A mixutre is prepared by mixing 10 gm `H_(2)SO_(4) and 40 gm SO_(3)` calculate,
(a) mole fraction of `H_(2)SO_(4)`
(b) % labelling of oleum

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the number of moles of H₂SO₄
The formula to calculate the number of moles is:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
For H₂SO₄:
- Mass = 10 g
- Molar mass of H₂SO₄ = 98 g/mol

Calculating the moles:
\[ \text{Moles of H₂SO₄} = \frac{10 \, \text{g}}{98 \, \text{g/mol}} \approx 0.102 \, \text{mol} \]

### Step 2: Calculate the number of moles of SO₃
For SO₃:
- Mass = 40 g
- Molar mass of SO₃ = 80 g/mol

Calculating the moles:
\[ \text{Moles of SO₃} = \frac{40 \, \text{g}}{80 \, \text{g/mol}} = 0.5 \, \text{mol} \]

### Step 3: Calculate the total number of moles in the mixture
\[ \text{Total moles} = \text{Moles of H₂SO₄} + \text{Moles of SO₃} = 0.102 \, \text{mol} + 0.5 \, \text{mol} = 0.602 \, \text{mol} \]

### Step 4: Calculate the mole fraction of H₂SO₄
The mole fraction (X) of a component is given by:
\[ X_{H₂SO₄} = \frac{\text{Moles of H₂SO₄}}{\text{Total moles}} \]
Calculating the mole fraction:
\[ X_{H₂SO₄} = \frac{0.102 \, \text{mol}}{0.602 \, \text{mol}} \approx 0.169 \]

### Step 5: Calculate the % labeling of oleum
The formula for % labeling of oleum is:
\[ \text{% labeling} = 100 + X \]
Where X is the mass of water that can be formed from the reaction of SO₃ with water.

From the reaction:
\[ \text{SO₃} + \text{H₂O} \rightarrow \text{H₂SO₄} \]
1 mole of SO₃ reacts with 1 mole of H₂O to produce 1 mole of H₂SO₄.

Given that we have 40 g of SO₃:
- Moles of SO₃ = 0.5 mol
- Therefore, the mass of water (H₂O) required = 0.5 mol × 18 g/mol = 9 g

Calculating the % labeling:
\[ \text{% labeling} = 100 + 9 = 109\% \]

### Final Answers:
(a) Mole fraction of H₂SO₄ = 0.169
(b) % labeling of oleum = 109%

---