Chapter 12 Exam Practice Problems (1) In which solvent would chromium acetate be most soluble? (a) ethanol, CH3CH2OH (b) methanol, CH3OH (c) benzene, C6H6 (d) hexane, C6H14 Chromium acetate is a polar solute and would therefore dissolve best in a polar solvent. (2) In which solvent would vegetable oil be most soluble? (a) ethanol, CH3CH2OH Vegetable oil is a non-polar solute (b) methanol, CH3OH and would therefore dissolve best (c) benzene, C6H6 in a non-polar solvent. (d) water, H2O (3) A solution of potassium hydroxide is made by dissolving 2.40 g of KOH in 8.72 g of water. The percent, by mass, of KOH in the solution is: % πππ π = πππ π πΎππ» 2.40 π × 100% = × 100% = ππ. π% π‘ππ‘ππ πππ π 2.40 π + 8.72 π (4) A glucose solution is prepared by dissolving 3.40 g of glucose, C6H12O6, in 115.5 g of water. What is the molality of the glucose solution? First calculate the number of moles of glucose that are present: 1 ππππ πππ’πππ π 3.40 π πππ’πππ π × = 0.01887 πππ πππ’πππ π 180.2 π πππ’πππ π Calculate the kg of solvent: 1 ππ π€ππ‘ππ 115.5 π π€ππ‘ππ × = 0.1155 ππ π€ππ‘ππ 1000 π π€ππ‘ππ Use these numbers to calculate the molality: πππ π πππ’π‘π 0.01887 πππ πππ’πππ π πππππππ‘π¦ = = = π. πππ π ππ π πππ£πππ‘ 0.1155 ππ π€ππ‘ππ (5) Would you expect the heat of solution for CaCl2 to be positive or negative? Explain your answer in a few sentences. The heat of solution should be positive (i.e. dissolution would be an endothermic reaction). You have to break very strong (ionic) bonds and are forming considerably weaker iondipole forces. (6) The vapor pressure of pure water is 92.5 mm Hg at 50 oC. If 32.78 g of glucose (C6H12O6) is stirred into 1.32 kg of water at 50 oC, what will the vapor pressure of the solution be? Step one is to calculate the mole fraction of the solvent (water). In order to do this we first calculate the number of moles of each component: 1 ππππ πππ’πππ π 32.78 π πππ’πππ π × = 0.1819 πππ πππ’πππ π 180.2 π πππ’πππ π 1.32 ππ π€ππ‘ππ × 1000 π π€ππ‘ππ 1 ππππ π€ππ‘ππ × = 73.25 πππ π€ππ‘ππ 1 ππ π€ππ‘ππ 18.02 π πππ’πππ π Now we can calculate the mole percent: πππ % π€ππ‘ππ 73.25 πππ ππ€ππ‘ππ = = = 0.9975 π‘ππ‘ππ πππ 73.25 πππ + 0.1819 πππ Finally, calculate the vapor pressure of the solution using Raoult’s Law: 0 0 ππ πππ = ππ πππ£πππ‘ ππ πππ£πππ‘ = (0.9975)(92.5 ππ π»π) = 92.3 ππ π»π (7) The density of a 6.10 M aqueous solution of NaOH is 1.22 g/mL. What is the molality of this solution? Remember that molality is defined as the number of moles of solute divided by the mass of the solvent in kg. Begin by assuming that you have 100.0 mL of solution. Calculate the number of moles of NaOH: (0.100 L) (6.10 M) = 0.610 moles NaOH This is the number of moles of solute. Now we need to calculate the mass of the solvent. We can calculate the total mass of the solution using the density: (100.0 mL)(1.22 g/mL) = 122 g Some of this 122 g that we’ve calculated is solute (NaOH), and some of it is solvent (water). We can calculate the mass of NaOH by multiplying the number of moles by the molar mass: π 0.610 πππππ ππππ» × 40.0 = 24.4 π ππππ» πππ The rest of the solution (122 g -24.4 g = 97.6 g= 0.976 kg) must be water. Now we have enough information to calculate the molality. 