A sample of oleum is labelled 109% . The % of free `SO_(3)` in the sample is

Options

A. `40%`
B. `80%`
C. `60%`
D. `9%`

Answer

To find the percentage of free SO₃ in a sample of oleum labeled as 109%, we can follow these steps:

### Step-by-Step Solution

1. **Understanding Oleum**:
Oleum is a solution of sulfur trioxide (SO₃) in sulfuric acid (H₂SO₄). The label "109%" indicates that if water were added to the oleum, the resulting solution would contain 109 grams of H₂SO₄ for every 100 grams of the solution.

2. **Assuming Mass of Oleum**:
Let's assume we have 100 grams of oleum. According to the label, this oleum would yield 109 grams of H₂SO₄ upon the addition of water.

3. **Setting Up the Equation**:
Let \( x \) be the mass of SO₃ present in the oleum. The remaining mass will be the mass of H₂SO₄ already present. Thus, the mass of H₂SO₄ before adding water can be expressed as:
\[ \text{Mass of H₂SO₄} = 100 - x \]

4. **Calculating the Mass of H₂SO₄ Formed**:
When SO₃ reacts with water, it forms H₂SO₄. The reaction is as follows:
\[ SO₃ + H₂O \rightarrow H₂SO₄ \]
The molar mass of SO₃ is 80 g/mol and the molar mass of H₂SO₄ is 98 g/mol. The moles of SO₃ can be calculated as:
\[ \text{Moles of SO₃} = \frac{x}{80} \]
Therefore, the mass of H₂SO₄ formed from \( x \) grams of SO₃ is:
\[ \text{Mass of H₂SO₄ formed} = \text{Moles of SO₃} \times \text{Molar mass of H₂SO₄} = \frac{x}{80} \times 98 = \frac{98x}{80} \]

5. **Total Mass of H₂SO₄ After Adding Water**:
The total mass of H₂SO₄ after adding water will be the sum of the H₂SO₄ already present and the H₂SO₄ formed from SO₃:
\[ \text{Total H₂SO₄} = (100 - x) + \frac{98x}{80} \]

6. **Setting Up the Equation for Total H₂SO₄**:
According to the label, this total must equal 109 grams:
\[ (100 - x) + \frac{98x}{80} = 109 \]

7. **Solving the Equation**:
Rearranging the equation gives:
\[ 100 - x + \frac{98x}{80} = 109 \]
Multiplying through by 80 to eliminate the fraction:
\[ 80(100 - x) + 98x = 8720 \]
Simplifying:
\[ 8000 - 80x + 98x = 8720 \]
\[ 18x = 8720 - 8000 \]
\[ 18x = 720 \]
\[ x = 40 \]

8. **Calculating the Percentage of Free SO₃**:
Now that we have \( x = 40 \) grams of SO₃, we can find the percentage of free SO₃ in the oleum:
\[ \text{Percentage of SO₃} = \left( \frac{x}{\text{Total mass of oleum}} \right) \times 100 = \left( \frac{40}{100} \right) \times 100 = 40\% \]

### Final Answer:
The percentage of free SO₃ in the sample of oleum is **40%**.