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(es24394) sung Homework 2 vanden bout (51640) This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. 001 10.0 points Both ammonia and phosphine (PH3 ) are soluble in water. Which is least soluble and why? 1. ammonia; it is too small to be hydrated by water molecules. 2. ammonia; it does not form hydrogen bonds with water molecules. 3. phosphine; the PH bonds are so strong that they cannot break to enable phosphine to hydrogen-bond with water. 4. phosphine; it does not form hydrogen bonds with water molecules. correct 5. ammonia; the NH bonds are so strong that they cannot break to enable the ammonia to hydrogen-bond with water. Explanation: PH3 is least soluble because it cannot form hydrogen bonds with water. 002 10.0 points Some distilled water is added to an empty beaker. A gram of copper(II) nitrate is added to the beaker with stirring. After a few minutes, what is in the beaker? 1. nitrogen gas, copper atoms, electrons, and water 2. a sticky material of copper hydronitrite 3. solid copper(II) nitrate and water 4. solid copper, nitrate ions, and water 5. copper ions, nitrate ions, and water correct Explanation: 1 All nitrates are soluble in water, so the copper(II) nitrate dissolves. Soluble ionic compounds are strong electrolytes, so the copper(II) nitrate also dissociates into its cation and anion parts: Cu+ and NO . 2 3 003 10.0 points Which is the best representation of the solvation of a sodium cation in water? oxygen hydrogen Na+ ion 1. 2. sung (es24394) Homework 2 vanden bout (51640) 2 (chloroethane), H2 O (water), C2 H5 F (uoroethane), C9 H20 (nonane). 1. C2 H5 Cl > C2 H5 F > H2 O > C9 H20 2. C2 H5 F > H2 O > C9 H20 > C2 H5 Cl 3. 3. H2 O > C9 H20 > C2 H5 Cl > C2 H5 F 4. C9 H20 > C2 H5 Cl > C2 H5 F > H2 O correct Explanation: Octane is a non-polar solvent and thus ranking in terms of decreasing miscibility requires ranking in terms of decreasing nonpolar character, i.e. from least to most polar. 006 10.0 points Octane (C8 H18 ) is a major component of gasoline. Which of the following would you expect to be the LEAST soluble in gasoline? 1. candle wax 2. motor oil 3. table salt correct correct Explanation: Since the water molecule has a partial negative charge on the O atom, it will orient itself to the O atom closest to the positively charged cation. 004 10.0 points When we put sugar in water it melts. 1. False correct 2. True Explanation: Sugar dissolves in water and melts when its temperature is raised above its melting point. 005 10.0 points Rank the following in terms of decreasing miscibility in C8 H18 (octane): C2 H5 Cl 4. table sugar 5. diesel fuel Explanation: Octane is a non-polar compound so it would be less likely to dissolve polar or ionic substances. 007 10.0 points Calculate the number of moles of oxygen that will dissolve in 45 L of water at 20 C if the partial pressure of oxygen is 0.21 atm. Henrys mol . constant for oxygen is 0.0013 L atm 1. 0.28 mol 2. 0.0062 mol 3. 0.00027 mol 4. sung (es24394) Homework 2 vanden bout (51640) 4. 0.0013 mol Correct answer: 4.2344 mol. 5. 0.012 mol correct Explanation: 008 10.0 points A decrease in temperature usually (increases, decreases, does not change) the solubility of common salts in water. 1. does not change 2. increases 3. decreases correct Explanation: Most common salts dissolve endothermically in water and therefore their solubility will increase with an increase in temperature. 009 10.0 points Which of the following is a restatement of Henrys Law? 1. As the partial pressure of a gas above a liquid increases, the gass solubility in that liquid also increases. correct 2. A mixture of two liquids will have a vapor pressure that isnt equal to either of the pure liquids vapor pressures. 3. Adding a non-volatile solute to a liquid will lower its vapor pressure. 