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How do you prepare 125 parts per million Fe from the salt Fe(NH4)2(SO4)2.6H2O?
Wiki User
∙ 8y ago
Molar mass Fe is 55.845 mg/mmol, so 125 mg represents 125 / 55.845 = 2.238 mmol Fe.
Since the molar mass of (NH4)2Fe(SO4)2·6H2O is 392.13 g/mol, 2.238 mmol of this Mohr's salt weights 392.13(mg/mmol) * 2.238 mmol = 878 mg Mohr's salt.
This has to be dissolved in 1.00 L water to give the desired 125 ppm (=mg/L) Fe2+ solution
Wiki User
∙ 8y ago
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