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Answer :
Final answer:
To meet the daily potassium requirement, 5.6 mL of potassium chloride needs to be added to the parenteral nutrition solution.
Explanation:
The student's question involves calculating the additional amount of potassium chloride to meet a patient's daily potassium requirement in a parenteral nutrition (PN) solution. First, we determine the amount of potassium already present in the PN: 7 mL x 4.4 mEq/mL = 30.8 mEq of potassium. Next, we subtract this amount from the daily requirement of 42 mEq to find the amount needed: 42 mEq - 30.8 mEq = 11.2 mEq.
Since potassium chloride (KCl) provides 2 mEq/mL, we divide the needed amount by this concentration to find the volume required: 11.2 mEq / 2 mEq/mL = 5.6 mL. Therefore, 5.6 mL of potassium chloride should be added to the PN.
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