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Answer :
Final answer:
The original concentration of Fe³+ in the water sample is approximately 8.00*10^3 mg/L.
Explanation:
To find the concentration of Fe³+ in the water sample, we first need to determine the amount of Fe³+ in the precipitate. The molecular mass of Fe(OH)3 is approximately 106.87 g/mol. So, 98.3g/106.87g/mol ≈ 0.92 mol of Fe(OH)3 was precipitated, and this includes 0.92 mol of Fe³+. The water volume is 6.42 L. So, the concentration of Fe³+ is 0.92 mol/6.42 L = 0.143 mol/L. Converting this to mg/L (since 1 mol of Fe³+ = 55.85 g = 55850 mg), the original concentration of Fe³+ is 0.143 mol/L * 55850 mg/mol ≈ 8.00*10^3 mg/L.
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