Solveeit Logo

Question

Question: Normalization of 10ml of '20 V' \({H_2}{O_2}\) is: A. 1.79 B. 3.58 C. 60.86 D. 6.086...

Normalization of 10ml of '20 V' H2O2{H_2}{O_2} is:
A. 1.79
B. 3.58
C. 60.86
D. 6.086

Explanation

Solution

In this question using the given values determine the number of moles present and then determine the normality with the help of normality and molarity relation.

Complete answer:
Now from the question
20V H2O2{H_2}{O_2} means one litre of H2O2{H_2}{O_2} solution will be producing 20 litres of oxygen
It is known that,
2H2O22H2O+O22{H_2}{O_2} \to 2{H_2}O + {O_2}
Now, 2 moles of H2O21moleO2{H_2}{O_2} \to 1mole{O_2}
20V H2O2{H_2}{O_2} produces 20litre O2{O_2}
or, 1 litre H2O2{H_2}{O_2} produces 20 litres O2{O_2}
1 litre H2O2{H_2}{O_2} will produce (20L22.4L)\left( {\dfrac{{20L}}{{22.4L}}} \right) moles of O2{O_2}
Now, number of moles in 1 litre of H2O2{H_2}{O_2}
=2×no.ofMolesofO2= 2 \times no.ofMolesof{O_2}
=2×2022.4= 2 \times \dfrac{{20}}{{22.4}}
=1.79moles= 1.79moles
Now, Molarity=moles of soluteVolume of Solution(Litre)Molarity = \dfrac{{\text{moles of solute}}}{{\text{Volume of Solution(Litre)}}}
=1.791=1.79M= \dfrac{{1.79}}{1} = 1.79M
Now Normality is given by:
The relation between Normality and Molarity
N=n×MN = n \times M
where 'n' is the number of equivalents
now from the question
nforH2O2=2nfor{H_2}{O_2} = 2
therefore,
Normality N=2×1.79N = 2 \times 1.79
Normality=3.58NNormality = 3.58N

Hence the correct option is B .

Note:
RELATION BETWEEN NORMALITY AND MOLARITY :-
Normality is defined as the number of gram equivalent of solute dissolved in one litre of solution. The unit of normality is N. It is mostly used in titration calculations. The Normality of normal solutions is one.
Normality=Molecular MassEquivalent Mass of SoluteNormality = \dfrac{{\text{Molecular Mass}}}{{\text{Equivalent Mass of Solute}}}
Molarity of a solution is defined as the number of moles of solute dissolved in one liter of the solution. The unit is M or mol L-1 . It is the most preferred unit of concentration.
Molarity=Number of moles of SoluteVolume of Solution(Litre)Molarity = \dfrac{{\text{Number of moles of Solute}}}{{\text{Volume of Solution(Litre)}}}
Molarity and Normality are related as follows:
Normality=Molarity×Molar MassEquivalent MassNormality = Molarity \times \dfrac{{\text{Molar Mass}}}{{\text{Equivalent Mass}}}
The normality of acids can be calculated with the following formula:
Normality = Molarity x Basicity
The value for basicity can be known by counting the number of H+{H^ + } ions an acid molecule can give.
The normality of bases can be calculated with the following formula:
Normality = Molarity x Acidity
The value for acidity can be known by counting the number of OHO{H^ - } ions a base molecule can give.