Question

Calculate molarity and normality of 25% (w/v) $N{{H}_{3}}$ solution:

Answer 1

Hint: Weight/volume (w/v) % of a component means the amount of the component dissolved in 100 mL of the solution. The expression used to calculate the percentage composition in w/v is
     \[(w/v)%=\dfrac{\text{mass of the component }}{\text{volume of the solution }}\times 100\]
Molarity (M) is the number of moles of a solute dissolved per litre of solution. It is given as
     \[\text{Molarity}=\dfrac{\text{mass of solute (in grams)}}{\text{molar mass}}\times \dfrac{1000}{\text{volume of solution (in mL)}}\]
Relationship between normality and molarity is
     \[\text{Normality = Molarity }\!\!\times\!\!\text{ n factor}\]

Complete answer:
25% of (w/v) $N{{H}_{3}}$ solution means that 25 g of $N{{H}_{3}}$ is dissolved in 100 mL of the solution.
To calculate molarity, we require molar mass of the component. So let us first calculate the molar mass of $N{{H}_{3}}$.
Molar mass of $N{{H}_{3}}$ = 14 + 3 = 17g$mo{{l}^{-1}}$
Given mass of $N{{H}_{3}}$= 25 g
Volume of the solution containing 25 g of $N{{H}_{3}}$= 100 mL
Substitute the above values in molarity expression, i.e.
     \[\text{Molarity}=\dfrac{\text{mass of solute (in grams)}}{\text{molar mass}}\times \dfrac{1000}{\text{volume of solution (in mL)}}\]
We obtain the molarity in terms of mol ${{L}^{-1}}$.
     \[\begin{align}
  & Molarity=\dfrac{25g}{17g\,mo{{l}^{-1}}}\times \dfrac{1000\text{ }mL\text{ }{{L}^{-1}}}{100\text{ }mL} \\
 & Molarity=\dfrac{250}{17}mol\text{ }{{L}^{-1}}=14.7M \\
 & \\
\end{align}\]
 Therefore, the molarity of 25% (w/v) $N{{H}_{3}}$ solution is 14.7 M.
Let us now calculate the equivalent mass of $N{{H}_{3}}$. Equivalent weight of a base is given as
     \[\text{Equivalent weight = }\dfrac{\text{Molar mass}}{\text{Acidity}}\]
Acidity of base is equal to the number of displaceable hydroxide ion ($O{{H}^{-}}$) one molecule of the base can give. Consider the ionization of $N{{H}_{3}}$ in solution, i.e.
     \[N{{H}_{3}}+{{H}_{2}}O\to NH_{4}^{+}+O{{H}^{-}}\]
Therefore, $N{{H}_{3}}$ is a base with acidity equal to 1. Molar mass of $N{{H}_{3}}$ = 17g$mo{{l}^{-1}}$
Therefore, equivalent mass of $N{{H}_{3}}$ will be = \[\dfrac{\text{Molar mass}}{\text{Acidity}}=\dfrac{17g\,mo{{l}^{-1}}}{1\,eq\,mo{{l}^{-1}}}=17g\,e{{q}^{-1}}\]
Normality and molarity of a solution are related by given relation
     \[\text{Normality = Molarity }\!\!\times\!\!\text{ n factor}\]
Where n factor = \[\dfrac{\text{molar}\,\text{mass}}{\text{equivalent mass}}\] which is equal to acidity for a base. Therefore, normality of $N{{H}_{3}}$will be equal to
     \[\begin{align}
  & \text{Normality (N) = Molarity(M) }\!\!\times\!\!\text{ n factor} \\
 & N=14.7mol\,{{L}^{-1}}\times \dfrac{\text{molar}\,\text{mass}}{\text{equivalent mass}} \\
\end{align}\]
 Substituting, molar mass of $N{{H}_{3}}$ = 17g$mo{{l}^{-1}}$and equivalent mass of $N{{H}_{3}}$ = $17g\,e{{q}^{-1}}$, we get
     \[\begin{align}
  & N=14.7mol\,{{L}^{-1}}\times \dfrac{17g\,mo{{l}^{-1}}}{17g\,e{{q}^{-1}}} \\
 & N=14.7eq\,{{L}^{-1}} \\
\end{align}\]
Hence, molarity of 25% (w/v) $N{{H}_{3}}$ solution is 14.7mol ${{L}^{-1}}$ or 14.7 M.
Normality of 25% (w/v) $N{{H}_{3}}$ solution is 14.7eq ${{L}^{-1}}$ or 14.7 N.

Note:
Note that normality of a solution is never less than molarity of the solution. It can only be equal to or greater than the molarity of the solution. Normality is equal to the molarity of the solution when molar mass is equal to the equivalent mass of the solute.