Answer :
Final answer:
The calculation determines the amount of sodium ions in meq in a 100.0 ml IV bag of normal saline. The calculated value is 14.92 meq based on the density provided. However, this value is not listed in the answer choices, implying a potential issue with the question or its options. option d is answer
Explanation:
The question involves calculating the amount of sodium ions in milliequivalents (meq) contained in a 100.0 ml IV bag of normal saline, which is known as a 0.9% NaCl solution. To start this calculation, we use the density of the solution (0.985 g/ml) to find the mass of the solution:
Mass of solution = Volume × Density = 100.0 ml × 0.985 g/ml = 98.5 g
Since normal saline is a 0.9% NaCl solution, it contains 0.9 g of NaCl per 100 g of solution. Thus, we can find the mass of NaCl in the IV bag:
Mass of NaCl = 98.5 g × 0.9% = 0.885 g
Sodium chloride (NaCl) contains one sodium ion per molecule. The molar mass of NaCl is 58.44 g/mol, and the molar mass of Na is 22.99 g/mol, which means each mole of NaCl gives one mole of sodium ions.
Now, we find the number of moles of sodium ions:
Moles of Na = Mass of NaCl / Molar mass of NaCl = 0.885 g / 58.44 g/mol = 0.01515 mol
Finally, since 1 mole equals 1000 meq, we can convert moles to meq:
meq of sodium ions = Moles of Na × 1000 meq/mol = 0.01515 mol × 1000 meq/mol = 15.15 meq
However, we've calculated this for the entire mass of the solution (100 g), and only needed to calculate it for the 98.5 g present in the 100.0 ml IV bag. Thus, scaling back for the actual mass:
meq of sodium ions (for 98.5 g of solution) = 15.15 meq × (98.5 g / 100 g) = 14.92 meq
Answer: The meq of sodium ions contained in the 100.0 ml IV bag of normal saline is approximately 14.92 meq, which is not one of the listed options. Therefore, it seems there may be a misunderstanding with the provided options or a mistake in the calculations. However, based on the calculations with the given density, the closest answer choice would be 6.02 meq (d).
