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 free `SO_(3)` and `H_(2)SO_(4)` in 112% `H_(2)SO_(4)` is

Options

A. 53.6, 46.4
B. 12.0, 88.0
C. 88.0, 12.0
D. 26.8, 73.2

Explanation

To solve the problem regarding the percentage of free SO₃ and H₂SO₄ in 112% H₂SO₄, we will follow these steps:

### Step 1: Understand the Composition of Oleum
Oleum, or fuming sulfuric acid, has the molecular formula H₂S₂O₇. When water is added to oleum, the free SO₃ reacts with water to produce more H₂SO₄. The reaction is as follows:
\[ \text{SO}_3(g) + \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{SO}_4(aq) \]

### Step 2: Determine the Total Mass of H₂SO₄
Given that we have a 100 g sample of oleum, and after dilution, we have a total of 112 g of H₂SO₄, we can calculate the mass of H₂SO₄ produced from the free SO₃ in the oleum.

### Step 3: Calculate the Mass of Water Added
Since the total mass of H₂SO₄ after dilution is 112 g, and we started with 100 g of oleum, we can find the mass of water added:
\[ \text{Mass of water added} = \text{Total mass of H}_2\text{SO}_4 - \text{Mass of oleum} \]
\[ \text{Mass of water added} = 112 \text{ g} - 100 \text{ g} = 12 \text{ g} \]

### Step 4: Calculate Moles of Water Added
To find the moles of water added, we use the formula:
\[ \text{Moles of water} = \frac{\text{Mass of water}}{\text{Molar mass of water}} \]
The molar mass of water (H₂O) is approximately 18 g/mol.
\[ \text{Moles of water} = \frac{12 \text{ g}}{18 \text{ g/mol}} = 0.67 \text{ moles} \]

### Step 5: Relate Moles of Water to Moles of Free SO₃
From the reaction, we see that 1 mole of SO₃ reacts with 1 mole of water. Therefore, the moles of free SO₃ will also be 0.67 moles.

### Step 6: Calculate the Mass of Free SO₃
To find the mass of free SO₃, we use its molar mass. The molar mass of SO₃ is:
\[ \text{Molar mass of SO}_3 = 32 \text{ g/mol (S)} + 3 \times 16 \text{ g/mol (O)} = 80 \text{ g/mol} \]
Now, we can calculate the mass of free SO₃:
\[ \text{Mass of free SO}_3 = \text{Moles of SO}_3 \times \text{Molar mass of SO}_3 \]
\[ \text{Mass of free SO}_3 = 0.67 \text{ moles} \times 80 \text{ g/mol} = 53.6 \text{ g} \]

### Step 7: Calculate the Mass of H₂SO₄
Now, we can find the mass of H₂SO₄ in the oleum:
\[ \text{Mass of H}_2\text{SO}_4 = \text{Total mass of H}_2\text{SO}_4 - \text{Mass of free SO}_3 \]
\[ \text{Mass of H}_2\text{SO}_4 = 112 \text{ g} - 53.6 \text{ g} = 58.4 \text{ g} \]

### Step 8: Calculate the Percentage Composition
Now we can calculate the percentage of free SO₃ and H₂SO₄ in the 112% H₂SO₄:
1. **Percentage of free SO₃**:
\[ \text{Percentage of free SO}_3 = \left( \frac{\text{Mass of free SO}_3}{\text{Total mass of H}_2\text{SO}_4} \right) \times 100 \]
\[ \text{Percentage of free SO}_3 = \left( \frac{53.6 \text{ g}}{112 \text{ g}} \right) \times 100 \approx 47.86\% \]

2. **Percentage of H₂SO₄**:
\[ \text{Percentage of H}_2\text{SO}_4 = \left( \frac{\text{Mass of H}_2\text{SO}_4}{\text{Total mass of H}_2\text{SO}_4} \right) \times 100 \]
\[ \text{Percentage of H}_2\text{SO}_4 = \left( \frac{58.4 \text{ g}}{112 \text{ g}} \right) \times 100 \approx 52.14\% \]

### Final Result
- Percentage of free SO₃: **47.86%**
- Percentage of H₂SO₄: **52.14%**