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Question: The volume strength of 1M \({{H}_{2}}{{O}_{2}}\) is: (Molar mass of \[{{H}_{2}}{{O}_{2}}=34gmo{{l}...

The volume strength of 1M H2O2{{H}_{2}}{{O}_{2}} is:
(Molar mass of H2O2=34gmol−1{{H}_{2}}{{O}_{2}}=34gmo{{l}^{-1}} )
A) 16.8
B) 11.35
C) 22.4
D) 5.6

Explanation

Solution

Hint Volume strength of H2O2{{H}_{2}}{{O}_{2}} is the amount or volume of O2{{O}_{2}} liberated during dissociation.
1 mole of H2O2{{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 H2O2{{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 H2O2{{H}_{2}}{{O}_{2}} molecule into its smaller constituents, H2O{{H}_{2}}O and O2{{O}_{2}}.
Now let’s write the balanced chemical equation for the decomposition reaction of the H2O2{{H}_{2}}{{O}_{2}}.
2H2O2→2H2O+O22{{H}_{2}}{{O}_{2}}\to 2{{H}_{2}}O+{{O}_{2}}
Here we can see that for two moles of H2O2{{H}_{2}}{{O}_{2}}, only one mole of O2{{O}_{2}} is liberated.
One mole of H2O2{{H}_{2}}{{O}_{2}} is equal to 22.4L22.4L in terms of volume at standard temperature and pressure conditions.
Since here for 2 moles of H2O2{{H}_{2}}{{O}_{2}} , 1 mole of O2{{O}_{2}} is liberated .So for one mole of H2O2{{H}_{2}}{{O}_{2}} only half mole of O2{{O}_{2}} will be liberated.
Now, convert the half mole of O2{{O}_{2}} into terms of volume, which is,
1 mole O2{{O}_{2}}=12×22.4\dfrac{1}{2}\times 22.4
1 mole O2{{O}_{2}}=11.2L
Volume strength ofH2O2= one mole ofO2×MolarityVolume\text{ }strength\text{ }of{{H}_{2}}{{O}_{2}}=\text{ }one\text{ }mole\text{ }of{{O}_{2}}\times Molarity
Volume strength ofH2O2= 11.2×1=11.2Volume\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,
2H2O2→2H2O+O22{{H}_{2}}{{O}_{2}}\to 2{{H}_{2}}O+{{O}_{2}}
Let, nH2O2=M×V{{n}_{{{H}_{2}}O}}_{_{2}}=M\times V, M is the concentration term and V is the volume and nH2O2{{n}_{{{H}_{2}}{{O}_{2}}}},number of moles of H2O2{{H}_{2}}{{O}_{2}}.
As V=1 and nH2O2{{n}_{{{H}_{2}}O}}_{_{2}}=M
Now,nO2=M2{{n}_{O}}_{_{2}}=\dfrac{M}{2} (for 1 mole only half mole of O2{{O}_{2}}is liberated)
Here volume of strength of O2{{O}_{2}} at STP (1 atm and 273K),
VO2=nO2×22.4L{{V}_{{{O}_{2}}}}={{n}_{O}}_{_{2}}\times 22.4L
VO2=M2×22.4L=12×22.4=11.2≈11.35L{{V}_{{{O}_{2}}}}=\dfrac{M}{2}\times 22.4L=\dfrac{1}{2}\times 22.4=11.2\approx 11.35L