Question
How many gram of Fe+3 are required to make 1.0 L of an aqueous solution of 10 ppm Fe+3? Assume density of water is 1.00 g/mL.
Answers
DrBob222
10 ppm = 10 g/10^6 g H2O. If water has density of 1.00 g/mL, then 10^6 grams H2O will be 1,000,000 mL.
10 ppm = 10 g/1,000,000 mL.
So divide the 1,000,000 ml by 1,000,(to obtain 1,000 mL total or 1L) and to keep things equal we divie 10 g by 1000 = 10/1000 = 0.01 g Fe^+3
10 ppm = 10 g/1,000,000 mL.
So divide the 1,000,000 ml by 1,000,(to obtain 1,000 mL total or 1L) and to keep things equal we divie 10 g by 1000 = 10/1000 = 0.01 g Fe^+3