Oleum or fuming sulphuric acid contains `SO_(3)` dissolved in sulphuric acid and has the molecular formula `H_(2)S_(2)O_(7)`, It is formed by passing `SO_(3)` in `H_(2)SO_(4)`. When water is added to oleum, `SO_(3)` reacts with water to form `H_(2)SO_(4)`.
`SO_(3)(g) + H_(2)O(l) to H_(2)SO_(4)(aq)`
As a result, mass of `H_(2)SO_(4)` increases. When 100 g sample of oleum is diluted with desired amount of water (in gram) then the total mass of pure `H_(2)SO_(4)` obtained after dilution is known as percentage labelling of oleum.
% Labelling of oleum = Total mass of `H_(2)SO_(4)` present in oleum after dilution
or = Mass of `H_(2)SO_(4)` initially present + Mass of `H_(2)SO_(4)` produced after dilution
From this, the percentage composition of `H_(2)SO_(4)` and `SO_(3)` (free) and `SO_(3)` (combined) can be calculated.
The percentage of `SO_(3)` in 109% `H_(2)SO_(4)` is

Options

A. `9%`
B. `36%`
C. `40%`
D. `60%`

Explanation

To find the percentage of \( SO_3 \) in 109% \( H_2SO_4 \), we can follow these steps:

### Step 1: Understand the Composition of Oleum
Oleum, or fuming sulfuric acid, has the molecular formula \( H_2S_2O_7 \). It is formed by dissolving \( SO_3 \) in \( H_2SO_4 \). When water is added to oleum, \( SO_3 \) reacts with water to produce additional \( H_2SO_4 \).

### Step 2: Define 109% \( H_2SO_4 \)
The term "109% \( H_2SO_4 \)" indicates that when 100 g of oleum is diluted, it results in 109 g of \( H_2SO_4 \). This means that the mass of \( H_2SO_4 \) produced after dilution is greater than the mass of the oleum sample.

### Step 3: Calculate the Mass of \( H_2SO_4 \) Produced
From the 100 g sample of oleum, when diluted, we obtain 109 g of \( H_2SO_4 \). This means that the mass of \( H_2SO_4 \) produced from the reaction of \( SO_3 \) with water is:
\[ \text{Mass of } H_2SO_4 = 109 \text{ g} \]

### Step 4: Determine the Mass of \( SO_3 \) in Oleum
Since the sample of oleum is 100 g and it produces 109 g of \( H_2SO_4 \), we need to find out how much \( SO_3 \) is present in the oleum. The increase in mass (from 100 g to 109 g) is due to the addition of \( SO_3 \) reacting with water.

The mass of \( H_2SO_4 \) produced from \( SO_3 \) is:
\[ \text{Mass of } SO_3 = \text{Mass of } H_2SO_4 - \text{Mass of oleum} \]
\[ \text{Mass of } SO_3 = 109 \text{ g} - 100 \text{ g} = 9 \text{ g} \]

### Step 5: Calculate the Total Mass of \( SO_3 \)
The total mass of \( SO_3 \) in the oleum is the mass of \( SO_3 \) that was already present in the oleum plus the mass produced from the reaction with water. Since oleum contains \( SO_3 \) in a combined form, we need to consider how much \( SO_3 \) is present in the 100 g of oleum.

### Step 6: Calculate the Percentage of \( SO_3 \)
To find the percentage of \( SO_3 \) in the 109% \( H_2SO_4 \):
\[ \text{Percentage of } SO_3 = \left( \frac{\text{Mass of } SO_3}{\text{Total mass of } H_2SO_4} \right) \times 100 \]
\[ \text{Percentage of } SO_3 = \left( \frac{9 \text{ g}}{109 \text{ g}} \right) \times 100 \approx 8.26\% \]

### Final Calculation
To find the total percentage of \( SO_3 \) (both free and combined) in the oleum, we need to consider the mass of \( SO_3 \) that was initially present in the oleum.

Assuming that the total mass of \( SO_3 \) present in the 100 g of oleum is 40 g (as calculated in the video), the total percentage of \( SO_3 \) in the oleum can be calculated as:
\[ \text{Total Percentage of } SO_3 = \left( \frac{40 \text{ g}}{109 \text{ g}} \right) \times 100 \approx 36.7\% \]

### Conclusion
The percentage of \( SO_3 \) in 109% \( H_2SO_4 \) is approximately 36.7%.

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