1. Draw pictures on the atomic level for a solid, a liquid, and a gas.

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1 EXTRA HOMEWORK 3A 1. Draw pictures on the atomic level for a solid, a liquid, and a gas. 2. What must be true about the kinetic energy of the particles making up a liquid if the liquid is to turn into a gas? 3. What causes the high surface tension and high viscosity of molasses? 4. Draw pictures on the atomic level of a crystalline solid and an amorphous solid. 5. Define ionic bond. 6. Define metallic bond. 7. Define covalent bond. 8. Identify the particles that make up each of the following types of matter, and identify the interparticle attractive forces found in each: (a) Cr (b) CaBr 2 (c) SiO 2 9. Draw atomic level pictures of the three substances in question Given unlabeled samples of the three substances in question 8, how could they be identified experimentally? *11. Identify each substance as metallic, ionic, or macromolecular. Substance Melting Electrical Conduction Water Point ( C) As A Solid As A Liquid Solubility no yes high yes yes low no no low

2 EXTRA HOMEWORK 3B 1. Define each of the following: (a) London dispersion force (b) Dipole-dipole interaction 2. Identify the particles that make up each of the following types of matter, and identify the interparticle attractive forces found in each: (a) CH 3 OH (b) S 8 (c) H 2 (d) Cl 2 O (e) U (f) NaNO 3 (g) B (h) CH 3 CH 2 F 3. Draw atomic level pictures of the eight substances in question Given unlabeled samples of solid C 12 H 22 O 11 and C 14 H 10, how could they be identified experimentally? *5. Identify each substance as metallic, ionic, macromolecular, polar molecular, or nonpolar molecular. Substance Melting Electrical Conduction Water Hexane Point ( C) As A Solid As A Liquid Solubility Solubility 1 45 no no low high no no high low no no low low no yes high low yes yes low low EXTRA HOMEWORK 3C 1. Define electrical conduction. 2. Draw a band diagram for each of the following. (a) a metallic network (c) an n-type conductor (b) a nonmetallic network (d) a p-type conductor *3. Aluminum antimonide is a semiconductor and is used as a laser. When energy is applied to an aluminum antimonide crystal, electrons are promoted from the valence band to the conduction band. When the electrons return to the valence band, the crystal emits light with a wavelength of 730. nm. Calculate the energy of the band gap in aluminum antimonide.

3 EXTRA HOMEWORK 3D 1. Predict the ionic substance in each pair that has the greater melting point, and explain why: (a) MgCl 2 or AlCl 3 (b) K 2 S or K 2 O (c) AlP or CsI 2. Predict the metallic substance in each pair that has the greater melting point, and explain why: (a) Mg or Sr (b) Sc or V (c) Mn or Co 3. Assuming that silicon and germanium form macromolecular substances, tell which would have the greater melting point, and explain why. 4. Draw Lewis structures and bond-orbital- models for each of the following molecular substances, and predict whether they are polar or nonpolar. (a) C 2 H 2 (b) C 2 H 4 (c) CH 3 CN (d) CH 3 NH 2 5. Predict the molecular substance in each pair that has the greater melting point, and explain why: (a) P 4 or As 4 (b) CO 2 or SO 2 (c) CH 3 NH 2 or CH 3 CN (d) Li 2 O or Br 2 O 6. Explain the boiling points of the following three substances: C 3 H 8-42ºC C 4 H 10-1ºC C 3 H 7 OH 97ºC 7. For each set of substances, rank them from lowest to highest melting points, and explain why based upon the type of particles and interparticle attractive forces found in each. (a) NaF, LiF, HF (b) C 6 H 6, C 6 H 5 CO 2 H, C 10 H 8 (c) H 2 O, H 2 S, H 2 Se (d) Li, Be, C *8. Predict the substance in each pair that has the greater melting point, and explain why: (a) BCl 3 or AlCl 3

