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Calculate the mass in grams of available oxygen in \[1\] litre of \[{{H}_{2}}{{O}_{2}}\]​ solution if \[10\text{ }mL\] of this solution is allowed to react completely with \[25\text{ }mL\] of​ $\dfrac{N}{20}KMn{{O}_{4}}$​ solution.
A.\[2.5\text{ }g\]
B.\[3\text{ }g\]
C.\[2\text{ }g\]
D.None of these

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Last updated date: 03rd Mar 2024
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IVSAT 2024
Answer
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Hint :We know that the volume of the mass of oxygen that is released from a reaction in which any chemical compound is converted to elemental oxygen is often defined as the available oxygen. We shall write the balanced equation and use it to mass.

Complete Step By Step Answer:
In the above reaction, hydrogen peroxide reacts with potassium permanganate to get converted to oxygen as potassium permanganate is a very strong oxidizing agent that itself gets reduced to manganese ion according to the following reaction;
Mass equivalent of \[{{H}_{2}}{{O}_{2}}\] in \[10~mL\] solution \[=~\] Mass equivalent of \[KMn{{O}_{4}}=25\times \dfrac{1}{20}=1.25\]
We have Mass equivalent of \[{{H}_{2}}{{O}_{2}}\] in \[1000~mL~=101.25\times 1000=125\] and thus, Mass equivalent of \[{{O}_{2}}\] ​ formed \[=125\]
\[KMnO_{4}^{{}}\] acts as a good oxidizing agent in acidic medium. If acid is not used \[KMnO_{4}^{{}}\] may be oxidized to \[Mn{{O}_{2}}\] giving a brown precipitate. Where on the left side are the volume and the strength of the hydrogen peroxide solution and on the right hand side are the volume and the strength of potassium permanganate.
$\therefore \dfrac{w}{\left( \dfrac{32}{2} \right)}\times 1000=125~[{{\left( {{O}_{2}} \right)}^{-1}}\xrightarrow{{}}{{\left( {{O}_{2}} \right)}^{0}}+2e]$
\[\Rightarrow {{w}_{{{O}_{2}}}}=2~g\]
Hence, the mass of available oxygen in in \[1\] litre of \[{{H}_{2}}{{O}_{2}}\] solution is \[2~g.\]

Therefore, the correct answer is option C.

Note :
Remember that the number of millie equivalents or equivalents of a chemical compound that is dissolved is one litre of the solution is called the normality of the solution. The molarity on the other hand is the number of moles of the solute present per litre of the solution.
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