Solveeit Logo

Question

Question: The normality of \[10\% \left( {w/v} \right)\] of acetic acid is: A. \[1N\] B. \[1.3N\] C. \...

The normality of 10%(w/v)10\% \left( {w/v} \right) of acetic acid is:
A. 1N1N
B. 1.3N1.3N
C. 1.7N1.7N
D. 1.9N1.9N

Explanation

Solution

Normality is defined as the measure of the concentration of a solution. Normality is expressed as the number of gram equivalents of the solute per litre of the solution.

Complete step by step answer:
Acetic acid is an organic carboxylic acid. It contains a methyl group attached o a carboxyl group. The chemical formula of acetic acid is CH3COOHC{H_3}COOH . The commercial glacial acetic acid is around 17N17N. The molar mass of acetic acid = 2 ×2{\text{ }} \times atomic mass of CC + 4 ×4{\text{ }} \times atomic mass of HH + 2 ×2{\text{ }} \times atomic mass of OO.
=2×12+4×1+2×16=60g/mol.= 2 \times 12 + 4 \times 1 + 2 \times 16 = 60g/mol.
The key points related to Normality are:
a. It is also known as equivalent concentration of the solute in solution.
b. It is the ratio of the number of grams equivalent of the solute per litre of the solution.
c. It is also used to determine the gram equivalent of the solute for a total volume of a solution.
d. It is abbreviated as NN and has the units ofeqL1eq{L^{ - 1}}.
The given solution is a 10%(w/v)10\% \left( {w/v} \right) acetic acid solution. The percentage indicates that 10g10g of acetic acid is dissolved in100mL100mL of water. The amount of acetic acid present in 1000mL1000mL or 1L1L of water =10×10=100g10 \times 10 = 100g.
The molar mass of acetic acid is 60g/mol60g/mol, i.e. 60g60g of acetic acid is present in 1mole1mole of acetic acid.
So the moles of acetic acid in 100g100g of acetic acid is
=10060=1.67moles.= \dfrac{{100}}{{60}} = 1.67moles.
Hence the moles of acetic acid present in 1L1L of water is 1.67moles1.67moles and so the solution is 1.67M1.67M .
The normality and the molarity of acetic acid are the same because the number of equivalents changes by ionization of acetic acid is 11. So the basicity of the acetic acid is one and hence the normality of acetic acids is equal to 1.67N1.67N.
Hence option C is the correct answer, i.e. the normality of 10%(w/v)10\% \left( {w/v} \right) of acetic acid is 1.7N1.7N.

Note:
The normality unit is primarily used to indicate the strength of a solution. It is vastly used in acid base chemistry for recording the concentration of unknown solutions.