Question

A 66 kg patient is in need of intravenous iron supplementation. She was ordered to receive a single intravenous 5 mg/kg dose
how many milliequivalents?


A 12% solution of potassium chloride (K CI- MW = 74.55 g/mol) has been prepared by a technician for direct intravenous inject
what is the osmolarity of this solution?
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Answer #1

1. Ordered dose of IV iron supplementation is 5mg /kg

Weight of the patient is 66 kg

Ordered dose - 5× 66= 330mg

Available 25ml of solution has 750mg of iron sulfate salt

750 mg of iron sulfate is present in 25 ml

1mg of iron sulfate is present in 25/750

Then 330 mg of iron sulfate is present in 25/750× 330

= 11ml

11ml of iron sulfate will be received by the patient as a single dose

2. Osmolarity of the solution is calculated by :-

Osmolarity = SUM OF ALL (molarity x n) OF EACH SOLUTE

n: number of particles that dissociated from the solute molecule.

Molarity of KCl is 74.5g/mol

number of particles of KCl is 2

Osmolarity = 74.5×2

= 149mOsmol/L

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