Question

The volume strength of 1M ${{H}_{2}}{{O}_{2}}$ is:
(Molar mass of \[{{H}_{2}}{{O}_{2}}=34gmo{{l}^{-1}}\] )
A) 16.8
B) 11.35
C) 22.4
D) 5.6

Answer 1

Hint Volume strength of ${{H}_{2}}{{O}_{2}}$ is the amount or volume of ${{O}_{2}}$ liberated during dissociation.
1 mole of ${{H}_{2}}{{O}_{2}}$ at STP will occupy a volume of 22.4L.

Complete step by step solution:
So here the question is about, to find the volume strength of 1 mole of ${{H}_{2}}{{O}_{2}}$ .So for that we should know, what does volume strength means.
Volume strength is the volume of oxygen liberated during the decomposition or dissociation of the ${{H}_{2}}{{O}_{2}}$ molecule into its smaller constituents, ${{H}_{2}}O$ and ${{O}_{2}}$.
Now let’s write the balanced chemical equation for the decomposition reaction of the ${{H}_{2}}{{O}_{2}}$.
\[2{{H}_{2}}{{O}_{2}}\to 2{{H}_{2}}O+{{O}_{2}}\]
Here we can see that for two moles of ${{H}_{2}}{{O}_{2}}$, only one mole of ${{O}_{2}}$ is liberated.
One mole of ${{H}_{2}}{{O}_{2}}$ is equal to $22.4L$ in terms of volume at standard temperature and pressure conditions.
Since here for 2 moles of ${{H}_{2}}{{O}_{2}}$ , 1 mole of ${{O}_{2}}$ is liberated .So for one mole of ${{H}_{2}}{{O}_{2}}$ only half mole of ${{O}_{2}}$ will be liberated.
Now, convert the half mole of ${{O}_{2}}$ into terms of volume, which is,
1 mole ${{O}_{2}}$=$\dfrac{1}{2}\times 22.4$
1 mole ${{O}_{2}}$=11.2L
\[Volume\text{ }strength\text{ }of{{H}_{2}}{{O}_{2}}=\text{ }one\text{ }mole\text{ }of{{O}_{2}}\times Molarity\]
\[Volume\text{ }strength\text{ }of{{H}_{2}}{{O}_{2}}=\text{ 11}\text{.2}\times \text{1=11}\text{.2}\], which is almost equal to 11.35.

So the correct option for this question is option (B).

Note: Alternate method-
Let’s write the balanced chemical equation,
\[2{{H}_{2}}{{O}_{2}}\to 2{{H}_{2}}O+{{O}_{2}}\]
Let, ${{n}_{{{H}_{2}}O}}_{_{2}}=M\times V$, M is the concentration term and V is the volume and ${{n}_{{{H}_{2}}{{O}_{2}}}}$,number of moles of ${{H}_{2}}{{O}_{2}}$.
As V=1 and ${{n}_{{{H}_{2}}O}}_{_{2}}$=M
Now,${{n}_{O}}_{_{2}}=\dfrac{M}{2}$ (for 1 mole only half mole of ${{O}_{2}}$is liberated)
Here volume of strength of ${{O}_{2}}$ at STP (1 atm and 273K),
${{V}_{{{O}_{2}}}}={{n}_{O}}_{_{2}}\times 22.4L$
${{V}_{{{O}_{2}}}}=\dfrac{M}{2}\times 22.4L=\dfrac{1}{2}\times 22.4=11.2\approx 11.35L$