Question

0.5 g of fuming sulphuric acid (H$_2$SO$_4$+SO$_3$), called oleum, is diluted with water. This solution is completely neutralised at 26.7 ml of 0.4 M NaOH. Find the percentage of free SO$_3$ in the sample solution.
A. 20.73%
B. 43.80%
C. 79.27%
D. 60.74%

Answer 1

Hint: This question is based on the mole concept. Find out the equivalence weight of molecules included in the composition of oleum. Then you can calculate the total number of equivalents, and then percentage can be calculated.

Complete step by step answer:
First, we know that oleum consists of sulphuric acid (H$_2$SO$_4$), and free SO$_3$. The amount of fuming sulphuric acid used to perform the reaction is given i.e. 0.5 g.
So, let us consider the mass of SO$_3$ in the given sample of oleum be x g.
Then, the mass of sulphuric acid will be equal to (0.5-x) g.
Now, if we calculate the equivalent weight of SO$_3$, then the molecular weight of SO$_3$ is 80. Thus, equivalent weight = 80/2 = 40
 Similarly, for sulphuric acid, the molecular weight is 98. Then, equivalent weight = 98/2 = 49
Thus, no. of g equivalents for SO$_3$ = $\dfrac{x}{40}$;
No. of g equivalents for H$_2$SO$_4$= $\dfrac{0.5-x}{49}$
Therefore, total no. of g equivalents = $\dfrac{x}{40}$ + $\dfrac{0.5-x}{49}$
According to the question, 26.7 ml of 0.4 N NaOH will have $\dfrac{0.4}{1000}$ $\times$ 26.7 equivalents of NaOH.
Now, if we equalise the no. of g equivalents of oleum, and NaOH, at the equivalence point; then $\dfrac{0.4}{1000}$ $\times$ 26.7= $\dfrac{x}{40}$ + $\dfrac{0.5-x}{49}$.
By calculating we get x = 0.1036
Therefore, % of free SO$_3$ = $\dfrac{0.1036}{0.5}$ $\times$ 100= 20.72 %
Thus, % of free SO$_3$ in the sample solution is 20.72%. The correct option is (A).

Note: Don’t get confused while calculating the percentage. Just move step by step as shown. The equivalents for the composition of oleum will be considered different in the terms of mass, as we have to calculate the percentage of SO$_3$.