4. Heating a solution will decrease the solubility of gases dissolved in it. Explanation: Henrys law, formulated by William Henry is 1803 states: At a constant temperature, the amount of a given gas dissolved in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid. 010 (part 1 of 2) 10.0 points How many moles of ions are contained in 1.58 L of a 1.34 M solution of KCl? Explanation: solute = KCl HO KCl(s) K+ (aq) + Cl (aq) 2 3 n=? Ions produced = 1 mol K+ + 1 mol Cl = 2 mol 011 (part 2 of 2) 10.0 points How many moles of ions are contained in 1.58 L of a 1.34 M solution of Mg(NO3 )2 ? Correct answer: 6.3516 mol. Explanation: solute = Mg(NO3 )2 n=? HO Mg(NO3 )2 (s) Mg2+ (aq) + 2NO (aq) 2 3 Ions produced = 1 mol Mg2+ + 2 mol NO 3 = 3 mol 012 10.0 points Raising the temperature of a sample of water will typically (decrease/increase) the solubility of (all/some) dissolved gases. 1. increase, some 2. increase, all 3. decrease, some 4. decrease, all correct Explanation: Because the dissolution of most all gases is an exothermic process, Le Chateliers principle suggests that all gases will become less soluble as the temperature of the solvent is raised. sung (es24394) Homework 2 vanden bout (51640) 4. 510 torr 013 10.0 points Theoretically, it would be harder to dissolve (NaCl/Al2S3 ) in water because the (higher/lower) the charge density, the lower the solubility. 1. Al2 S3 , higher correct 2. NaCl, higher 3. Al2 S3 , lower 4. NaCl, lower Explanation: Al2 S3 has a higher charge density (Al3+ , 2 S ) than NaCl (Na+ , Cl ). High charge density corresponds to high lattice energy and thus much more endothermic (i.e. less spontaneous) dissolution. 014 10.0 points To increase the solubility of a gas in water, you should 1. increase the surface area of the particles. 2. increase the pressure of the gas. correct 3. decrease the pressure of the gas. 4. increase the temperature of the water. Explanation: 015 10.0 points If acetic acid has a pure vapor pressure of 20 torr at 30 C and acetaldehye has a pure vapor pressure of 1000 torr at 30 C, a mixture of 1 moles of acetic acid and 4 moles of acetaldehyde would have what total vapor pressure? 1. 816 torr 1. III only 2. 804 torr correct 2. II only correct 3. 216 torr 3. I and III 5. 204 torr Explanation: Ptotal = a Pa + b Pb + ... Ptotal = 0.2 20 torr + 0.8 1000 torr Ptotal = 804 torr 4 016 10.0 points What is the vapor pressure the of solvent in an aqueous solution at 25 C of 0.05 m urea (CO(NH2 )2 ), a nonelectrolyte? The vapor pressure of water at 25 C is 23.76 Torr. Correct answer: 23.7386 Torr. Explanation: Pvp = 23.76 Torr T = 25 C MW = 18.02 g/mol A 0.05 m solution will contain 0.05 mol urea per 1000 g H2 O, so the number of moles of H2 O is 1000 g nH 2 O = = 55.4939 mol 18.02 g/mol xH 2 O = nH 2 O nH2 O + nurea 55.4939 mol = 55.4939 mol + 0.05 mol = 0.9991 . P = xsolvent Ppure solvent = 0.9991 (23.76 Torr) = 23.7386 Torr . 017 10.0 points Which of the following would raise the vapor pressure of a sample of aniline in a closed container? I) increasing the size of the container II) increasing the temperature of the sample III) increasing the applied pressure sung (es24394) Homework 2 vanden bout (51640) 1. 215 grams 4. II and III 2. 3.9 grams 5. I, II and III 3. 8.0 grams 6. I only 4. 4.0 grams 7. I and II Explanation: Because evaporation is always an endothermic process, a substances vapor pressure is always directly proportional to its temperature. Adding a non-volatile solute to any substance will decrease its vapor pressure. Factors such as sample size and container size do not inuence he vapor pressure. 018 10.0 points What will be the boiling point of a solution of 8 moles of sodium dichromate (Na2 Cr2 O7 ) dissolved in 8 kg of water? Use the following values: Kb = 0.512 K m1 Kf = 1.86 K m1 1. 380.1 K 4. 105.6 C correct 2. 378.6 K 5. 95.9 C 3. 373.8 K 6. 101.5 C 4. 377.6 K 5. 374.5 K correct Explanation: 8 moles me = 3 = 3 mol kg1 8 kg Tb = Kb me boiling point = 374.5 K 019 10.0 points Your friend decides to decrease cooking time for spaghetti noodles by adding table salt to 2.0 L water. How much table salt (NaCl) must be added to the water in order to raise the boiling point by 4.0 C? (Kf = 1.86 C/m, Kb = 0.512C/m) 5. 460 grams correct 5 Explanation: There are 15.625 moles of particles in the solution, which corresponds to 7.8125 moles of NaCl, or 456.6 grams NaCl. 020 10.0 points Over what temperature range would a 4 m solutiond of NaCl remain a liquid? (for water kb = 0.512 and kf = 0.186) 1. 