Asked by Nico
How many milliliters of a solution containing 2 mEq of potassium chloride (KCl) per milliliter should be used to obtain 2.98 g of potassium chloride.
Answers
Answered by
DrBob222
1 mol KCl = 74.6 g
1 eqivalent KCl = 74.6
1 mEq = 74.6 mg = 0.0746 g.
2 mEq/mL = 2*74.6 mg/mL = 149.2 mg/mL or 0.149 grams/mL KCl.
You want 2.98 g KCl so
0.149 grams KCl/mL x # mL = 2.98 g.
Solve for # mL.
1 eqivalent KCl = 74.6
1 mEq = 74.6 mg = 0.0746 g.
2 mEq/mL = 2*74.6 mg/mL = 149.2 mg/mL or 0.149 grams/mL KCl.
You want 2.98 g KCl so
0.149 grams KCl/mL x # mL = 2.98 g.
Solve for # mL.
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