A mmonia is highly soluble gas in water and gives a alkaline solution of `NH_(4) OH`. What volume of `NH_(3)` gas at `STP` will be required to the passed in `100 mL` of `0.5 M H_(2) SO_(4)` to bring down its strength to `0.25 M`?
(For titrations with aqueous `NH_(3)`, it is assumed that `NH_(4) OH` dissociates to 100% extent)
a. 2.24 L b. 1.68 L c. 1.12 L d. 0.56 L`

Explanation

To solve the problem, we need to determine the volume of ammonia gas (NH₃) required to neutralize a given amount of sulfuric acid (H₂SO₄) and reduce its concentration from 0.5 M to 0.25 M.

### Step-by-Step Solution:

1. **Calculate the initial moles of H₂SO₄**:
- We know the concentration (Molarity) and volume of the H₂SO₄ solution.
- Moles of H₂SO₄ = Molarity × Volume (in liters)
- Given: Molarity = 0.5 M, Volume = 100 mL = 0.1 L
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