Question

A client with vitamin deficiencies is admitted to the acute care setting due to signs of dehydration. The nurse's assessment of laboratory values reveals a potassium of \(2.9 \mathrm{mEq} / \mathrm{L}\). Which dietary selection would be best for this client? A. Banana B. Apricot C. Spinach D. Artichoke

Short answer

Answer: C. Spinach.

Step-by-step solution

Research potassium content in mentioned foods

Look up the potassium content of the given food items: banana, apricot, spinach, and artichoke.

Present the potassium content of each food item

Based on research, the potassium content for each food item per 100 grams is as follows: A. Banana - \(358 \mathrm{mg}\) B. Apricot - \(259 \mathrm{mg}\) C. Spinach - \(558 \mathrm{mg}\) D. Artichoke - \(370 \mathrm{mg}\)

Compare and determine the best food source of potassium

Compare the potassium content of each food item: A. Banana: \(358 \mathrm{mg} / 100 \mathrm{g}\) B. Apricot: \(259 \mathrm{mg} / 100 \mathrm{g}\) C. Spinach: \(558 \mathrm{mg} / 100 \mathrm{g}\) D. Artichoke: \(370 \mathrm{mg} / 100 \mathrm{g}\) In this case, Spinach (Option C) has the highest potassium content with \(558 \mathrm{mg}\) per 100 grams. Thus, it is the best dietary selection for the client with dehydration and low potassium levels. Answer: C. Spinach.