4 EXTRA HOMEWORK 3E 1. Explain how an increase in each of the following would affect an equilibrium between liquid water and water vapor in a closed container: (a) temperature (b) surface area 2. Calculate the amount of heat it would take to convert 50.0 g of water at 20.0 C to steam at C. The specific heat capacity for ice is 2.22 J/gC, the specific heat capacity for water is 4.18 J/gC, and the specific heat capacity for steam is 2.01 J/gC. The heat of fusion for ice is 335 J/g and the heat of vaporization for water is 2260 J/g. *3. Calculate the amount of heat it would take to convert 50.0 g of methanol (CH 3 OH) at 20.0 C to C. The specific heat capacity for solid methanol is 48.5 J/molC, the specific heat capacity for liquid methanol is 79.5 J/molC, and the specific heat capacity for gaseous methanol is 52.3 J/molC. The heat of fusion for methanol is 4.28 kj/mol and the heat of vaporization for methanol is 37.4 kj/mol. Methanol has a normal melting point of -97.0ºC and a normal boiling point of 64.7ºC EXTRA HOMEWORK 3F 1. Consider Phase Diagram 1 for iodine below. Answer each of the following questions: (a) What is the normal boiling point for iodine? (b) What is the melting point for iodine at 1 atm? (c) What phase is present at room temperature and normal atmospheric pressure? (d) What phase is present at 186ºC and 1.0 atm? Phase Diagram 1 Phase Diagram 2 2. Consider Phase Diagram 2 for sulfur above. The rhombic and monoclinic phases are two solid allotropic phases. Answer each of the following questions: (a) Below what pressure will solid sulfur sublime? (b) Which of the two solid phases of sulfur is most dense?

5 EXTRA HOMEWORK 3G 1. Indicate which solvent, toluene (C 7 H 8 ) or methanol (CH 3 OH), would be more suitable for dissolving each of the following: (a) CH 3 CH 2 CH 2 OH (b) LiNO 3 (c) CCl 4 (d) I 2 2. For each of the following pairs, indicate which substance would be more soluble in water: O O (a) CH 3 C H or CH 3 C OH (b) CH 4 or CH 3 F (c) vitamin A or vitamin C 3. Why does C 5 H 11 OH dissolve both polar and nonpolar substances? 4. Indicate each as a strong electrolyte, a weak electrolyte, or a nonelectrolyte: (a) HCl (b) KI (c) C 2 H 5 OH (d) HF 5. State whether each of the following is a strong electrolyte, weak electrolyte, or nonelectrolyte, and complete each equation. Showing how the solute would be expressed accurately in solution: (a) Fe(NO 3 ) 3 (s) (c) H 2 SO 4 (g) 6. Complete the following table: (b) CH 2 OHCH 2 OH (l) (d) H 3 PO 4 (g) Soluble in H 2 O (Yes or No) Ionizes or Dissociates in H 2 O (No, a little, a lot) C 3 H 6 O 3 C 3 H 8 LiOH NaI HCN HClO 3

6 EXTRA HOMEWORK 3H 1. Calculate the mass percent of solute in each of the following solutions. (a) 15.3 grams of sodium chloride dissolved in grams of water. (b) 19 grams of sulfur dissolved in 158 grams of benzene. 2. Calculate the mole fraction of solute in each of the following solutions. (a) 0.25 moles of sucrose dissolved in moles of water. (b) 27.0 grams of ethanol (C 2 H 5 OH) dissolved in grams of water. 3. Calculate the molarity of solute in each of the following solutions. (a) 2.36 moles of dextrose dissolved in enough water to make 2.50 liters of solution. (b) 18.5 g of sodium acetate trihydrate dissolved in enough water to make 750. milliliters of solution. 4. Calculate the molality of solute in each of the following solutions. (a) 2.35 moles of naphthalene dissolved in kilograms of carbon tetrachloride. (b) 17.2 g of ethylene glycol (C 2 H 6 O 2 ) dissolved in 500. grams of water. 5. Calculate the molarity of each ion present in each of the following solutions. (a) 1.30 M potassium carbonate (b) M aluminum sulfate 6. Calculate the number of grams of nickel (II) chloride hexahydrate needed to prepare ml of a M nickel (II) chloride solution. 7. A solution is prepared by dissolving 5.0 g of toluene (C 7 H 8 ) in 225 g of benzene (C 6 H 6 ). The density of the solution is g/ml. Fill out the unshaded regions in the table below, then calculate the following concentration units Mass (g) Quantity of Matter (mol) Volume (ml) Solute Solvent Solution (total) (a) mass percent of toluene (c) molality of toluene (b) molarity of toluene (d) mole fraction of toluene (continued on next page)

