Question

100 mL of ozone at STP were passed through 100 mL of 10 volume ${H_2}{O_2}$ solution. What is the volume strength of ${H_2}{O_2}$ after the reaction?
a.) 9.5
b.) 9.0
c.) 4.75
d.) 4.5

Answer 1

Hint: The volume strength may be defined as the volume of Oxygen gas liberated out from the ${H_2}{O_2}$ solution. It can be found by the formula as-

Volume Strength = $\dfrac{{Volume{\text{ of }}{{\text{O}}_2}}}{{Volume{\text{ of }}{{\text{H}}_2}{{\text{O}}_2}}}$

Complete answer:
First let's understand what is given to us and the reaction happening and what we need to find out.
So, we are given that volume of ozone = 100 ml
This ozone is passed through ${H_2}{O_2}$.
Volume of ${H_2}{O_2}$= 100 ml
Volume strength of ${H_2}{O_2}$= 10 volume
And now we have to find out the volume strength of ${H_2}{O_2}$ after the reaction.
First, let us write the reaction.
The first reaction is the breaking of ozone into oxygen gas and oxygen atom.
${O_3} \to {O_2} + O$
The second reaction could be the splitting of ${H_2}{O_2}$ as-
${H_2}{O_2} \to {H_2}O + O$
The third reaction can be that the oxygen atoms from first reaction and second reaction combine forming the oxygen molecule as-
$O + O \to {O_2}$
Thus, when we observe the first reaction, we get that
100 ml of ozone at STP will give 100 ml of ${O_2}$ and 100 ml of ‘O’ atom. This 100 ml of ‘O’ will combine with another 100 ml of ‘O’ atom forming the ${O_2}$ molecule which will come after the reaction with ${H_2}{O_2}$.
Thus, ${H_2}{O_2}$ will contribute equally to formation of this 100 ml of ${O_2}$ molecule and So, 50 ml of Oxygen comes from ${H_2}{O_2}$.
From this, we can calculate the volume of ${O_2}$ produced at STP by the formula as-
Volume of ${O_2}$ produced at STP = Volume strength $\times$ Volume of ${H_2}{O_2}$
Volume of ${O_2}$ produced at STP = 10 $\times$ 100
Volume of ${O_2}$ produced at STP = 1000 ml
Thus, the total amount of Volume of ${O_2}$ produced at STP is 1000 ml. But 50 ml of Oxygen is given by ${H_2}{O_2}$.
Thus, we have ${O_2}$ = 1000 - 50 = 950 ml
Now, again using formula we can find volume strength after the reaction as-

Volume Strength = $\dfrac{{Volume{\text{ of }}{{\text{O}}_2}}}{{Volume{\text{ of }}{{\text{H}}_2}{{\text{O}}_2}}}$

Volume Strength = $\dfrac{{950}}{{100}}$

Volume Strength = 9.5
Thus, the volume strength of ${H_2}{O_2}$ after the reaction comes out to be 9.5

So, the option a.) is the correct answer.

Note: We should do the numerical step by step to avoid any confusion.
Further, we should take care that the oxygen is liberated when the reaction takes place with the ${H_2}{O_2}$ molecule not itself. There is an equal contribution of ${H_2}{O_2}$ in the formation of oxygen molecules with the ozone.
The volume strength will be calculated after subtracting the change that occurred during the reaction.