Im coming at this problem with no knowledge of ppm, but as im told if you are ge

Question

Im coming at this problem with no knowledge of ppm, but as im told if you are getting dissolve ~2.02 in solution, it is not correct.. How do you prepare, 1L of 5000ppm Fe^3+, using Iron Nitrate Nonahydrate [ Fe(NO3)3 • 9H2O ] Im coming at this problem with no knowledge of ppm, but as im told if you are getting dissolve ~2.02 in solution, it is not correct.. How do you prepare, 1L of 5000ppm Fe^3+, using Iron Nitrate Nonahydrate [ Fe(NO3)3 • 9H2O ] Im coming at this problem with no knowledge of ppm, but as im told if you are getting dissolve ~2.02 in solution, it is not correct.. How do you prepare, 1L of 5000ppm Fe^3+, using Iron Nitrate Nonahydrate [ Fe(NO3)3 • 9H2O ]

Explanation / Answer

Molar mass of [ Fe(NO3)3 • 9H2O ] = 404 g/mol

1 ppm = 1mg / 1 L

1 mg = 0.001 g

Molar mass of Fe3+ = 55.84 g/mol

So 1g of [ Fe(NO3)3 • 9H2O ] = (404 / 55.84) g of Fe3+

5000 mg = 5 g

1L of 5000 ppm Fe3+ = 1L of 5 g of Fe3+ = 5* (404 / 55.84) => 36.175 g of  [ Fe(NO3)3 • 9H2O ] contains 5 g of Fe3+ which is equal to 5 ppm of 1 Ltr solution.

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