7 *8. A solution is prepared by dissolving 15.0 g of sodium nitrate in enough water to make the solution s volume ml. The solution s density was determined to be 1.12 g/ml. Calculate the (a) mass percent of sodium nitrate (c) molality of sodium nitrate (b) molarity of sodium nitrate (d) mole fraction of sodium nitrate *9. The term proof is defined as twice the percent by volume of pure ethanol (C 2 H 5 OH) in solution. Thus, a solution that is 95% ethanol by volume is 190 proof. What is the molarity of ethanol in a 151 proof ethanol/water solution? The density of pure ethanol is 0.79 g/ml and the density of pure water is 1.00 g/ml. *10. A solution is prepared by dissolving 516 mg of oxalic acid (H 2 C 2 O 4 ) to make ml of solution. A ml portion of this solution is diluted to ml. The density of the diluted solution is 1.07 g/ml. Fill out the table and then solve for the concentration units below. Mass (g) Quantity of Matter (mol) Volume (ml) Solute Solvent Solution (total) (a) What is the molarity of the oxalic acid in the diluted solution? (b) What is the mass percent of the oxalic acid in the diluted solution? (c) What is the mole fraction of the oxalic acid in the diluted solution? (d) What is the molality of the oxalic acid in the diluted solution? (e) Suppose you wanted to make another batch of the oxalic acid solution with the same molality as the diluted solution above, but you only had 45.0 mg of oxalic acid. What mass of water would you have to use?

8 EXTRA HOMEWORK 3I 1. Calculate the vapor pressure of each of the following solutions at 24 C if the vapor pressure of pure water is 22.4 torr at 24 C. (a) 0.45 moles of fructose (C 6 H 12 O 6 ), a nonelectrolyte, dissolved in moles of water. (b) 0.45 moles of aluminum chloride dissolved in moles of water. 2. Calculate the vapor pressure of each of the following solutions at 22 C if the vapor pressure of pure water is 19.8 torr at 22 C. (a) 21.5 grams of fructose (C 6 H 12 O 6 ), a nonelectrolyte, dissolved in 175 g water. (b) 21.5 grams of aluminum chloride dissolved in 175 g water. 3. Heptane (C 7 H 16 ) and octane (C 8 H 18 ) form an ideal solution. At 25 C, the vapor pressures of heptane and octane are 45.8 torr and 10.9 torr, respectively. A solution is made by mixing 50.0 ml of heptane (density = g/ml) with 50.0 ml of octane (density = g/ml). (a) Calculate the mole fractions of heptane and octane in the solution (b) Calculate the equilibrium vapor pressure of the solution 4. Consider the following four liquids and/or solutions. pure water a 0.1 m solution of sucrose in water a 0.1 m solution of potassium nitrate in water a 0.1 m solution of calcium nitrate in water Place the liquids and/or solutions in order of lowest vapor pressure to highest vapor pressure *5. A solution is made by dissolving 28.5 grams of glycerin (C 3 H 8 O 3 ), a nonelectrolyte, in 125 ml methanol (CH 3 OH). Calculate the vapor pressure of the solution at 21.2 C if the vapor pressure of pure methanol is 100. torr at 21.2 C, and methanol s density is g/ml. *6. Calculate the vapor pressure at 25ºC of an aqueous solution that is 5.50% sodium chloride by mass. The vapor pressure of pure water is 23.8 torr at 25 C. * grams of a nonelectrolyte is dissolved in 75.0 grams of water at 22 C, and the vapor pressure of the solution is 18.6 torr. If the vapor pressure of pure water at 22 C is 19.8 torr, calculate the molar mass of the nonelectrolyte. *8. Anthracene is 94.3% carbon and 5.7% hydrogen by mass. When 8.25 grams of anthracene is dissolved in grams of hexane (C 6 H 12 ) at 25 C, and the vapor pressure of the solution is 137 torr. If the vapor pressure of pure hexane at 25 C is 151 torr, determine the molar mass and the molecular formula of anthracene.

