Question

A 71 kg patient is in need of intravenous iron supplementation. She was ordered to receive a single intravenous 9 mg/kg dose of iron (III) sulfate (ferric sulfate - Fe2(SO4)3 - MW=399.88 g/mol). Iron (III) sulfate was prepared as a stabilized aqueous solution for this patient by dissolving 975 mg of stabilized iron sulfate salt in water to produce 15 mL of solution. How many milliequivalents of iron (Fe 3+) will the patient receive as a single dose? Assume the salt completely disassociates in solution.

answer for this question should be in MEq not ml

answer should be in units of mOsmol/L

Answer 1

1. The question here is slightly ambiguous.

Iron prescribed for the patient(71kg) is 9mg/kg of Fe₂(SO4)₃  = 71 x 9 = 639 mg
Iron solution actually prepared = 975 mg of Fe₂(SO4)₃  
Lets assume this patient receives the dose as prescribed so he receives 639 mg of Fe₂SO_4.
MW of Fe₂(SO4)₃is 399.88 grams which means 399.88 mg of Fe₂(SO4)₃ =1 millimole
So 639 mg of Fe₂(SO4)₃ =  (1/399.88) x 639 = 1.6 millimole
Each millimole of Fe₂(SO4)₃ on dissolution gives 3 MEq of Fe³⁺,
so 1.6 millimole of Fe₂(SO4)₃ on dissolution produces 1.6 x 3 = 4.8 MEq of Fe³⁺

2. Osmolarity of a solution can be determined using the following equation:

mOsm/L = [wt. of the substance(g/l) / Mol. wt .(g)] x no.of particles x 1000

wt. of substance in this case = 23.4 % x 10 = 234 g/l
Mol. wt. = 58.44 g
no. of particles = 2

So mOsm/L = ( 234/58.44 ) x 2 x 1000 = 8008.21