It is given that the purity of \[{H_2}{O_2}\] in a given sample is \[85\% \].
Mass of \[85\% \]\[{H_2}{O_2}\] required = \[\dfrac{{34}}{{200}} \times \dfrac{{100}}{{85}}\]
Mass of \[85\% \]\[{H_2}{O_2}\] required = \[0.2g\]

The correct answer is option (B).

Note:
You need to remember the formula used in this question i.e. Equivalence is the product of normality and volume. Normality is defined as the product of molarity and n- factor. Also, Number of moles of a substance is the ratio of its given weight and its molar mass.