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Question: Calculate the strength of \(10\) volume solution of hydrogen peroxide. A. \(3%\) B. \(4%\) C. ...

Calculate the strength of 1010 volume solution of hydrogen peroxide.
A. 33%
B. 44%
C. 11%
D. 22%

Explanation

Solution

Volume strength of hydrogen peroxide is the volume of oxygen released on decomposition of one volume of hydrogen peroxide at standard temperature and pressure (STP). So, 1010 volume hydrogen peroxide on reduction will release ten times the volume of the sample at STP.

Complete step by step solution:
Given that,
We have to find the strength of a 1010 volume solution of hydrogen peroxide.
So, for finding the volume strength we have to write a balanced equation of hydrogen peroxide i.e. H2O2{{H}_{2}}{{O}_{2}} decomposition to form water and oxygen.
So, here is the balanced chemical reaction of H2O2{{H}_{2}}{{O}_{2}} decomposition to form water and oxygen:
2H2O22H2O+O22{{H}_{2}}{{O}_{2}}\to 2{{H}_{2}}O+{{O}_{2}}
2×34g=68g2\times 34g=68g 22.4latSTP22.4 l at STP
As we can see that, in H2O2{{H}_{2}}{{O}_{2}}oxidation state of oxygen is 1-1, because when we apply the rule that sum of oxidation number of all the atoms in a neutral compound is zero and here oxidation state of hydrogen atom is +1+1, so the oxidation state of oxygen atom will be 1-1.
Similarly, in H2O{{H}_{2}}Ooxidation state of oxygen will be +2+2, and in O2{{O}_{2}}oxidation state is 00.
So, from the balanced reaction of H2O2{{H}_{2}}{{O}_{2}}, we can see that two moles of H2O2{{H}_{2}}{{O}_{2}}is giving one mole of O2{{O}_{2}}.
As we know that one mole of any gas at STP has volume 22.422.4 litres.
Therefore, one mole of H2O2{{H}_{2}}{{O}_{2}} at STP will give 22.422.4 litres of oxygen.
As per the given question,
1010 volume solution of H2O2{{H}_{2}}{{O}_{2}}means that one litre of this solution will give 1010 litres of oxygen at STP.
On the basis of the above equation, 22.422.4 litres of oxygen are produced from 68g68g H2O2{{H}_{2}}{{O}_{2}} at STP.
So, 1010 litres of Oxygen at STP is produced from 68×1022.4=30.3630g\dfrac{68\times 10}{22.4}=30.36\approx 30g H2O2{{H}_{2}}{{O}_{2}}.
Therefore, strength of H2O2{{H}_{2}}{{O}_{2}}in 1010 volume H2O2{{H}_{2}}{{O}_{2}} solution is 30g30g per litre i.e. 33% H2O2{{H}_{2}}{{O}_{2}} solution.

Hence, the correct option is A.

Note: Volume strength of hydrogen peroxide is the concentration of H2O2{{H}_{2}}{{O}_{2}} in terms of volumes of oxygen gas based on its decomposition to form water and oxygen. So here, 1010 volume solution of H2O2{{H}_{2}}{{O}_{2}} means that one litre of this solution will give 1010 litres of oxygen at STP. Possibly, you may make mistakes in balancing the decomposition reaction.