Question

5 ml of 1N HCl, 20 ml of $\dfrac{N}{2}$ ${{H}_{2}}S{{O}_{4}}$ and 30 ml of $\dfrac{N}{9}$ $HN{{O}_{3}}$ are mixed together and the volume is made 1 L. The normality of the resulting solution is:
(A)- $\dfrac{N}{5}$
(B)- $\dfrac{N}{10}$
(C)- $\dfrac{N}{20}$
(D)- $\dfrac{N}{40}$

Answer 1

Hint: When three different solutions having different solute of normality are mixed together to form a solution in volume V, then the normality of the resulting solution is given as:
     \[NV={{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}+{{N}_{3}}{{V}_{3}}\]


Complete answer:
Given normality of HCl solution, ${{N}_{1}}=1N$
Volume of HCl taken, ${{V}_{1}}=5\,ml$
Normality of ${{H}_{2}}S{{O}_{4}}$ solution, ${{N}_{2}}=\dfrac{N}{2}$
Volume of ${{H}_{2}}S{{O}_{4}}$ taken, ${{V}_{2}}=20\,ml$
Normality of $HN{{O}_{3}}$ solution, ${{N}_{3}}=\dfrac{N}{3}$
Volume of $HN{{O}_{3}}$ taken, ${{V}_{3}}=30\,ml$
All the three solutions of given normality are mixed together in respective volumes to form the solution in one litre.
Volume of the resulting solution, $V=1\,L=1000\,ml$
Let the normality of the solution be formed to be ‘${{N}_{v}}$’. Therefore, normality of the resulting solution can be calculated as
\[\begin{align}
  & {{N}_{v}}V={{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}+{{N}_{3}}{{V}_{3}} \\
 & {{N}_{v}}\times 1000\,ml=1N\times 5\,ml+\dfrac{N}{2}\times 20\,ml+\dfrac{N}{3}\times 30\,ml \\
 & {{N}_{v}}\times 1000\,ml=1N\times 5\,ml+N\times 10\,ml+N\times 10\,ml \\
 & {{N}_{v}}\times 1000\,ml=(1N\times 5+N\times 10+N\times 10)ml \\
\end{align}\]

Cancelling ml from both the sides, we get
\[\begin{align}
  & {{N}_{v}}\times 1000=5N+10N+10N \\
 & {{N}_{v}}\times 1000=25N \\
 & {{N}_{v}}=\dfrac{25}{1000}N=\dfrac{N}{40} \\
\end{align}\]
Thus, the normality of the solution formed in one litre by mixing ml of 1N HCl, 20 ml of $\dfrac{N}{2}$ ${{H}_{2}}S{{O}_{4}}$ and 30 ml of $\dfrac{N}{9}$ $HN{{O}_{3}}$ is $\dfrac{N}{40}$.
So, the correct answer is “Option D”.

Additional Information:
Normality of a solution is equal to the number of gram equivalents of the solute dissolved per litre of the solution. It is usually denoted as N and its unit is $eq\,{{L}^{-1}}$.
\[\begin{align}
& Normality \text{ }=\text{ }\dfrac{number\,of\,gram\,equivalents\,of\,solute}{volume\,of\,solution\,(in\,litre)} \\
& or\,N\text{ }=\text{ }\dfrac{mass\,of\,solute\,dissolved\,(in\,gram)\text{ }\times \text{ }1000}{Equivalent\,weight\,of\,solute\text{ }\times \text{ }volume\,of\,solution\,(in\,ml)} \\
\end{align}\]

Note: Note that the volume of solutions mixed is taken in millilitre and volume of the resulting solution is given in litre. Convert all the units for volume in either litre or millilitre and use the same unit throughout the calculation to avoid errors.
If concentration of the solutions mixed in different volumes was expressed in terms of molarity, then the expression of for the normality of the resulting solution would be \[nMV={{n}_{1}}{{M}_{1}}{{V}_{1}}+{{n}_{2}}{{M}_{2}}{{V}_{2}}+{{n}_{3}}{{M}_{3}}{{V}_{3}}\].