9 EXTRA HOMEWORK 3J 1. Consider the following four liquids and/or solutions. pure water a 0.3 m solution of glucose in water a 0.2 m solution of lithium bromide in water a 0.1 m solution of magnesium bromide in water Place the liquids and/or solutions in order of (a) lowest freezing point to highest freezing point (b) lowest boiling point to highest boiling point (c) lowest osmotic pressure to highest osmotic pressure 2. A solution is made by dissolving 0.75 moles of lithium bromide in 0.30 kilograms of water. Calculate (a) the freezing point of the solution (b) the boiling point of the solution. 3. A solution is made by dissolving 21.5 grams of copper (II) chloride in 425 grams water. Calculate (a) the freezing point of the solution (b) the boiling point of the solution. 4. An organic solid is found to be 54.5% carbon, 9.2% hydrogen, and 36.3% oxygen by mass. A solution containing grams of the organic solid dissolved in 115 grams of carbon disulfide has a boiling point of 50.2 C. Determine the molecular formula of organic solid. 5. A 1.00 g sample of an organic compound is dissolved in 8.50 g of benzene, and the freezing point of the solution is 0.3 C. If the compound is known to be 62.0% carbon, 10.4% hydrogen, and 27.6% oxygen by mass, calculate the molecular formula of the compound. *6. A g sample of an organic compound is dissolved in 25.0 g water, and the freezing point of the solution is C. (a) If the compound is known to be 53.31% carbon, 11.18% hydrogen, and 35.51% oxygen by mass, calculate the molecular formula of the compound. (b) Draw a Lewis structure for the compound with this formula that is capable of forming hydrogen bonds (c) Draw a Lewis structure compound with this formula that is not capable of forming hydrogen bonds *7. Cyclohexane freezes at 6.5ºC, and a solution of 3.57 grams of styrene (C 8 H 8 ) dissolved in grams of cyclohexane freezes at -18.4ºC. A second solution is prepared by dissolving 4.23 grams of phenylacetylene in grams of cyclohexane, and this solution freezes at -25.1ºC. Phenylacetylene is 94.1% carbon and 5.9% hydrogen by mass. Determine the (a) molal freezing point constant for cyclohexane (b) molecular formula of phenylacetylene *8. A aqueous 2.00 m hydrofluoric acid solution has a normal boiling point of C. Find the percent ionization of hydrofluoric acid in the solution.

10 EXTRA HOMEWORK 3K 1. A solution is prepared by dissolving 8.92 grams of potassium bromide in enough water to make the total volume ml. Calculate the osmotic pressure of this solution at 27ºC. 2. A 5.87 milligram sample of a protein is dissolved in water at 25 C to make 10.0 ml of solution. The osmotic pressure of the solution is 2.45 torr. Calculate the molar mass of the protein. 3. Before refrigeration was common, many fruits were preserved by mixing them with large amounts of sugar. Explain how sugar acts as preservatives. *4. Explain how the Tyndall effect can be used to distinguish between a colloidal suspension and a true solution. EXTRA 3A ANSWERS 1. solid liquid gas 2. The kinetic energy of the particles must be greater than the attractive forces between the particles 3. Strong attractive forces between the particles that make up molasses (the sugar molecules and the water molecules) 4. crystalline amorphous 5. The electrostatic attraction between positive and negative ions 6. The attraction between the nuclei of metal atoms and the valence electrons in delocalized molecular orbitals (continued on next page)

11 7. The attraction between the nuclei of nonmetal atoms and the valence electrons in molecular orbitals 8. (a) atoms; metallic bonds (b) positive and negative ions; ionic bonds (b) atoms; covalent bonds 9. (a) (b) (c) 10. Cr is the only conductor as a solid because it is metallic, CaBr 2 is the only water soluble substance because it is ionic, and the other will be SiO 2 which is macromolecular *11. 1 ionic, 2 metallic, 3 macromolecular EXTRA 3B ANSWERS 1. (a) the attractive force that exists between molecules, caused by the polarizing of electron clouds in neighboring molecules, producing temporary dipole moments (b) the attractive force that exists between polar molecules, caused by the attraction of the permanent positive and negative ends of neighboring molecules 2. (a) molecules; hydrogen bonds (b) molecules; London dispersion forces (c) molecules; London dispersion forces (e) atoms; metallic bonds (g) atoms; covalent bonds (d) molecules dipole-dipole interactions (f) positive and negative ions; ionic bonds (h) molecules; dipole-dipole interactions 3. (a) (b) (c) (d) (e) (f) (g) (h) (continued on next page)

