Unit 4 Practice Problems Key Solubility Homogeneous mixture, parts
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1. Unit 4 Practice Problems Key Solubility Homogeneous mixture, parts indistinguishable. Solid-liquid= salt water, gas-liquid=O2 in water, gas-gas= CO2 in air. 2. The ionic crystal is solvated by the polar water molecules through ion-dipole bonding. The cation is surrounded by the negative side of the water molecule and the anion is surrounded by the positive side of the water molecule. 3. The sucrose molecule has multiple sites of H-boding where polar water molecules can attach and solvate the molecule. 4. Solubility is determined by the extent to which the interacting particles (solvent-solute) have similar types of intermolecular bonding. If the solute-solvent IMF’s are stronger than the solvent-solvent and solute-solute interactions than the substance will be soluble. 5. a) lattice energy is greater because for the solution to cool the energy required (absorbed) to break apart the lattice is greater than the energy released when the solute and solvent particles bond. βπ»π πππ’π‘πππ = πππ ππ‘ππ£π, πππππ‘βπππππ b) Solubility increases with increasing temperature. Increased temp=increased KE, therefore more energy available to break apart the KNO3 lattice. 6. a) Methanol. NaCl is ionic, so it dissolves through ion-dipole forces. Both methanol and 1-propanol have a single polar OH group, but the hydrocarbon portion interacts only weakly with the ions and 1-propanol has a longer hydrocarbon portion than methanol. b) Water. Ethylene glycol molecules have two OH groups, so they interact with each other through H-bonding. H bonds formed with water can replace these H-bonds between molecules better than dipole-induced dipole forces with hexane can. c) Ethanol. Diethyl ether molecules interact through dipole-dipole and dispersion forces. They can form H bonds to water or to ethanol. But ethanol can also interact with the diethyl ether effectively through dispersion forces because it has a hydrocarbon chain. 7. No. The dipole-dipole forces attracting sugar molecules is stronger than the weak LDF for the nonpolar polish remover. 8. No. There are no, or very few attachment points for the polar water molecule to solvate the oil. 9. HCl is a polar molecule that can easily attract to the polar water molecule. Propane uses LDF’s to attract, polar water molecule has no points of attachment for solvation. 10. a) dipole-induced dipole b) dipole-dipole c) ion-dipole 11. Gas solubility decreases with increasing temperature, so less dissolved oxygen for aquatic critters to breath. 12. Gas solubility increases with increasing pressure according to Henry’s Law S gas=K Pgas 13. Solute CO2(g) NaCl(s) increasing temp decrease increase decreasing temp increase decrease increasing pressure increase no effect decreasing pressure decrease no effect 14. Sg = (3.1 x 10-2 M/atm)(4.0 atm) = 0.12 mol/L 15. Sg = kPg= (6.8 x 10-4 M/L-atm)(0.78 x 2.50 atm) 16. . mass (grams) solute 5g moles (mol) .083 mol Sg = 1.3 x 10-3 mol/L volume (L) 5π × 1 πππ 60.06 π = .083 πππ solvent 95 g 5.27 mol .095 L solution 100g (.083 + 5.27) = 5.35 mol .099 L 17. mole H2O mass % 1πππ 18.02 π .005 L 1ππΏ 1πΏ × = .099πΏ 1.01π 1000ππΏ 1ππΏ 1πΏ 95π π»2 π × × = .095πΏ 1.0π 1000ππΏ 100π π πππ × mole NaOH 95 π × mol = m/MM 123 g NaOH/40.0 g = 3.08 mol NaOH mol = m/MM 289 g H2O/18.02 g = 16.04 mol H2O % = (msolute/mtotal) x 100 [123 g H2O/(289 + 123 g)]100 = 29.9 % = 5.27 πππ X = molsolute/moltotal X = 3.08 mol/(3.08 mol + 16.04 mol) =.16 0.161 M = molsolute/Vsolution(L) M = 3.08 mol NaOH/0.300 L = 10.3 mol/L mole fraction molarity 123g/(123g + 289g) x 106= 299,000 ppm PPM 18. Solution (in 100g H2O) Sat, Unsat, Supersat 40 g of KClO3 at 50oC 110 g NaNO3 at 45oC supersaturated extrapolate graph, saturated unsaturated 70 g KNO3 at 60oC 70 g NH4Cl at 70oC 19. a)12 πππ π»πΆπ × b) 37% = c) π = 36.46 π 1 πππ 437.52 π +/- how many g to make saturated? -21 - -18 approx. 4 = 437.52 π πππ π ππ π πππ’π‘πππ 1182.5π 1000ππΏ +/- how many °C to make saturated? +33 - × 100% = 1182.5π = 1.18π/ππΏ 20. a) this isn’t strictly an M1V1=M2V2 problem. First, find molar mass of Ca(OH)2, then multiply by desired concentration and then by desired volume. Notice how the units cancel except for mass. 74.1π .127πππ × 1πππ 1πΏ × .250πΏ 1 = 2.35π ππ π πππ’π‘π. You would add 2.35 g of solid Ca(OH)2 to about 200mL of water then mix to dissolve and then top it off to a final volume of 250mL. b) This is a standard dilution type problem, M1V1=M2V2 π£1 = (.127π)(.250πΏ) 1.00π = 3.18π₯10−3 πΏ ππ 31.8 ππΏ measure 31.8mL of 1.00M Ca(OH)2 add it to a flask until a volume of 250mL 21. a) same procedure as question 20 (a). b) M1V1=M2V2 π£1 = (.125π)(0.1πΏ) 3.00π (.2π)(0.1πΏ) 22. M1V1=M2V2 π£1 = 23. M1V1=M2V2 π2 = .487π = 1.05π ππ π πππ’π‘π = 4.2ππΏ, πππ π‘π π€ππ‘ππ πππ π πππππ π£πππ’ππ ππ 0.1πΏ (100ππΏ) = .0411 πΏ ππ 41.1ππΏ (.4π)(0.025πΏ) .075πΏ 84.01π .125πππ .1πΏ × 1πΏ × 1 1πππ = 0.133π πππ First determine moles of NH4NO3 πΏ .914πππ π= = 1.09π .835πΏ 24. π = 73.2 πππ»4 ππ3 × 1πππ = .914 πππππ 80.06 π 25. a) This isn’t a dilution problem, rather a problem to set up conversion factors. 0.7πππ ππ2 ππ4 × 1πΏ = 2.8πΏ πππ πππ’π‘πππ .250πππ b) 0.8 πΏ ππ2 ππ4 × .250πππ 142.05π × 1πππ = 1πΏ 28.4 π ππ2ππ4 c) 157π ππ2 ππ4 × 1πππ × 1πΏ = 142.05π .250πππ 4.42 πΏ ππ2ππ4
