the normal concentration of sodium chloride in blood is 0.9%(w/v). therefore assuming a FW for NaCl of 58.4 g/mole, an isotonic 0.9% (w/v) saline solution would have how many mili-equivilents per liter (mEq/L) of Na+ ions?
To find the milliequivalents per liter (mEq/L) of Na+ ions in a 0.9% (w/v) saline solution, you need to follow a few steps:
1. Calculate the molarity of NaCl:
- First, convert the percentage concentration to grams per 100 mL (w/v):
0.9 grams/100 mL = 0.9 g/100 mL
- Since the question asks for mEq/L, convert the volume to liters:
100 mL = 0.1 L
- Convert grams to moles by dividing by the molar mass of NaCl (58.4 g/mol):
0.9 g / 58.4 g/mol = 0.015 mol
- Finally, calculate the molarity by dividing the moles by the volume in liters:
0.015 mol / 0.1 L = 0.15 M
2. Calculate the milliequivalents of Na+:
- Since NaCl dissociates in water, one mole of NaCl produces one mole of Na+ ions.
- Therefore, the molarity of Na+ ions will be the same as the molarity of NaCl, which is 0.15 M.
- Since 1 mol of a monovalent ion is equal to 1 milliequivalent, the concentration remains the same:
0.15 mEq/L
So, an isotonic 0.9% (w/v) saline solution would have 0.15 mEq/L of Na+ ions.