Oleum is considered as a solution of `SO_(3)` in `H_(2)SO_(4)`, which is obtained by passing `SO_(3)` in solution of `H_(2)SO_(4)` When 100 g sample of oleum is diluted with desired mass of `H_(2)O` then the total mass of `H_(2)SO_(4)` obtained after dilution is known is known as % labelling in oleum.
For example, a oleum bottle labelled as 109% `H_(2)SO_(4)` means the 109 g total mass of pure `H_(2)SO_(4)` will be formed when 100 g of oleum is diluted by 9 g of `H_(2)O` which combines with all the free `SO_(3)` present in oleum to form `H_(2)SO_(4)` as `SO_(3)+H_(2)O to H_(2)SO_(4)`
What is the % of free `SO_(3)` in an oleum that is labelled as 104.5% `H_(2)SO_(4)` ?

Options

A. 10
B. 20
C. 40
D. None of these

Answer

To find the percentage of free \( SO_3 \) in an oleum that is labelled as 104.5% \( H_2SO_4 \), we can follow these steps:

### Step 1: Understand the Reaction
The reaction between \( SO_3 \) and \( H_2O \) to form \( H_2SO_4 \) is given by:
\[ SO_3 + H_2O \rightarrow H_2SO_4 \]
From the stoichiometry of the reaction, we know that 80 g of \( SO_3 \) reacts with 18 g of \( H_2O \).

### Step 2: Determine the Amount of Water
The oleum is labelled as 104.5% \( H_2SO_4 \). This means that when 100 g of oleum is diluted, it produces 104.5 g of \( H_2SO_4 \). The excess over 100 g indicates the amount of water that reacts with the free \( SO_3 \) in the oleum:
\[ \text{Excess} = 104.5 \, \text{g} - 100 \, \text{g} = 4.5 \, \text{g of } H_2O \]

### Step 3: Calculate the Amount of \( SO_3 \)
Using the stoichiometric ratio from the reaction, we can find out how much \( SO_3 \) corresponds to 4.5 g of \( H_2O \):
\[ \text{Using the ratio: } \frac{80 \, \text{g } SO_3}{18 \, \text{g } H_2O} = \frac{x \, \text{g } SO_3}{4.5 \, \text{g } H_2O} \]
Cross-multiplying gives:
\[ x = \frac{80 \times 4.5}{18} \]
Calculating this:
\[ x = \frac{360}{18} = 20 \, \text{g of } SO_3 \]

### Step 4: Calculate the Percentage of Free \( SO_3 \)
Now that we know there are 20 g of free \( SO_3 \) in 100 g of oleum, we can calculate the percentage of free \( SO_3 \):
\[ \text{Percentage of } SO_3 = \left( \frac{20 \, \text{g}}{100 \, \text{g}} \right) \times 100 = 20\% \]

### Final Answer
The percentage of free \( SO_3 \) in an oleum that is labelled as 104.5% \( H_2SO_4 \) is **20%**.

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