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.
What is the amount of free `SO_(3)` in an oleum sample labelled as '118%'.

Options

A. 0.4
B. 0.5
C. 0.7
D. 0.8

Explanation

To find the amount of free \( SO_3 \) in an oleum sample labeled as '118%', we can follow these steps:

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

### Step 2: Determine the Amount of \( H_2SO_4 \) and Water
The percentage labeling of '118%' means that for every 100 grams of \( H_2SO_4 \), there are 18 grams of free \( SO_3 \) (since 118% = 100% + 18%). Therefore, in 100 grams of oleum:
- Mass of \( H_2SO_4 \) = 100 g
- Mass of \( SO_3 \) = 18 g

### Step 3: Calculate the Number of Moles of Water
To convert \( SO_3 \) to \( H_2SO_4 \), we need to add water. The reaction is:
\[ H_2O + SO_3 \rightarrow H_2SO_4 \]
From the labeling, we know that the amount of water required to convert the free \( SO_3 \) to \( H_2SO_4 \) is equal to the additional percentage over 100%. Thus, we have:
- Mass of water = 18 g

The number of moles of water can be calculated using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
The molar mass of water (\( H_2O \)) is 18 g/mol. Therefore:
\[ \text{Number of moles of water} = \frac{18 \, \text{g}}{18 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 4: Determine the Number of Moles of \( SO_3 \)
From the reaction, we see that 1 mole of \( H_2O \) reacts with 1 mole of \( SO_3 \). Therefore, the number of moles of \( SO_3 \) present will also be:
\[ \text{Number of moles of } SO_3 = 1 \, \text{mol} \]

### Step 5: Calculate the Mass of \( SO_3 \)
The molar mass of \( SO_3 \) is 80 g/mol. Therefore, the mass of \( SO_3 \) can be calculated as:
\[ \text{Mass of } SO_3 = \text{Number of moles} \times \text{Molar mass} = 1 \, \text{mol} \times 80 \, \text{g/mol} = 80 \, \text{g} \]

### Conclusion
The amount of free \( SO_3 \) in the oleum sample labeled as '118%' is **80 grams**.

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