Question

How many grams of sodium chloride (NaCl) should be used in preparing 20 liters of a solution containing 154 mEq/L?

Answer 1

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