An oleum sample labelled as `104.5%` in 10 g of this sample 90 mg water is added then which is/are correct for resulting solution.

Options

A. Solution contain 10.09 g `H_2SO_4`
B. Solution contain 15.86% free `SO_3`
C. Solution contain 8.49 g `H_2SO_4`
D. Solution contain 20% free `SO_3`

Explanation

To solve the problem step by step, we need to analyze the oleum sample and the addition of water to it. Here’s how we can break it down:

### Step 1: Understand the oleum sample
The oleum sample is labeled as `104.5%`. This means that in 100 grams of oleum, there are 104.5 grams of H₂SO₄ (sulfuric acid) produced when water is added.

### Step 2: Determine the amount of H₂SO₄ in the sample
Given that we have a 10-gram sample of oleum, we can calculate the amount of H₂SO₄ it can produce.

Using the ratio from the label:
- For 100 g of oleum, H₂SO₄ produced = 104.5 g
- For 10 g of oleum, H₂SO₄ produced = (104.5 g / 100 g) × 10 g = 10.45 g

### Step 3: Calculate the amount of water required
To produce the 104.5 g of H₂SO₄, the oleum requires 4.5 g of water. Therefore, for our 10 g sample:
- Water required = (4.5 g / 100 g) × 10 g = 0.45 g

### Step 4: Add the additional water
We are adding an additional 90 mg of water to the sample. Convert this to grams:
- 90 mg = 0.09 g

### Step 5: Calculate the total water added
Total water = water required + additional water = 0.45 g + 0.09 g = 0.54 g

### Step 6: Calculate the amount of SO₃ in the sample
From the oleum, we can calculate the amount of SO₃ present. The reaction of SO₃ with water to form H₂SO₄ can be represented as:
\[ \text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 \]

From the 10 g sample, we can find the amount of SO₃:
- Molar mass of SO₃ = 80 g/mol
- Molar mass of H₂O = 18 g/mol

Using the amount of water required:
- Amount of SO₃ available = (80 g / 18 g) × 0.45 g = 2 g

### Step 7: Calculate the amount of SO₃ that reacts with the added water
Now, we need to calculate how much SO₃ will react with the additional 0.09 g of water:
- Amount of SO₃ that reacts = (80 g / 18 g) × 0.09 g = 0.4 g

### Step 8: Calculate the remaining SO₃
Remaining SO₃ after the reaction:
- Remaining SO₃ = Initial SO₃ - SO₃ that reacted = 2 g - 0.4 g = 1.6 g

### Step 9: Calculate the percentage of free SO₃ in the solution
To find the percentage of free SO₃ in the resulting solution:
- Total mass of the solution = mass of oleum + total water added = 10 g + 0.54 g = 10.54 g
- Percentage of free SO₃ = (Remaining SO₃ / Total mass of solution) × 100 = (1.6 g / 10.54 g) × 100 ≈ 15.18%

### Step 10: Calculate the amount of H₂SO₄ in the solution
The amount of H₂SO₄ present in the solution:
- H₂SO₄ = Total mass of oleum + Total water - Remaining SO₃ = 10 g + 0.54 g - 1.6 g = 8.94 g

### Final Results
- Free SO₃ = 15.18%
- Amount of H₂SO₄ = 8.94 g