Question

The normality of $10\% $(mass/volume) acetic acid is:
(A) $1N$
(B) $10N$
(C) $1.7N$
(D) $0.83N$

Answer 1

Hint: Normality is the measure of concentration equal to the gram equivalent weight per litre of solution gram equivalent weight is the measure of reactive capacity of the molecule.

Complete step by step answer:
$10\% $ (w/v) acetic acid is given in the question.
$10\% $ (w/v) acetic acid means $10$g acetic acid in $100$ml of water.
Thus, we need $1000$ml of water to calculate normality of the acetic acid.
$10$g acetic acid is in $100$ml of water. Therefore, $100$g acetic acid is present in $1000$ml of water.
We know that,
Morality $ = \dfrac{{{\text{Number of mass of solute }}\left( n \right)}}{{{\text{Volume of solution}}\left( l \right)}}$
It can be defined as morality is equal to the number of moles of solute present in one litre of solution.
So, Number of moles of solute $ = \dfrac{{{\text{Given mass}}}}{{{\text{Molar mass}}}} = \dfrac{{100g}}{{600g/mol}} = 1.67mol$
As we know that,
${\text{n}} \times {\text{molarity}} = {\text{normality}}$
${\text{n}} \to $n factor/acidity/basicity
In case of acetic acid $\left( {C{H_3}COOH} \right)$
So, the n factor is equal to \[1\]
Normality is equal to morality
Normality = Morality
N = \[1.67 \approx 1.7N\]
Thus, $10\% $ the normality of acetic acid solution is \[1.7N\],

So, the correct answer is Option C .

Note:
To make $10\% $ acetic solution, dilute the solute in sufficient solvent to produce the final volume of solution desired. For example, to prepare \[100\]ml of a $10\% $ by (w/v) solution of acetic acid, dilute \[10\]gm of acetic acid with distilled or deionized water to make \[100\]ml of sodium.Acetic acid can be hazardous chemical if not used in a safe and appropriate manner. This liquid is highly corrosive to skin and eyes, that’s why it must be handled with extreme care. Acetic acid can also be damaging to the internal organs of ingested or in the case of vapor inhalation.