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

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.