Question

Percentage purity of commercial sulphuric acid is $ 98\% ({\text{w}}/{\text{w}}) $ and its specific gravity is $ 1.8 $ . Its normality is:
A) $ 98 $
B) $ 79.4 $ $ $
C) $ 36.8 $
D) $ 18 $

Answer 1

Hint: One of the expressions used to measure the concentration of a solution is normality in chemistry. It is abbreviated as 'N' and is sometimes known as a solution's equivalent concentration. It is mainly used as a measure of reactive species in a solution and in situations involving acid-base chemistry during titration reactions especially. We shall calculate the moles of sulphuric acid present in 100g and then the volume using specific gravity. Then, we shall calculate the molarity and thus the normality.

Formula Used
Molarity $ {\text{ = }}\dfrac{{{\text{moles of solute}}}}{{{\text{volume of solution in litres}}}} $
 $ N = M \times basicity $
Where
 $ N $ is the normality of the solution
 $ M $ is the molarity of the solution.

Complete Step-by-Step Solution
According to the question, the following information is provided to us:
In a $ 100g $ solution, the weight of sulphuric acid is $ 98g $
Molecular weight of $ {H_2}S{O_4} = 98g $
Now, we will find out the number of moles of sulphuric acid in the above sample
Number of moles of $ {H_2}S{O_4} = \dfrac{{{\text{weight}}}}{{{\text{molecular weight}}}} = \dfrac{{98}}{{98}} $
So, the number of moles of $ {H_2}S{O_4} $ is $ 1 $ mole of $ {H_2}S{O_4} $
Now, we have to find the volume of $ {H_2}S{O_4} $
This can be calculated by $ \dfrac{{{\text{mass}}}}{{{\text{density}}}} $
The mass of the sample is $ 100g $
And the specific density is known to be $ 1.8 $
Now, we will substitute these values in the above equation to get the volume if the sulphuric acid
 $ \dfrac{{100}}{{1.84}} = 54.34{\text{ml}} $
Which can be rewritten as $ 5.43 \times {10^{ - 2}}{\text{L}} $
Now, we will calculate the molarity of the solution
The molarity of sulphuric acid $ {\text{ = }}\dfrac{{{\text{moles of solute}}}}{{{\text{volume of solution in litres}}}} $
 $ = \dfrac{1}{{5.43 \times {{10}^{ - 2}}}} = 18.4{\text{M}} $
So, the molarity of the sulphuric acid solution is $ 18.4{\text{M}} $
Now, we know that the basicity of $ {H_2}S{O_4} $ is $ 2 $
Normality is given by $ N = M \times {\text{basicity}} $
So, the normality of $ {H_2}S{O_4} $ $ = 18.4 \times 2 = 36.8{\text{N}} $
Hence, the correct option is (C.)

Note
There are many other ways to approach this question. Every student thinks differently and there can be multiple thinking styles. Just remember the formula correctly and that will solve half your question. Next step is to substitute and find the answer