Mole fraction of `K_(2)CO_(3)` in a mixture of `K_(2)CO_(3)` and `KHCO_(3)` is 0.5. What will be the volume of 0.1 N HCl required to neutralize 1.252 g of the mixture?

Explanation

To solve the problem, we need to find the volume of 0.1 N HCl required to neutralize 1.252 g of a mixture of \( K_2CO_3 \) and \( KHCO_3 \) with a mole fraction of \( K_2CO_3 \) equal to 0.5.

### Step-by-Step Solution:

1. **Determine the Composition of the Mixture**:
- Given that the mole fraction of \( K_2CO_3 \) is 0.5, the mole fraction of \( KHCO_3 \) is also 0.5.
- This means that in the mixture, the moles of \( K_2CO_3 \) and \( KHCO_3 \) are equal.

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