Answer:
<em>The amount of potassium chloride contained in the remaining intravenous would be 5.1 mEq</em>
Explanation:
Given
that 1000 mL of Dextrose in water contains 1.5 mEq of potassium chloride per 100 mL
Calculating the amount of potassium contained per mL
If 1.5 mEq of potassium chloride is contained in 100 mL of Dextrose in water, 1 mL would be;
per mL = 1.5 m Eq / 100 mL
= 0.015 mEq /mL
So the amount of potassium chloride per mL is 0.015 mEq /mL
Calculating the amount of liquid remaining.
Given the start time = 0700 Hrs
the end time = 1300 Hrs
the time the intravenous lasted = 1300 Hrs - 0700 Hrs = 6 Hrs
So the intravenous lasted for 6 hours.
Therefore at an infusion rate of 110 mL/hr the amount of intravenous infused would be;
6 hrs x 110 mL/hr = 660 mL
Therefore the amount of intravenous infused is 660 mL
The intravenous that remained would be;
1000 mL - 600 mL = 340 mL
The intravenous that would remain after 6 Hours would be 340 mL
Calculating the amount of potassium chloride contained in the remaining intravenous;
Given that the amount of potassium chloride contained in 1 mL of the intravenous is 0.015 mEq;
The amount of potassium chloride contained in the intravenous remaining would be;
0.015 mEq/mL x 340 mL = 5.1 mEq
Therefore the amount of potassium chloride contained in the remaining intravenous would be 5.1 mEq
