Question

Given,0.5 M ${{H}_{2}}S{{O}_{4}}$ is diluted from 1 liter to 10 liters, normality of the resulting solution is:
A. 1 N
B. 0.1 N
C. 10 N
D. 11 N

Answer 1

Hint: There is a relation between molarity and normality of the solution and it is as follows.
N = (n) (M)
Here, N = Normality of the solution
M = Molarity of the solution
n = Basicity of the acids.

Complete answer:
- In the question it is given that 0.5 M sulphuric acid is diluted from 1 lit to 10 lit and we have to find the normality of the resulting solution.
- Before we are going to calculate the normality of the resulting mixture we should calculate the normality of the 0.5 M sulphuric acid and it is as follows.
N = (n) (M)
Here, N = Normality of the solution
M = Molarity of the solution = 0.5 M
n = Basicity of sulphuric acid = 2
N = (2) (0.5)
N = 1.0 N
- Now we have to calculate the normality of the diluted solution and it is as follows.
\[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\]
Here ${{N}_{1}}$ = Normality of the sulphuric acid = 1N
${{V}_{1}}$ = Volume of the sulphuric acid = 1 L
${{N}_{2}}$ = Normality of the diluted sulphuric acid
${{V}_{2}}$ = Volume of the diluted sulphuric acid = 10 L
- Substitute all the known values in the above formula to get the normality of the diluted sulphuric acid.
\[\begin{align}
  & {{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}} \\
 & {{N}_{2}}=\frac{{{N}_{1}}{{V}_{1}}}{{{V}_{2}}} \\
 & =\frac{1.0\times 1}{10} \\
 & =0.1N \\
\end{align}\]
- Therefore the normality of the resulting solution (diluted sulphuric acid) is 0.1 N.

So, the correct option is B.

Note:
If we are going to add a large amount of water to the known concentrated solution of the acid, the normality of the solution is going to decrease because concentration is inversely proportional to the volume of the solution.