12 4. Solid C 12 H 22 O 11 will dissolve in water because it is a polar molecular substance with hydrogen bonding, C 14 H 10 will not because it is a nonpolar molecular substance *5. 1 nonpolar molecular, 2 polar molecular, 3 macromolecular, 4 ionic, 5 metallic EXTRA 3C ANSWERS 1. The movement of charged particles in one direction. 2. (a) (b) (c) (d) * J EXTRA 3D ANSWERS 1. (a) AlCl 3 - Both are ionic substances with ionic bonds between the ions. The aluminum ion has a higher charge than the magnesium ion, so it attracts more strongly to the chloride ions. (b) K 2 O - Both are ionic substances with ionic bonds between the ions. The oxide ion is smaller than the sulfide ion, so it can get closer to the potassium ion, attracting more strongly. (c) AlP - Both are ionic substances with ionic bonds between the ions. The aluminum and phosphide ions have higher charges than the cesium and iodide ions, so they attract more strongly. And, the aluminum and phosphide ions are smaller than the cesium and iodide ions, so they can get closer together, attracting more strongly. 2. (a) Mg Both are metallic substances with metallic bonds between the atoms. They both have the same number of bonding electrons per atom (2), but Mg atoms are smaller, so the bonding electrons are in smaller molecular orbitals, which attract to the bonding nuclei more strongly (b) V - Both are metallic substances with metallic bonds between the atoms. V has more net bonding electrons per atom than Sc (5 to 3), which attract to the bonding nuclei more strongly making the metallic bonds stronger. (c) Mn - Both are metallic substances with metallic bonds between the atoms. Mn has more net bonding electrons per atom than Co (Mn has 6 bonding electrons 1 antibonding electron = 5 net bonding electrons per atom, Co has 6 bonding electrons 3 antibonding electrons = 3 net bonding electrons per atom), which attract to the bonding nuclei more strongly making the metallic bonds stronger. (continued on next page)

13 3. Si If both are macromolecular substances, their attractive forces would be covalent bonds. Si atoms are smaller, so the bonding electrons are in smaller molecular orbitals, which attract to the bonding nuclei more strongly 4. (a) (b) (c) (d) nonpolar nonpolar polar polar 5. (a) As 4 Both are nonpolar molecular substances, with only London dispersion forces between the molecules. As 4 has more electrons, so it is more polarizing, making its London dispersion forces stronger. (b) SO 2 SO 2 is a polar molecular substance, with London dispersion forces and dipole-dipole interactions between the molecules. CO 2 is a nonpolar molecular substance, having only London dispersion forces between the molecules. (c) CH 3 NH 2 CH 3 NH 2 is a polar molecular substance, with London dispersion forces, dipole-dipole interactions, and hydrogen bonding between the molecules. CH 3 CN is a polar molecular substance, but only has London dispersion forces and dipole-dipole interactions between the molecules. (d) Li 2 O Li 2 O is an ionic substance with strong ionic bonds between the ions. Br 2 O is a polar molecule substance, with only London dispersion forces and dipole-dipole interactions between the molecules. 6. Both C 3 H 8 and C 4 H 10 are nonpolar molecular, with only London dispersion forces. C 4 H 10 has more electrons, so it is more polarizing, making its London dispersion forces stronger. C 3 H 7 OH is polar molecular, with London dispersion forces, dipole-dipole interactions, and hydrogen bonding. (continued on next page)

14 7. (a) HF < NaF < LiF HF is a polar molecular substances, with weak London dispersion forces, dipole-dipole interactions, and hudrogen bonds between the molecules. Both NaF and LiF are ionic substances with ionic bonds between the ions. The lithium ion is smaller than the sodium ion, so it can get closer to the fluoride ion, attracting more strongly. (b) C 6 H 6 < C 10 H 8 < C 6 H 5 CO 2 H C 6 H 6 and C 10 H 8 Both are nonpolar molecular substances, with only London dispersion forces between the molecules, but C 10 H 8 has more electrons, so it is more polarizing, making its London dispersion forces stronger. C 6 H 5 CO 2 H is a polar molecular substance, with London dispersion forces, dipole-dipole interactions, and hydrogen bonding between the molecules. (c) H 2 S < H 2 Se < H 2 O H 3 S amd H 2 Se are polar molecular substances, with London dispersion forces and dipole-dipole interactions between the molecules, but H 2 Se has more electrons, so it is more polarizing, making its London dispersion forces stronger. H 2 O is a polar molecular substance, with London dispersion forces, dipole-dipole interactions, and hydrogen bonding between the molecules. (d) Li < Be < C Both Li and Be are metallic substances with metallic bonds between the atoms, but Be has more net bonding electrons per atom than Li (2 to 1), which attract to the bonding nuclei more strongly making the metallic bonds stronger. C is a macromolecular substance, and because of the low number of net bonding electrons per atom for the two metals, the covalent bonds in C are stronger. *8. AlCl 3 > BCl 3 AlCl 3 is an ionic network with strong ionic bonds, BCl 3 is nonpolar molecular, with weak London dispersion forces EXTRA 3E ANSWERS 1. (a) An increase in temperature would increase the rate of evaporation. As more water vapor accumulates in the closed container, the rate of condensation would increase. Eventually, the rate of condensation will equal the rate of evaporation, and equilibrium is reestablished. This new equilibrium will now have a greater amount of water vapor, so its equilibrium vapor pressure would now be higher. (b) An increase in surface area would increase the rate of evaporation and condensation equally, so there would be no change in the equilibrium ,000 J *3. 69,200 J

