The percentage labelling (mixture of `H_(2)SO_(4)` and `SO_(3))` refers to the total mass of pure `H_(2)SO_(4)`.The total amount of `H_(2)SO_(4)` found after adding calculated amount of water to `100 g` oleum is the percentage labelling of oleum. Higher the percentage lebeling of oleum higher is the amount of free `SO_(3)` in the oleum sample.
The percent free `SO_(3)` is an oleum is 20%. Label the sample of oleum in terms of percent `H_(2) SO_(4)`.

Options

A. 1.135
B. 1.045
C. 1.0675
D. 1.2

Explanation

To solve the problem of labeling the oleum in terms of percent \( H_2SO_4 \), we will follow these steps:

### Step 1: Understand the Composition of Oleum
Oleum is a mixture of sulfuric acid (\( H_2SO_4 \)) and sulfur trioxide (\( SO_3 \)). The percentage labeling refers to the total mass of pure \( H_2SO_4 \) present in the oleum.

### Step 2: Determine the Mass of Free \( SO_3 \)
We are given that the percent free \( SO_3 \) in the oleum is 20%. This means that in 100 g of oleum, there are 20 g of \( SO_3 \).

### Step 3: Calculate the Moles of \( SO_3 \)
To find the number of moles of \( SO_3 \), we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
The molar mass of \( SO_3 \) is 80 g/mol. Thus, the number of moles of \( SO_3 \) is:
\[ \text{Moles of } SO_3 = \frac{20 \, \text{g}}{80 \, \text{g/mol}} = 0.25 \, \text{moles} \]

### Step 4: Calculate the Amount of Water Needed
From the reaction \( H_2O + SO_3 \rightarrow H_2SO_4 \), we see that 1 mole of \( SO_3 \) requires 1 mole of \( H_2O \). Therefore, for 0.25 moles of \( SO_3 \), we need 0.25 moles of \( H_2O \).

Now, calculating the mass of water needed:
\[ \text{Mass of } H_2O = \text{Number of moles} \times \text{Molar mass of } H_2O \]
The molar mass of \( H_2O \) is 18 g/mol:
\[ \text{Mass of } H_2O = 0.25 \, \text{moles} \times 18 \, \text{g/mol} = 4.5 \, \text{g} \]

### Step 5: Calculate the Total Mass of \( H_2SO_4 \)
The total mass of \( H_2SO_4 \) in the oleum after adding water is the mass of \( H_2SO_4 \) originally present plus the mass of \( H_2SO_4 \) formed from \( SO_3 \):
- The mass of \( H_2SO_4 \) formed from 20 g of \( SO_3 \) is also 20 g, because each mole of \( SO_3 \) produces one mole of \( H_2SO_4 \).
- Therefore, the total mass of \( H_2SO_4 \) after adding water is:
\[ \text{Total } H_2SO_4 = 100 \, \text{g (original oleum)} + 20 \, \text{g (from } SO_3\text{)} = 120 \, \text{g} \]

### Step 6: Calculate the Percentage of \( H_2SO_4 \)
The percentage labeling of oleum in terms of \( H_2SO_4 \) is given by:
\[ \text{Percentage of } H_2SO_4 = \frac{\text{mass of } H_2SO_4}{\text{total mass}} \times 100 \]
The total mass after adding water is \( 100 \, \text{g (oleum)} + 4.5 \, \text{g (water)} = 104.5 \, \text{g} \):
\[ \text{Percentage of } H_2SO_4 = \frac{120 \, \text{g}}{104.5 \, \text{g}} \times 100 \approx 114.8\% \]

### Conclusion
The percentage labeling of the oleum in terms of \( H_2SO_4 \) is approximately **114.8%**.