Question

The primary health care provider prescribes 1000 mL of 5% dextrose in water containing 1.5 mEq of potassium chloride (KCl) per 100 mL to infuse at a rate of 110 mL/hr. The intravenous starts at 0700. At 1300, how many mEq will be left in the remaining amount of fluid?

Answer 1

The amount of potassium chloride contained in the remaining intravenous would be 5.1 mEq

Step-by-step 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