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Question: The normality of \({\text{10}}\) volume \({H_2}{O_2}\) is nearly? A) \({\text{2}} \cdot {\text{1}}...

The normality of 10{\text{10}} volume H2O2{H_2}{O_2} is nearly?
A) 2⋅1{\text{2}} \cdot {\text{1}}
B) 3⋅4{\text{3}} \cdot {\text{4}}
C) 1⋅7{\text{1}} \cdot {\text{7}}
D) 5⋅1{\text{5}} \cdot {\text{1}}

Explanation

Solution

Try to write a balanced chemical reaction for the decomposition reaction of hydrogen peroxide. From that reaction, one can find out the number of moles used of hydrogen peroxide and then can calculate the gram equivalent of the hydrogen peroxide. Normality can be calculated as the gram equivalence divided volume and at last select the correct choice.

Complete step by step answer:

  1. First of all, as we need to find out the normality of the hydrogen peroxide, we need to write a balanced reaction for the decomposition of the hydrogen peroxide as below,
    2H2O2(l)→Heat2H2O(l)+O2(g)2{H_2}{O_{2(l)}}\xrightarrow{{Heat}}2{H_2}{O_{(l)}} + {O_{2(g)}}
    In the above reaction when two moles of hydrogen peroxide decompose in presence of heat, they form two molecules of water and one molecule of oxygen.
  2. So, from the above-balanced reaction equation we have found out the number of moles of hydrogen peroxide to decompose into its components. Now we can calculate the gram equivalent of the hydrogen peroxide as below,
    Gram equ. of H2O2=Molar massn factor{\text{Gram equ}}{\text{. of }}{H_2}{O_2} = \dfrac{{{\text{Molar mass}}}}{{{\text{n factor}}}}
    The molar mass of hydrogen peroxide =2×1+2×16=34g = 2 \times 1 + 2 \times 16 = 34g
    n factor of hydrogen peroxide is two as two moles needed.
    Now let us put these values in the above equation we get,
    Gram equ⋅ wt⋅ of H2O2=342=17{\text{Gram equ}} \cdot {\text{ wt}} \cdot {\text{ of }}{H_2}{O_2} = \dfrac{{34}}{2} = 17
  3. Now as we got the value of gram equivalents of hydrogen peroxide, we can calculate the normality by using the following formula,
    Normality of H2O2=Gram equivalent weightVolume{\text{Normality of }}{{\text{H}}_2}{O_2} = \dfrac{{{\text{Gram equivalent weight}}}}{{Volume}}
    As we know the volume of hydrogen peroxide is given in the question as 10 litres{\text{10 litres}} and the gram equivalent value is 17{\text{17}}, we get the above equation as,
    Normality of H2O2=1710{\text{Normality of }}{{\text{H}}_2}{O_2} = \dfrac{{17}}{{10}}
    By doing the calculation of the above equation we get,
    Normality of H2O2=1⋅7N{\text{Normality of }}{{\text{H}}_2}{O_2} = 1 \cdot 7N
    Therefore, the normality of 10{\text{10}} volume H2O2{H_2}{O_2} is 1⋅7{\text{1}} \cdot {\text{7}} which shows option C as the correct choice.

Note:
The gram equivalent weight of a substance is the mass of one equivalent substance that is present in the reaction. One should remember to divide the molar mass by the number of moles of that substance to calculate the equivalent weight. Normality gives the idea about the number of moles of reactive units present in a solution.