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Genrish500 [490]
3 years ago
6

If you digested 25 mg of pure ferrocene in nitric acid and then diluted the resulting solution up to 500 mL with 1% aqueous hydr

ochloric acid, what would be the concentration of iron in solution (provide an answer in both mg/L and mol/L)?
Chemistry
1 answer:
Vilka [71]3 years ago
6 0

Answer:

1.345*10⁻⁴ mol/L

15.023 mg/L

Explanation:

The chemical <u>formula of ferrocene</u> is Fe(C₅H₅)₂, thus its molecular weight is:

55.845 g/mol + 10*12g/mol + 10 *1g/mol = 185.845 g/mol

  • The moles of Fe contained in 25 mg (or 0.025 g) of ferrocene are:

0.025gFerrocene*\frac{1molFerrocene}{185.845g} *\frac{1mol Fe}{1molFerrocene}=1.345*10^{-4} molFe

  • The final volume is 500 mL, or 0.5 L. So the iron concentration in mol/L is:

\frac{1.345*10^{-4}molFe}{0.5L}= 2.69*10^{-4} mol/L

  • We can convert that value into mg/L:

2.69*10^{-4} \frac{molFe}{L} *\frac{55.845g}{1molFe}*\frac{1000mg}{1g}=15.023 mg/L

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