1 (99)

Prévia do material em texto

4 Aqueous Reactions Solutions to Exercises 0.150 M 25.0 mL = 0.054348 = 0.0543 M 69.0 mL 0.0543 M 0.0543 M (c) Plan. Calculate concentration of and CI⁻ due to the added solid. Then sum to get total concentration of Solve. 75.0 3.60 mL g KCl so 74.55 1 mol g KCl 1000 mL = 0.6439 = 0.644 M 0.250 M CaCl₂; 2(0.250 M) = 0.500 M Cl⁻ total = 0.644 M + 0.500 M = 1.144 M 0.644 M 0.250 M 4.73 Analyze/Plan. Follow the logic of Sample Exercise 4.14. Solve. (a) = 16.89 = 16.9 mL 14.8 M 14.8 M Check. (b) 14.8 M x10.0 = 0.296 M Check. 150/500 ≈ 0.30 M 4.74 (a) 0.500 0.110L = 0.00917 = 9.2 mL 6.0 M 6.0 M (b) = 0.240 M 250 mL 4.75 (a) Plan/Solve. Follow the logic in Sample Exercise 4.13. The number of moles of sucrose needed is 0.250 mol 0.250 = 0.06250 = 0.0625 mol 1L Weigh out 0.0625 mol = Add this amount of solid to a 250 mL volumetric flask, dissolve in a small volume of water, and add water to the mark on the neck of the flask. Agitate thoroughly to ensure total mixing. (b) Plan/Solve. Follow the logic in Sample Exercise 4.14. Calculate the moles of solute present in the final 350.0 mL of 0.100 M C₁₂H₂₂O₁₁ solution: = 0.100 mol 1L C₁₂H₂₂O₁₁ x0.3500 = 0.0350 mol C₁₂H₂₂O₁₁ Calculate the volume of 1.50 M glucose solution that would contain 0.0350 mol C₁₂H₂₂O₁₁: 93