An oleum sample contains `10 g SO_(3)` and `15 g H_(2) SO_(4)`
Answer the following questions on the basis of above information :
`%` labelling of oleum sample is .

Options

A. `27.25%`
B. `106%`
C. `109%`
D. `118%`

Explanation

To find the percentage labeling of the oleum sample containing 10 g of SO₃ and 15 g of H₂SO₄, we can follow these steps:

### Step 1: Calculate the molecular weight of SO₃ and H₂SO₄
- **Molecular weight of SO₃**:
- Sulfur (S) = 32 g/mol
- Oxygen (O) = 16 g/mol
- SO₃ = 32 + (3 × 16) = 32 + 48 = **80 g/mol**

- **Molecular weight of H₂SO₄**:
- Hydrogen (H) = 1 g/mol
- Sulfur (S) = 32 g/mol
- Oxygen (O) = 16 g/mol
- H₂SO₄ = (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = **98 g/mol**

### Step 2: Calculate the total mass of the oleum sample
- Total mass = Mass of SO₃ + Mass of H₂SO₄
- Total mass = 10 g + 15 g = **25 g**

### Step 3: Calculate the percentage of SO₃ in the oleum sample
- Percentage of SO₃ = (Mass of SO₃ / Total mass) × 100
- Percentage of SO₃ = (10 g / 25 g) × 100 = **40%**

### Step 4: Calculate the percentage labeling of the oleum sample
- The labeling of oleum is based on the total mass of SO₃ present in the sample.
- Since we have 10 g of SO₃ in the sample, we can express this in terms of 100 g of the sample.
- If 25 g of the sample contains 10 g of SO₃, then 100 g of the sample would contain:
- (10 g / 25 g) × 100 g = **40 g of SO₃**

### Step 5: Calculate the total mass of the oleum sample for labeling
- The total mass of the oleum sample for labeling = Mass of H₂SO₄ + Mass of SO₃
- Total mass for labeling = 15 g + 40 g = **55 g**

### Step 6: Calculate the percentage labeling of the oleum sample
- Percentage labeling = (Total mass of SO₃ / Total mass for labeling) × 100
- Percentage labeling = (40 g / 55 g) × 100 = **72.73%**

### Final Answer
The percentage labeling of the oleum sample is approximately **72.73%**.

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