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How many meq of sodium ions are contained in a 100.0 mL IV bag of normal saline (0.900 wt% aqueous NaCl)? The density of the solution is 0.985 g/mL.

A. 1.71 meq
B. 17.1 meq
C. 0.171 meq
D. 1.71 x 10² meq

Answer :

Final answer:

To find the milliequivalents of sodium ions in a 100 ml normal saline IV bag, we calculate the mass of NaCl based on the solution's weight percentage and density, convert to moles, and then to meq. The result, 15.15 meq, does not match the provided options.

Explanation:

The question requires calculating the milliequivalents (meq) of sodium ions in a 100.0 ml IV bag of normal saline which has a concentration of 0.9% NaCl by weight and a density of 0.985 g/ml.

First, we calculate the mass of NaCl in 100.0 ml of the solution:

  • Volume of solution = 100.0 ml
  • Density of solution = 0.985 g/ml
  • Mass of solution = Volume × Density = 100.0 ml × 0.985 g/ml = 98.5 g

Since normal saline is a 0.9% NaCl solution, the mass of NaCl in 100.0 ml is:

  • Mass of NaCl = 0.9% of total mass = 0.009 × 98.5 g = 0.885 g

We convert grams of NaCl to moles, knowing the molar mass of NaCl is 58.44 g/mol, and then to meq:

  • Molar mass of NaCl = 58.44 g/mol
  • Moles of NaCl = Mass of NaCl / Molar mass of NaCl = 0.885 g / 58.44 g/mol
  • Since 1 eq of Na+ is 1 mol of Na+, and there are 1000 meq in 1 eq, the meq of Na+ = Moles of NaCl × 1000 meq/eq

Therefore, the meq of sodium ions can be calculated as follows:

  • meq of Na+ = (0.885 g / 58.44 g/mol) × 1000 meq/eq

The result gives us the meq of sodium ions in the IV bag. After doing the math, we find that the correct answer is 15.15 meq, which is not one of the provided options.

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