Question

$HN{{O}_{3}}$ used as a reagent has a specific gravity of 1.42 g/mL and contains 70 % by strength of $HN{{O}_{3}}$.
Calculate the normality of acid.
A) 15.78
B) 10
C) 20
D) None of the above

Answer 1

Hint: The answer to this question is dependent on the calculation of molarity of the solution using the given data which is given by the formula, $Normality=\dfrac{percentage~by~weight\times 10\times d}{Eq.wt.}$ and this facts leads you to the required answer.

Complete Solution :
- In the lower classes of chemistry in the physical chemistry part, we have studied the basic measurement parameters of the solutions such as molarity, molality, normality and also mole fraction of the solutions.
- Now, let us see in detail how normality is calculated according to the data that is provided in question.
We know that normality of the solution is given by the formula $Normality = \dfrac{percentage~by~weight\times 10\times d}{Eq.wt.}$ and this can also be written as $Normality = \dfrac{no.of~equivalence}{Eq.wt.}$
Here, we have to firstly find the equivalent weight in order to calculate the normality of the solution.
thus, equivalent weight is given by the formula:
\[Eq.wt.=\dfrac{Mol.wt.}{V.factor}\]
- where V – factor is nothing but the valency factor which is defined as factor which helps us to identify the number of valence electrons present in the valence shell of a given element of compound.
Now, since the nitric acid only one proton can be replaced, the valency factor will be one.
Therefore the above formula will be equal to the molecular weight.
Thus, the equivalent weight$ = 1 + 4 + 16\times 3 = 63g$
Now, since molarity is the number of equivalence present in one litre of the solution, the given density value is to be converted into litres and therefore, the density will be $d = 1.42g/ml\times 1000 = 1420g/L$
- Now, mass of nitric acid in one litre of the solution will be $Mass=1420\times 70% = 1420\times \dfrac{70}{100} = 994g$
- Now, the number of equivalence of nitric acid will be$ = \dfrac{994}{63} = 15.78$
Since the molarity of the solution according to equation $Normality = \dfrac{no.of.equivalence}{volume(litres)}$
we have, $Normality = \dfrac{15.78}{1} = 15.78N$
Therefore, normality of the solution is $15.78N$
So, the correct answer is “Option A”.

Note: Note that equivalent weight and molecular weight are different terms and do not be confused about this factor as equivalent weight is proportional mass of chemical entities which combines or displaces other chemical entities and molecular weight is the mass of the molecule present in the compound.