HW key _3 S problems part b

# HW key _3 S problems part b - S6 A 1.37 M aqueous solution...

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S6. A 1.37 M aqueous solution of citric acid (H 3 C 6 H 5 O 7 ) has a density of 1.10 g/cm 3 . What are the mass percent, molality, and mole fraction of the citric acid? If we have 1.00 L of solution then: 1.37 mol citric acid × 192.1 g mol = 263 g citric acid 1.00 × 10 3 mL solution × 1.10 g mL = 1.10 × 10 3 g solution mass % citric acid = 263 g 1.10 × 10 3 g × 100 = 23.9% In 1.00 L of solution, we have 263 g of citric acid and 1.10 × 10 3 263 = 840 g of H 2 O molality = 1.37 mol citric acid 0.84 kg H 2 O = 1.6 mol/kg 840 g H 2 O × 1 mol 18.0 g = 47 mol H 2 O χ citric acid = 1.37 47 + 1.37 = 0.028 S7. A solution is made by dissolving 25 g of NaCl in enough water to make 1.00 L of solution. Assume that the density of the solution is 1.00 g/cm 3 . Calculate the mass percent, molarity, molality, and mole fraction of NaCl. 1.00 L × 1000 mL L × 1.00 g mL = 1.00 × 10 3 g solution mass % NaCl= 25 g NaCl 1.00 × 10 3 g solution × 100 = 2.5% molarity= 25 g NaCl 1.00 L solution × 1 mol NaCl 58.44 g NaCl = 0.43 M 1.00 × 10 3 g solution contains 25 g NaCl and 975 g H

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HW key _3 S problems part b - S6 A 1.37 M aqueous solution...

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