Answer :
Final answer:
There are approximately 154 mEq of Na+ in one liter of normal saline. This is calculated based on the fact that normal saline contains 9.0 grams of NaCl per liter and the molecular weight of NaCl being roughly 58.44 g/mol.
Explanation:
The student's question pertains to the amount of sodium ions (Na+) present in one liter of normal saline. Normal saline typically contains 9.0 grams of sodium chloride (NaCl) per liter. To find out how many milliequivalents (mEq) of Na+ are in 1 liter of normal saline, we need to first understand that normal saline is a 0.9% NaCl solution, which translates to 0.9 grams of NaCl per 100 milliliters of water, or 9 grams per liter.
To convert grams of NaCl to mEq of Na+:
- Calculate the amount of moles of NaCl in 9 grams, knowing that the molecular weight of NaCl is approximately 58.44 g/mol.
- Since one mole of NaCl releases one mole of Na+ ions, the same number of moles will be present for Na+.
- Finally, considering that 1 mole of a substance contains Avogadro's number of particles, we use the conversion that 1 mole of Na+ is equivalent to 1000 mEq to find the number of mEq/L.
By using these conversions, we find that the formula weight of NaCl is approximately 58.44 g/mol, and since normal saline contains 9.0 g of NaCl per liter, there are approximately 154 mEq of Na+ in one liter of normal saline, because each mole of NaCl dissociates into one mole of Na+, which is equivalent to 1000 mEq.
