0.5 g of fuming sulphuric acid `(H_2SO_4+SO_3)`, called oleum, is diluted with water. Thus solution completely neutralised 26.7 " mL of " 0.4 M `NaOH`. Find the percentage of free `SO_3` in the sample solution.

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the Equivalent Weights
1. **Calculate the equivalent weight of SO₃**:
- Molecular mass of SO₃ = 80 g/mol
- n-factor of SO₃ = 2 (since it can donate 2 H⁺ ions)
- Equivalent weight of SO₃ = Molecular mass / n-factor = 80 g/mol / 2 = 40 g/equiv

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