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250 milliliters of a concentrated iron (II) ion solution are sitting on the lab bench in need of analysis to determine its exact concentration of Fe2+ ions. As part of the analysis a 15 mL sample of this solution was diluted to a final volume of 100 mL. The iron (II) ion concentration of the diluted solution was found to be 1.35 x 10-3 mg/mL Fe2+. What was the total number of milligrams of iron (II) that were contained in the original solution?

Note that this question is not asking for how many mg/mL of iron (II) where in the original solution, but the total mass of iron contained in the original solution. Hint: You must first calculate the mg/mL of iron (II) in the original solution.

User Crazyscot
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1 Answer

7 votes

Answer:

2,25mg of Fe²⁺

Step-by-step explanation:

A chemical concentration is defined as an amount of solute per amount of solution. The diluted solution has an iron (II) concentration of 1,35x10⁻³ mg/mL. As the dilution was of 15 mL in 100 mL, the concentration of the original solution is:

1,35x10⁻³ mg/mL ×
(100mL)/(15mL) = 9,00x10⁻³ mg/mL

As the total volume of the concentrated iron solution is 250 mL, the total mass of this solution is:

250 mL ×
(9,00x10^(-3)mg)/(1mL) = 2,25 mg of Fe²⁺

I hope it helps!

User Ajxs
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