0.610 πππ πππππππ‘π¦ = = 6.25 π 0.976 ππ (8) At 20.0 oC, the vapor pressure of acetone (CH3COCH3) is 181.7 mm Hg and the vapor pressure of water is 17.5 mm Hg. A solution is made by mixing 128.4 g water and 97.60 g acetone. What is the vapor pressure of the solution? Like problem #6, this is a Raoult’s Law problem. It differs from #6 in that both of the components of the system have a vapor pressure; we need to calculate the vapor pressure for each of the components. Start by calculating the number of moles of each: 128.4 π π»2 π × 1 πππ π»2 π = 7.129 πππ π»2 π 18.02 π π»2 π 97.60 π ππππ‘πππ × 1 πππ ππππ‘πππ = 1.680 πππ ππππ‘πππ 58.09 π ππππ‘πππ Now we can calculate the mole fraction of each component: ππ€ππ‘ππ = πππ % π€ππ‘ππ 7.129 πππ = = 0.8093 π‘ππ‘ππ πππ 7.129 πππ + 1.680 πππ πππππ‘πππ = πππ % ππππ‘πππ 1.680 πππ = = 0.1907 π‘ππ‘ππ πππ 7.129 πππ + 1.680 πππ The vapor pressure of the solution is equal to the sum of the partial pressures of each component: 0 0 ππ πππ = ππ€ππ‘ππ + πππππ‘πππ = ππ€ππ‘ππ ππ€ππ‘ππ + πππππ‘πππ πππππ‘πππ ππ πππ = (0.8093)(17.5 ππ π»π) + (0.1907)(181.7 ππ π»π) = 14.16 ππ π»π + 34.65 ππ π»π = ππ. ππ ππ π―π (9) A 5% w/w aqueous solution of ammonia (NH3) has a density of 0.9597 g/mL. For this solution, calculate: (a) molarity of ammonia (b) molality of ammonia (c) mole fraction of ammonia Assume that you have have 100.0 g of solution. By definition, for a 5% w/w solution of ammonia, 5% of the 100.0 g (= 5.000 g) is ammonia, and the other 95.00 g must be water. a. molarity is defined as moles of solute divided by volume of solution. Moles of ammonia can be easily calculated using the mass of ammonia: 1 πππ ππ»3 5.000 π ππ»3 × = 0.293 πππ ππ»3 17.04 π ππ»3 To calculate the volume of solution, use the density of the solution in conjunction with the mass: 1 ππΏ π πππ’π‘πππ 100.0 π π πππ’π‘πππ × = 104.2 ππΏ = 0.1042 πΏ 0.9597 π π πππ’π‘πππ Now calculate the molarity: 0.293 πππ πππππππ‘π¦ = = π. ππ π΄ 0.1042 πΏ b. molality is defined as moles of solute divided by mass of solvent At the beginning, we deduced that 95.00 g of the solution is water (the solvent). This is 0.095 kg. πππππππ‘π¦ = 0.293 πππ = π. πππ π 0.095 ππ c. To calculate the mole fraction you first need to know the total number of moles. The number of moles of ammonia was calculated earlier so you just need to calculate the number of moles of water: 95.00 π π€ππ‘ππ × 1 πππ π€ππ‘ππ = 5.271 πππ π€ππ‘ππ 18.02 π π€ππ‘ππ Now you can calculate the mole fraction of ammonia: πππ»3 0.293 πππ»3 = = = π. ππππ πππ»3 + ππ»2π 0.293 + 5.271 (10) When 124 g of a sugar is added to 1.10 kg of water, the freezing point is lowered by 2.4 oC. Kf for water is 1.86 oC/m. What is the approximate molecular weight of the compound? You can use the freezing point change to calculate the molality of the solution. (Since this is a sugar it will not dissociate in solution, so i=1.) βππ = ππΎπ π so βπ 2.4 β π= = = 1.29 π ππΎπ (1) (1.86 β ) π Remember that molality is defined as moles of