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A.P. Chemistry Solutions Chapter 11 ↑ ↑ ↓ ↑ ↑ ↑ Solute Solvent M m + +
Separating them… Chromatography Chapter 11 making Homogeneous Mixtures Colligative Properties (Where Conc. Affects Them) Changes Phase No Set Ratio B.P. Elevation Solute parts Solvent V.P. Lowering F.P. Depression Solvation Process Concentration Descriptions Raoult’s Law V.P.(sol’n) = C(solv) V.P.(solv) + C(solute) V.P.(solute) DT = Kmi M 1. Expand Solute volatile solutes Ideal 2. Expand Solvent m DH= + 3. Intermix DH= Non-volatile solutes + C van’t Hoff Factor, i Non-Ideal DH= - DH= hydration DHhyd= mass % Small Deviations Large Deviations +/- Neg. Deviations DH= Pos. Deviations DH= Low [ ] High [ ] - + Like Dissolves Like (generally) Low charges High charges Heat of solution DHsol Phases Involved Effect (Generally) ↑ ↑ Sol. in Liq. T sol. (Dictates the degree to which you can dissolve a solute in a solvent) Temp. ↓ + ↑ - DH= + DH= DH= Gas in Liq. T sol. affected by: large Entropy driven ↑ ↑ Press. Gas in Liq. P sol. ( ) solubility From the Gas P = kC → Henry’s Law insoluble soluble (≥0.10 M)
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End of Chapter Problem:
You’re in the lab…in front of you are four substances. Two are labeled solutes…they are 25.0 grams of TTE (C2Cl3F3) and 2.69 grams of CsOH. Two are labeled solvents…they are grams of benzene (C6H6) and grams of water. Which substances would you combine to make solutions in which the solute would be considered soluble in the solvent? Looking at the one that produces a water solution, calculate the heat of hydration if the heat of solution is -72 kJ/mol and the lattice energy for the solute is -724 kJ/mol. Calculate the molarity, molality, and mole fraction (with respect to the solvent) of this solution given that the density of the solution is 1.03 g/mL Find the freezing pt. of this solution assuming ideal behavior. Find the vapor pressure of this solution. basic, non-volatile, ionic solute mol 6.83 mol TTE in benzene (both nonpolar) CsOH in water DH= + or – and small DH1= 724 kJ DH2 + DH3 = DHhyd= DHsol – DH1= -796 kJ C = mol/(6.83 mol mol) = 0.995 Vol of sol’n = g/(1.03 g/mL) M = mol/0.122 L = molar m = nol/ kg = molal =122 mL = L -0.543°C DT = Kfmi = (1.86 K kg/mol)(0.146 mol/kg)(2)= 0.543 K Would the measured VP be higher or lower than 23.7? Lower, DH is neg; negative deviation VPsolution = Csolvent VPsolvent= (0.995)(23.8 torr)= 23.7 torr
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