6.0 g of a steel containing sulphur as an impurity was burnt in excess of oxygen, where sulphur is oxidised to `SO_2`. The `SO_2` evolved was oxidised to `SO_4^(2-)` ions by the action of `H_2O_2` solution in the presence of 30 mL solution of 0.04 M NaOH. 22.48 " mL of " 0.024 M HCl was required to neutralise the excess of NaOH after the above oxidation. Calculate the percentage of sulphur in the given sample of steel (Atomic mass of S is 32).

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the milliequivalents of NaOH added
We know the volume and molarity of the NaOH solution used. The formula for milliequivalents (mEq) is:
\[ \text{mEq} = \text{Volume (mL)} \times \text{Molarity (M)} \]
Given:
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