In a sample of sodium carbonate some sodium sulphate is also mixed. 1.25 g of this sample is dissolved and the volume made up to 250 mL. 25 mL of this solution neutralises 20 mL of `(N)/(10)` sulphuric acid. Calculate the percentage of sodium carbonate in the sample.
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### Step 1: Determine the amount of sulfuric acid used in equivalents
The normality of the sulfuric acid is given as \( \frac{N}{10} \).
- Normality (N) = \( \frac{1}{10} \) N
- Volume of sulfuric acid used = 20 mL = 0.020 L
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