`5 mL` of `8 N HNO_(3), 4.8 mL` of `5N HCl` and a certain volume of `17 M H_(2)SO_(4)` are mixed together and made upto `2 "litre"`. `30 mL` of this acid mixture exactly neutralizes `42.9 mL` of `Na_(2)CO_(3)` solution containing `1 g Na_(2)CO_(3)`. `10 H_(2)O "in" 100 mL` of water. Calculate the amount of sulphate ions in `g` present in solution.

Explanation

To solve the problem step-by-step, we will first calculate the milliequivalents of each acid present in the mixture, determine the total milliequivalents of acid in the solution, and then find the amount of sulfate ions present.

### Step 1: Calculate the milliequivalents of HNO₃
- Normality (N) of HNO₃ = 8 N
- Volume of HNO₃ = 5 mL

Milliequivalents of HNO₃ = Normality × Volume (in L)
\[ \text{Milliequivalents of HNO₃} = 8 \, \text{N} \times 0.005 \, \text{L} = 0.04 \, \text{equivalents} = 40 \, \text{milliequivalents} \]
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