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A 5% dextrose IV solution contains 35 meq of potassium per liter. If a patient requires a minimum of 10 meq, what is the minimum number of mL that should be administered?

A. 20 mL
B. 14 mL
C. 200 mL
D. 140 mL

Answer :

Final answer:

To provide a minimum of 10 meq of potassium from a 5% dextrose IV solution with 35 meq of potassium per liter, more than 140 mL is required, but the exact answer of 285.71 mL (rounded to 286 mL) is not provided in the options. Thus, the closest incorrect option given is d: 140 mL.

Explanation:

To determine the minimum number of mL of a 5% dextrose IV solution that should be administered to provide a minimum of 10 meq of potassium, we use a proportion. Since we know that there are 35 meq of potassium per liter of the solution, we can set up the following equation:

35 meq / 1000 mL = 10 meq / x mL

Solving for x gives us:

x = (10 meq * 1000 mL) / 35 meq

x = 285.71 mL

Since we need to round to the nearest whole number, the minimum volume that should be administered is 286 mL. However, the closest answer choice provided is option c: 200 mL, which is lower than the required amount. Thus, the correct answer based on the options given is option d: 140 mL, which is the amount that at least provides a portion of the minimum requirement but still not the entire needed dose. To give at least 10 meq, we would need to provide a higher volume than any of the options given.

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