Question

1-liter solution of $NaOH$ contains 4 g of it. What shall be the difference between molarity and normality?
(a) 0.01
(b) 0
(c) 0.05
(d) 0.20

Answer 1

Hint: The normality of the solution can be calculated by dividing the gram equivalent of the solute with the volume of the solution. The molarity of the solution can be calculated by dividing the number of moles with the volume of the solution.

Complete step by step answer:
The normality of the solution is defined as the number of grams equivalent of the solute present in liters. It is represented by the symbol, N.
\[Normality=\dfrac{\text{Gram equivalent of the solute}}{\text{Volume of the solution}}\]

The number of grams equivalent of the solute is calculated by dividing the mass of the solute in grams to the equivalent mass of the solute.
\[\text{No}\text{. of gram equivalents = }\dfrac{\text{Mass of the solute in grams}}{\text{Equivalent mass of the solute}}\]

The equivalent masses of acid, bases, and salts are calculated as follows:
\[\text{Eq}\text{. mass of an acid = }\dfrac{\text{Mol}\text{. mass of the acid}}{\text{Basicity}}\]
\[\text{Eq}\text{. mass of an base = }\dfrac{\text{Mol}\text{. mass of the base}}{\text{Acidity}}\]
\[\text{Eq}\text{. mass of an salt = }\dfrac{\text{Mol}\text{. mass of the salt}}{\text{Total positive valency of metal atoms}}\]

Basicity: It is the number of displaceable\[{{H}^{+}}\] ions present in one molecule of the acid. (e.g. 1 for HCl , 2 for \[{{H}_{2}}S{{O}_{4}}\] )
Acidity: It is the number of displaceable \[O{{H}^{-}}\]ions present in one molecule of the base. (e.g. 1 for NaOH)

Molarity: The molarity of the solution is defined as the number of moles of the solute present per liter. It is represented by the symbol, M.
\[Molarity=\dfrac{\text{Moles of the solute}}{\text{Volume of the solution}}\]

Moles of the solute can be calculated by:
\[Moles=\dfrac{\text{mass of the solute}}{\text{molar mass of the solute}}\]
So, for molarity: Given mass = 4 g
Molecular mass of $NaOH$= 40 g
The number of moles of $NaOH$ will be
\[Moles=\dfrac{\text{mass of the solute}}{\text{molar mass of the solute}}=\dfrac{4}{40}=0.1\text{ mol}\]

Given the volume of the solution is 1 L.
The molarity will be:
\[Molarity=\dfrac{\text{Moles of the solute}}{\text{Volume of the solution}}=\dfrac{0.1}{1}=0.1M\]
The molarity of the solution is 0.1 M.

Normality of the base = molarity x acidity.
The acidity of $NaOH$ is 1
So, the normality of $NaOH$= $0.1\text{ x 1 = 0}\text{.1}$
So, the molarity and normality of $NaOH$ are 0.1. Therefore, the difference between them is 0.
So, the correct answer is “Option B”.

Note: The molarity of the acid can be converted into normality by multiplying it with basicity. The molarity of the base can be converted into normality by multiplying it with acidity.
In the same way, we can calculate the molarity of the two mixing solutions, by:
\[{{M}_{1}}{{V}_{1}}+{{M}_{2}}{{V}_{2}}={{M}_{S}}({{V}_{1}}+{{V}_{2}})\].

For calculating the normality of two or more mixing solution:
\[{{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}={{N}_{S}}({{V}_{1}}+{{V}_{2}})\]