How many grams of sodium chloride (NaCl) should be used in preparing 20 liters of a solution containing 154 mEq/L?
To determine the grams of sodium chloride (NaCl) needed, we need to convert mEq/L to grams/L.
1 mole of NaCl = 58.44 g
1 mole of NaCl = 1 mole of Na+ ions + 1 mole of Cl- ions
So, the molecular weight of Na+ ions = 22.99 g/mol
And, the molecular weight of Cl- ions = 35.45 g/mol
1 mole of NaCl = 1 mole of Na+ ions + 1 mole of Cl- ions
1 mole of NaCl = 22.99 g + 35.45 g = 58.44 g
To convert mEq/L to grams/L, we need to multiply the mEq/L by the molecular weight of NaCl.
154 mEq/L * 58.44 g/mole = 9006.96 g/L
So, in 20 liters of a solution containing 154 mEq/L, the amount of sodium chloride needed is 9006.96 g/L * 20 L = <<9006.96*20=180139.2>>180139.2 g
To determine the number of grams of sodium chloride (NaCl) needed, follow these steps:
Step 1: Convert milliequivalents per liter (mEq/L) to moles per liter (mol/L).
Since NaCl is a 1:1 electrolyte, the number of moles of NaCl is equivalent to the number of milliequivalents (mEq) of NaCl. Therefore, 154 mEq/L is equivalent to 154 mmol/L.
Step 2: Convert moles per liter (mol/L) to moles.
Since the solution volume is given in liters, multiplying the concentration (in mol/L) by the volume (in liters) gives the number of moles in the solution. In this case, multiplying 154 mmol/L by 20 L gives 3080 mmol.
Step 3: Convert moles to grams.
To convert moles to grams, you need to know the molar mass of NaCl, which is 58.44 g/mol. Multiply the number of moles (3080 mmol) by the molar mass (58.44 g/mol) to get the number of grams.
3080 mmol * 58.44 g/mol = 179,827.2 g
Therefore, you would need approximately 179,827.2 grams of sodium chloride (NaCl) to prepare 20 liters of a solution containing 154 mEq/L.