solute divided by mass of solvent. Given the molality and the mass of solvent), the number of moles of solute can be calculated: (molality)(mass solvent) = (1.29 m)(1.10 kg) = 1.42 moles compound Use the number of moles, in conjunction with the given mass of sugar, to calculate the molar mass: 124 g/1.42 moles = 87.3 g/mol (11) The Henry’s law constant for hydrogen gas (H2) is 7.8 x 10-4 M/atm at 25 oC. A sealed container of water has a partial pressure of 2.0 atm H2 in the headspace above water. What is the concentration of H2 in the water? Henry’s Law states that the concentration of a gas in solution is the Henry’s law constant times the pressure above the solvent: πΆπ»2 = ππ»π»2 ππ»2 = (7.8 × 10−4 π ) (2.0 ππ‘π) = π. π × ππ−π π΄ ππ‘π (12) The density of apple juice is 1.04 g/mL. A 16-oz. (480 mL) bottle contains 52 g of sucrose, C11H22O11 . Calculate the percentage by mass of sugar of apple juice. By definition, the mass percent of a substance is the mass of that substance divided by the total mass of the solution. Here, the mass of sucrose is given, but the total mass of the solution needs to be calculated by using the density: msoln =(1.04 g/mL)(480 mL) = 499.2 g ππ π’ππππ π 52 π πππ π % = × 100% = × 100% = ππ. π% πππππππ ππ πππ 499.2 π (13) Cyclohexane has a normal boiling point of 80.74oC and a Kb value of 2.79 oC/m. If 2.60 g of glucose (C6H12O6) is added to 284 g of cyclohexane, what would the new boiling point be? Start by calculating the number of moles of glucose added to the solution: 2.60 π πππ’πππ π × 1 πππ πππ’πππ π = 0.0144 πππ πππ’πππ π 180.1 ππππ’πππ π Now calculate the molality of the solution; remember that molality is defined as moles of solute divided by mass of solvent: πππππππ‘π¦ = 0.0144 πππ = 0.0507 π 0.284 ππ We can use this to calculate the change in boiling point: βππ = ππΎπ π = (1) (2.79 β ) (0.0507) = 0.141 β π The new boiling point would be: 80.74 oC + 0.141 oC = 80.88 oC (14) Helium has a solubility of 5.55 x 10-4 M at 1.50 atm. What would its solubility be at 9.76 atm? πΆ1 πΆ2 = π1 π2 5.55 × 10−4 π πΆ2 = 1.50 ππ‘π 9.76 ππ‘π C2 = 3.61 x 10-3 M (15) What is the molarity of an 0.80 m solution of CH3COOH? The density of the solution is 1.0055 g/mL. Remember that molality is defined as moles of solute divided by mass of solvent; thus, an 0.80 m solution is 0.80 moles in 1.0 kg of solvent. Molarity is defined as moles of solute divided by volume of solution. You don’t know the volume of the solution but since you are given the density, you can calculate the volume using the mass. The mass of the solution would be the mass of the solvent (here, 1.0 kg or 1.0 x 103 g) plus the mass of the solute, which you need to calculate: 0.80 πππ πΆπ»3 πΆπππ» × 60.06 ππΆπ»3 πΆπππ» = 48.048 π πΆπ»3 πΆπππ» 1 πππ πΆπ»3 πΆπππ» The total mass of the solution will be: mtot= msolvent+ msolute=1000g + 48.048 g = 1048.048 g You can use this to calculate the volume of the solution: 1 ππΏ 1πΏ π = 1048.048 π ( )( ) = 1.042 πΏ 1.0055 π 1000 ππΏ Use this to calculate the molarity: π= πππππ ππ π πππ’π‘π 0.80 πππ = = π. ππ π΄ πΏ ππ π πππ’π‘πππ 1.042 πΏ