Question

The volume strength of $1.5N{H_2}{O_2}$ solution is :
A. $16.8L$
B. $8.4L$
C. $4.2L$
D. $5.2L$

Answer 1

Hint : First of all we should understand about the term volume strength of hydrogen peroxide. The volume strength of hydrogen peroxide can be defined as a term that is used to express the concentration of ${H_2}{O_2}$ in terms of volumes of oxygen. When hydrogen peroxide decomposes it forms water and oxygen gas. So this term is based on the decomposition of hydrogen peroxide. The reaction of decomposition of hydrogen peroxide will be: $2{H_2}{O_2} \to 2{H_2}O + {O_2}$ with the help of this reaction we will calculate volume strength of $1.5N{H_2}{O_2}$ solution.

Complete step by step solution:
Let suppose that in one litre solution there are $M$ moles of hydrogen peroxide. From the stoichiometry of the reaction we got to know that $2$ moles of hydrogen peroxide gives one mole of oxygen or we can say that $22.4L$ of oxygen is produced at STP from $2$ moles of hydrogen peroxide.
Hence one mole of ${H_2}{O_2}$ will produce oxygen gas $= \dfrac{{22.4L}}{2} \times 1 = 11.2L$
And $M$ moles of ${H_2}{O_2}$ will produce oxygen gas $= (11.2 \times M)L$,We can say that one litre of hydrogen peroxide will produce $= (11.2 \times M)L$ of oxygen gas.
Therefore volume strength of ${H_2}{O_2}$ $= 11.2 \times M$ but we have to convert this term to in terms of normality. We know that n factor of ${H_2}{O_2}$ is two so $M = \dfrac{N}{2}$ $= \dfrac{{1.5}}{2}$ (value of normality is given in the problem that is $1.5$ $N$)
Thus the volume strength of Hydrogen peroxide $= 11.2 \times \dfrac{{1.5}}{2} = 8.4$ litre

Here option B is correct. That is $8.4L$ answer to this problem.

Note : We have approached this problem with the stoichiometry of the decomposition reaction, we calculated the value of the oxygen gas produced by $M$ moles of hydrogen peroxide which is $11.2 \times M$ litre and calculated the volume strength of hydrogen peroxide.