Calculate the concentration of Fe in an iron (II) solution in units of mg/L. Sol
ID: 555509 • Letter: C
Question
Calculate the concentration of Fe in an iron (II) solution in units of mg/L.
Solution is FeH14O11S, 0.10g of FeH14O11S is used and dissolved in 100ml of distilled water. FeH14O11S has a molecular weight of 278.006g/mol. Fe has a molecular weight of 55.85g/mol.
1. What is concentration of Iron in this 100ml solution in mg/L?
This solution is then diluted, this is done by using a pipette to draw 10ml from the iron (II) solution and is then added to a seperate 100mL flask. This seperate 100mL flask is then filled to 100mL with distilled water.
2. What is the concentration of Iron in this diluted 100mL solution in mg/L?
Hope someone can help me with an answer soon, thank you!
Explanation / Answer
Ans. #1. Moles of FeH14O11S = Mass / Molar mass
= 0.10 g / (278.006 g/ mol)
= 3.597045 x 10-4 mol
# 1 mol FeH14O11S consists of 1 mol Fe-atom.
So,
Moles of Fe-atom = Moles of Moles of FeH14O11S
Hence,
Moles of Fe-atom = 3.597045 x 10-4 mol
Mass of Fe in sample taken = Moles of Fe x its Molar mass
= 3.597045 x 10-4 mol x (55.85 g/ mol)
= 0.02009 g
= 20.09 mg
# Given, volume of solution = 100.0 mL = 0.100 L
Now,
[Fe] = Mass of Fe in mg / Volume of solution in liters
= 20.09 mg/ 0.100 L
= 200.9 mg/ L
#2. Using C1V1 (stock solution) = C2V2 (diluted solution)
Or, 200.9 mg L-1 x 10.0 mL = C2 x 100.0 mL
Or, C2 = (200.9 mg L-1 x 10.0 mL) / 100.0 mL
Hence, C2 = 20.09 mg L-1
Therefore, [Fe] in diluted solution = 20.09 mg / L
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