Question

Strength of oleum is expressed in percentage. What is the percentage of free $SO_3$ in a sample of oleum labelled as 101.125%?

Answer 1

5%

Answer 2

For oleum, the percentage strength is defined as the weight of $H_2SO_4$ produced per 100 g of oleum after dilution with water. In oleum, the composition is taken as $H_2SO_4 \cdot SO_3$ so that 100 g oleum yields 100 g $H_2SO_4$ before any free $SO_3$ reacts with water. When water is added, the free $SO_3$ reacts:

$SO_3 + H_2O \rightarrow H_2SO_4$

A sample labelled as 101.125% means 100 g oleum produces 101.125 g $H_2SO_4$ upon dilution, indicating an extra 1.125 g of $H_2O$ has combined with the free $SO_3$.

Using the stoichiometry:

• 18 g of $H_2O$ reacts with 80 g of $SO_3$ to form $H_2SO_4$.

Thus, the mass of free $SO_3$ in 100 g oleum is:

$\text{Free } SO_3 = \left(\dfrac{80}{18}\right) \times 1.125 = \dfrac{80 \times 1.125}{18} = \dfrac{90}{18} = 5 \text{ g}$

So, the percentage of free $SO_3$ is:

$5\%$

Core Explanation:
Subtract the inherent acid (100 g) from the total acid after dilution (101.125 g) to get 1.125 g water. Then, using the ratio from the reaction $(SO_3 + H_2O \rightarrow H_2SO_4)$ (80 g $SO_3$ per 18 g $H_2O$), compute free $SO_3$ as $(80/18) \times 1.125 = 5$ g, i.e. 5%.