15 EXTRA 3F ANSWERS 1. (a) 184.4ºC (b) 113.6ºC (c) solid (d) gas 2. (a) mm Hg (b) rhombic EXTRA 3G ANSWERS 1. (a) methanol (b) methanol (c) toluene (d) toluene 2. (a) CH 3 COOH both oxygens and the hydrogen bonded to the oxygen are capable of hydrogen bonding to water molecules. CH 3 CHO only has one oxygen capable of hydrogen bonding to water molecules. (b) CH 3 F the molecule is polar, and the fluorine is capable of hydrogen bonding to water molecules. CF 4 is a nonpolar molecule. (c) Vitamin C more OH groups to hydrogen bond to water molecules 3. The molecule has a nonpolar end (C 5 H 11 -) that attracts nonpolar solvents and a polar end (-OH) that attracts polar solvents 4. (a) strong electrolyte (b) strong electrolyte (c) nonelectrolyte (d) weak electrolyte 5. (a) strong electrolyte Fe(NO 3 ) 3 (s) Fe 3+ (aq) + 3NO 3 - (aq) (b) nonelectrolyte (c) strong electrolyte (d) weak electrolyte CH 2 OHCH 2 OH (l) CH 2 OHCH 2 OH (aq) H 2 SO 4 (g) 2H + (aq) + SO 4 2- (aq) H 3 PO 4 (g) H 3 PO 4 (aq) 6. C 3 H 6 O 3 Yes No C 3 H 8 No - LiOH Yes A Lot NaI Yes A Lot HCN Yes A Little HClO 3 Yes A Lot

16 EXTRA 3H ANSWERS 1. (a) 8.98% (b) 11% 2. (a) (b) (a) M (b) M 4. (a) 3.13 m (b) m 5. (a) 2.60 M K +, 1.30 M CO 3 2- (b) M Al 3+, M SO g 7. (a) 2.2% toluene (b) 0.21 M toluene (c) 0.24 m toluene (d) toluene *8. (a) 13.4% NaNO 3 (b) 1.76 M NaNO 3 (c) 1.82 m NaNO 3 (d) NaNO 3 * M C 2 H 5 OH *10. (a) M H 2 C 2 O 4 (b) H 2 C 2 O 4 (c) H 2 C 2 O 4 (d) m H 2 C 2 O 4 (e) 233 g H 2 O EXTRA 3I ANSWERS 1. (a) 21.9 torr (b) 20.6 torr 2. (a) 19.6 torr (b) 18.6 torr 3. (a) heptane, octane (b) 29.3 torr 4. (a) 0.1 m calcium nitrate, 0.1 m potassium nitrate, 0.1 m sucrose, pure water * torr * torr *7. 47 g/mol *8. C 14 H 10

17 EXTRA 3J ANSWERS 1. (a) 0.2 m lithium bromide, 0.3 m glucose and 0.1 m magnesium bromide (tie), pure water (b) pure water, 0.3 m glucose and 0.1 m magnesium bromide (tie), 0.2 m lithium bromide (c) pure water, 0.3 m glucose and 0.1 m magnesium bromide (tie), 0.2 m lithium bromide 2. (a) -9.3ºC (b) 102.6ºC 3. (a) -2.10ºC (b) ºC 4. C 4 H 8 O 2 5. C 6 H 12 O 2 *6. (a) C 4 H 10 O 2 (b) (c) *7. (a) 20.1 C /m (b) C 8 H 6 *8. 10% (7% without rounding to the correct number of significasnt figures) EXTRA 3K ANSWERS atm 2. 4,450 g/mol 3. The large amount of sugar dissolves in water that is in contact with fruit, making a concentrated sugar solution. Bacteria cells come in contact with this concentrated sugar solution, and through the cell membranes, water flows out of the bacteria cells into the sugar solution faster than water flows into the bacteria cells from the sugar solution. This dehydrates the bacteria and kills them. *4. A beam of light passing through a colloidal suspension will be scattered, but the same beam of light passing through a true solution will not be scattered.

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