0.5 gm of fuming `H_(2)SO_(4)` (Oleum) is diluted with water. This solution is completely neutralised by 26.7 ml of 0.4 M `NaOH` solution. Calculate the percentage of free `SO_(3)` in the given sample. Give your answer excluding the decimal places.

Options

A. Mass of `SO_(3)` is 0.104 g
B. % of free `SO_(3) = 20 : 7`
C. Normality of `H_(2) SO_(4)` for neutralization is 0.2 N
D. Weight of `H_(2) SO_(4)` is 0.104 g

Answer

To solve the problem, we need to calculate the percentage of free SO₃ in the given sample of fuming H₂SO₄ (oleum). Here’s a step-by-step solution:

### Step 1: Define Variables
Let:
- \( x \) = mass of SO₃ in the oleum
- Mass of H₂SO₄ = \( 0.5 - x \) grams

### Step 2: Calculate Equivalent Weights
- **For H₂SO₄**:
- Molar mass of H₂SO₄ = 98 g/mol
- n-factor of H₂SO₄ = 2 (since it can donate 2 protons)
- Equivalent weight of H₂SO₄ = \( \frac{98}{2} = 49 \) g/equiv

- **For SO₃**:
- Molar mass of SO₃ = 80 g/mol
- n-factor of SO₃ = 2 (since it can react with 2 moles of NaOH)
- Equivalent weight of SO₃ = \( \frac{80}{2} = 40 \) g/equiv

### Step 3: Calculate Gram Equivalents
The total gram equivalents of H₂SO₄ and SO₃ must equal the gram equivalents of NaOH used in the reaction.

- **Gram equivalents of NaOH**:
- Normality of NaOH = 0.4 N
- Volume of NaOH = 26.7 mL = 0.0267 L
- Gram equivalents of NaOH = Normality × Volume = \( 0.4 \times 0.0267 = 0.01068 \) equivalents

### Step 4: Set Up the Equation
The equation for the sum of gram equivalents is:
\[ \frac{0.5 - x}{49} + \frac{x}{40} = 0.01068 \]

### Step 5: Solve for \( x \)
Multiply through by the least common multiple (LCM) of the denominators (49 and 40) to eliminate the fractions:
\[ 40(0.5 - x) + 49x = 0.01068 \times 1960 \]
Calculating \( 0.01068 \times 1960 \):
\[ 0.01068 \times 1960 = 20.9768 \]
Now substituting back into the equation:
\[ 20 - 40x + 49x = 20.9768 \]
Combine like terms:
\[ 20 + 9x = 20.9768 \]
Solving for \( x \):
\[ 9x = 20.9768 - 20 \]
\[ 9x = 0.9768 \]
\[ x = \frac{0.9768}{9} \approx 0.1085 \text{ grams} \]

### Step 6: Calculate Percentage of Free SO₃
Now, we can find the percentage of free SO₃ in the sample:
\[ \text{Percentage of free SO₃} = \left( \frac{x}{0.5} \right) \times 100 \]
Substituting the value of \( x \):
\[ \text{Percentage of free SO₃} = \left( \frac{0.1085}{0.5} \right) \times 100 \approx 21.7\% \]

### Final Answer
The percentage of free SO₃ in the given sample is approximately **22%** (excluding decimal places).

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