What weight of `Na_(2)CO_(3)` of 93% purity would be required to neutralise `45.6 mL` of `0.235 N` acid?

Explanation

To find the weight of Na₂CO₃ of 93% purity required to neutralize 45.6 mL of 0.235 N acid, we can follow these steps:

### Step 1: Calculate the milliequivalents of the acid
The formula for calculating milliequivalents (mEq) is:
\[ \text{mEq} = \text{Normality (N)} \times \text{Volume (mL)} \]
Given:
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