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[117(g NaCl) / 58.5(g NaCl/mol NaCl)] / 40.0(L solution) = [117/58.5]/40.0 = 2.00(mol NaCl) / 40.0(L) = 0.0500 mol NaCl / L solution = 0.0500 M

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Q: What is the molarity of a solution containing 117 grams of sodium chloride in enough water to make a 40.0 L solution?
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