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Solution Chemistry

Solution Chemistry. Colligative Properties. Differences between pure solvents and solutions that relate only to the concentration of solute as opposed to the identity of the solute. . Solvent vs. Solution Boiling Point. 1.0M NaCl , boils at 101°C. Pure H 2 O, boils at 100 ° C.

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Solution Chemistry

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  1. Solution Chemistry

  2. Colligative Properties • Differences between pure solvents and solutions that relate only to the concentration of solute as opposed to the identity of the solute.

  3. Solvent vs. Solution Boiling Point 1.0M NaCl, boils at 101°C • Pure H2O, boils at 100°C

  4. Boiling Point Elevation

  5. Concentration & Colligative Properties • Electrolytes will have a greater affect on boiling point/ freezing point due to their dissociation in a solvent. • Examples: • 1M CH3F(aq)  1M CH3F(aq) • 1M NaCl(aq)  1M Na+(aq) + 1M Cl-(aq) • 1M H2SO4(aq)  2M H+(aq) + 1M SO42-(aq)

  6. Salt to the Pasta Water Adds Flavor AND • A possibly higher boiling temperature, therefore a quicker cooking time!

  7. Calculating Boiling Point Elevation • ΔT = iKbmsolute • ΔT: change in temperature • i: number of molecules or ions formed upon addition to the solvent • Kb: constant that is characteristic of the solvent; molal boiling-point elevation constant • msolute: molality of the solute in the solution

  8. Molal Boiling Point/Freezing Point Constants

  9. Boiling Point Elevation • Determine the boiling point of a solution containing 45.0g NaCl in 200.0g of ethanol. • ΔT = iKbmsolute • ΔT: ? • i: 2 (NaCl Na+ + Cl-) • Kb: 1.22 °C/m • msolute: 3.87m • ΔT= 9.44°C; 78.4°C + 9.44°C = 87.84°C

  10. Determining molar mass from Boiling Point Elevation • A solution contains 3.75g of a nonvolatile pure hydrocarbon in 95.0g acetone. The boiling points of pure acetone and the solution are 55.95°C and 56.50°C, respectively. The molal boiling point constant of acetone is 1.71°C-m. What is the molar mass of the hydrocarbon?

  11. Determining molar mass from Boiling Point Elevation • ΔT = iKbmsolute • ΔT: 56.6 – 55.95 = 0.55°C • i: 1 (hydrocarbons are molecules) • Kb: 1.71 °C/m • msolute: ? • 0.55 = 1 x 1.71 x m; m = 0.322 • 0.322 moles/ kg = x moles/ 0.095kg • Moles = 0.03059 = 3.75grams • Molar mass x 0.03059 moles = 3.75 grams • Molar Mass = 122.6g/mol

  12. Freezing Point Depression

  13. Freezing Point Depression

  14. Freezing Point Depression

  15. Calculating Freezing Point Depression • ΔT = iKfmsolute • ΔT: change in temperature • i: number of molecules or ions formed upon addition to the solvent • Kf: constant that is characteristic of the solvent; molal freezing-point depression constant • msolute: molality of the solute in the solution

  16. Calculating Freezing Point Depression • Determine the freezing point of a solution containing 25.8g of AlCl3 in 100.0g of chloroform. • ΔT = iKfmsolute • ΔT: ? • i: 4 (AlCl3 Al3++ 3Cl-) • Kf: 4.68 °C/m • msolute: 1.95m • ΔT = 36.5°C; -63.5°C + 36.5°C = -27.0°C

  17. Determining Molar Mass from Freezing-Point Depression • A solution of 2.50g of a compound having the empirical formula C6H5P in 25.0g of benzene is observed to freeze at 4.3°C. Calculate the molar mass of the solute and its molecular formula.

  18. Determining Molar Mass from Freezing-Point Depression • ΔT = iKfmsolute • ΔT: 5.5 – 4.3 = 1.2°C • i: 1 (molecular compound) • Kf: 5.12 °C/m • msolute: ? • 1.2 = 1 x 5.12 x m; m = 0.234 • 0.234moles/ kg = x moles/ 0.025kg • Moles = 0.005859 = 2.50g • Molar mass x 0.005859g = 2.50g • Molar mass = 426.69g/mol • Molecular formula: C24H20P4

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