Question

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

Answer 1

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 find the equivalents of oxygen reacting and convert it to mass.

Complete step by step solution:
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,
${\text{5}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}{\text{ + 2KMn}}{{\text{O}}_{\text{4}}}{\text{ + 3}}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} \to 5{{\text{O}}_{\text{2}}}{\text{ + }}{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + 2MnS}}{{\text{O}}_{\text{4}}}{\text{ + 8}}{{\text{H}}_{\text{2}}}{\text{O}}$
As per the law of titrimetry,
The Milliequivalents of hydrogen peroxide in 10 mL solution = Milliequivalents of potassium permanganate or,
${{\text{V}}_{\text{1}}}{{\text{S}}_{\text{1}}}{\text{ = }}{{\text{V}}_{\text{2}}}{{\text{S}}_{\text{2}}}$
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, ${\text{10}} \times {{\text{S}}_{\text{1}}}{\text{ = 25}} \times \dfrac{1}{{20}} = \dfrac{5}{4}$
Or, ${{\text{S}}_{\text{1}}} = \dfrac{5}{4} \times 10 = 1.25$ , therefore $1.25$ milliequivalents of hydrogen peroxide is present in 10 mL solution.
Therefore the amount of hydrogen peroxide in 1000 mL solution = 125 milliequivalents.
As the source of oxygen is hydrogen peroxide, hence the moles of oxygen formed is equal to the moles of hydrogen peroxide present in the reaction medium = 125 milliequivalents.
To convert the milliequivalents to mass,
1 milliequivalent = $\dfrac{{{\text{equivalent weight}}}}{{{\text{1000}}}}$ = $\dfrac{{\dfrac{{{\text{weight of the substance}}}}{{{\text{molecular weight}}}} \times 1000}}{2} = \dfrac{{\dfrac{{\text{w}}}{{32}} \times 1000}}{2}$ =125
Therefore, weight of oxygen formed = 2 grams, as calculated from the earlier equation.
Hence, the correct option is option C.

Note:
The number of milliequivalents 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.