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13 Properties of Solutions Solutions to Exercises 1750 kg Zn 1000 kg X 65.39 1 mol = 26,762.5 = 2.68 X 10⁴ mol Zn m = 2.676 7000 10⁴ kg mol Cu = 3.82 Zn (b) Plan. M = mol Zn/L brass. Use mol Zn from part (a). Change 1 m³ -> L brass and calculate M. Solve. 1m³ (10)³ m³ X = 1000 L M = 2.676 10⁴ mol = = 26.8 M Zn 1000 L brass 13.58 (a) 0.0500 mol C₈H₁₀N₄O₂ 194.2 mol C₈H₁₀N₄O₂ C₈H₁₀N₄O₂ = 9.7100 1kg CHCl₃ 9.710 mass (b) 1000 g 1 mol CHCl₃ = 8.375 = 8.38 mol CHCl₃ 0.0500 13.59 Analyze. Given: 4.6% by volume (in air), 1 atm total pressure. Find: partial pressure and molarity of in air. Plan. 4.6% by volume means 4.6 mL of could be isolated from 100 mL of air, at the same temperature and pressure. According to Avogadro's Law, equal volumes of gases at the same temperature and pressure contain equal numbers of moles. By inference, the volume ratio of to air, 4.6/100 or 0.046, is also the mole ratio. Solve. Pco₂ = xPₜ = 0.046 atm) = 0.046 atm M = mol air = n/V. M = n/V = P/RT = RT = 0.046 310 atm 0.08206 mol K atm = 1.8 10⁻³ M 13.60 (a) For gases at the same temperature and pressure, volume % = mol %. The volume and mol % of in this breathing air is 4.0%. (b) Pco₂ = xPₜ = 0.040 (1atm) = 0.040 atm = RT = 0.040 310 0.08206 mol K atm = 1.6 10⁻³ M Colligative Properties (section 13.5) 13.61 freezing point depression, = boiling point elevation, = osmotic pressure, П = M RT; vapor pressure lowering, PA = 378