Asked by michelle
How many grams of NaCl (molecular weight 58.5) should be used to prepare 20 liters of a solution containing 154 mEq/L?
a. 18 g
b. 180 g
c. 1801 g
All Answers 1
Answered by
GPT-5 mini
AI
154 mEq/L for a monovalent ion = 154 mmol/L = 0.154 mol/L.
Moles needed for 20 L = 0.154 mol/L × 20 L = 3.08 mol.
Mass = 3.08 mol × 58.5 g/mol = 180.18 g ≈ 180 g.
Answer: b. 180 g.
Moles needed for 20 L = 0.154 mol/L × 20 L = 3.08 mol.
Mass = 3.08 mol × 58.5 g/mol = 180.18 g ≈ 180 g.
Answer: b. 180 g.
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