104.1 C 2. 94.4 C 3. 98.5 C Explanation: Pure water will remain a liquid between 0 C and 100 C. A 4 m NaCl solution would have a freezing point depression of 2 4 0.186 = 1.488 C, and a new freezing point of 1.488 C. It would have a boiling point elevation of 2 4 0.512 = 4.096 C, and a new boiling point of 104.096 C. The solution would thus be stable from 104.096 to 1.488 C, a range of 105.6 C. 021 10.0 points Which of the following has the lowest freezing point? Assume that all the species are mixed as shown and are completely dissociated. 1. 1.5 m lithium sulfate sung (es24394) Homework 2 vanden bout (51640) 2. 1.0 m magnesium phosphate 3. 2.0 m sodium perchlorate 4. 1.5 m aluminum nitrate correct 5. 1.0 m potassium phosphate Explanation: The aluminum nitrate has an ideal vant Ho factor (i) of 4 which will scale the stated molality (1.5) up to a 6.0 for the eective molality. This eective molality is the highest in the group and therefore will have the lowest fp, the highest bp, and the maximum osmotic pressure. 022 10.0 points The freezing point of seawater is about 1.85C. If seawater is an aqueous solution of sodium chloride, calculate the molality of seawater. The kf for water is 1.86 K/m. 1. 0.995 m 2. 3.70 m 3. 1.99 m 4. 3.44 m 5. 0.497 m correct Explanation: 023 10.0 points How many moles of CO(NH2 )2 are present in 150 g of water, if the freezing point of the solution is 4.02 C? kf (water) = 1.86 K kg/mol. 1. 0.313 mol 2. 0.324 mol correct 3. None of these 4. 0.249 mol 5. 0.259 mol Tf = kf m nCO(NH2 )2 = kf msolvent Tf msolvent nCO(NH2 )2 = kf (4.02 K) (0.15 kg) = 1.86 K kg/mol = 0.324194 mol . 6. 0.27 mol Explanation: kf = 1.86 K kg/mol Tf = 4.02 C = 4.02 K msolvent = 150 g = 0.15 kg 6 024 10.0 points Rank the following aqueous solutions from lowest to highest boiling point: 0.5 m NaCl, 1 m KCl, 0.5 m BaCl2, 1 m Ba(NO3 )2 ). 1. Ba(NO3 )2 < 0.5 m NaCl < 0.5 m BaCl2 < 1 m KCl 2. 0.5 m NaCl < 0.5 m BaCl2 < 1 m KCl < 1 m Ba(NO3 )2 correct 3. 0.5 m BaCl2 < 1 m KCl < 1 m Ba(NO3 )2 < 0.5 m NaCl 4. 1 m KCl < 1 m Ba(NO3)2 < 0.5 m NaCl < 0.5 m BaCl2 Explanation: All of the solutions ( 0.5 m NaCl, 1 m KCl, 0.5 m BaCl2, 1 m Ba(NO3 )2 ) are composed of strong electrolytes and thus their vant Ho coecients must be taken into account. These are 2, 2, 3 and 3, respectively. Their eective molalities are then 1, 2, 1.5 and 3, respectively. 025 10.0 points A semi-permeable membrane can withstand an osmotic pressure of 0.75 atm. What molarity of aqueous magnesium bromide solution would reach the limit for this membrane? (assume RT = 25 L atm mol1 ). sung (es24394) Homework 2 vanden bout (51640) 1. 0.01 mM 2. 0.03 mM 3. 1 M 4. 0.03 M 5. 0.01 M correct Explanation: Use the formula for osmotic pressure. Remember that magnesium bromide will dissociate into 3 particles, so there will be a Vant Ho factor of 3 for this calculation. 026 10.0 points Catalase (a liver enzyme) dissolves in water. A 16 mL solution containing 0.166 g of catalase exhibits an osmotic pressure of 1.2 Torr at 20 C. What is the molar mass of catalase? 1. 1.58 105 g/mol correct 2. 1.33 105 g/mol 3. 2.53 105 g/mol 4. 2.3 105 g/mol 5. 2.11 10 g/mol 6. 2.81 105 g/mol Explanation: V = 16 mL = 0.016 L T = 20 C + 273 = 293 K R = 0.08206 L atm/K/mol =n m = 0.166 g = 1.2 Torr 5 7 027 10.0 points Why is salting meat a good idea if you want your meat preserved? 1. Salting causes bacteria on meat to incorporate a dangerous amount of salt into their proteins. 2. Salting causes bacteria on meat to dehydrate. correct 3. Salting causes bacteria on meat to swell and burst with water. Explanation: 028 10.0 points Which of the following is a possible combination of values for Hlattice , Hhydration and Hsolution , respectively, for a salt whose dissolution is endothermic. 1. 450, +400, and 50 kJ mol1, 2. +550, 480, and + 1030 kJ mol1 , 3. +640, 620, and + 20 kJ mol1 , correct 4. 900, 900, and 1800 kJ mol1 , Explanation: Hsolution = Hlattice + Hhydration - the problem stipulates that Hsolution be positive (endothermic) and this limits the values the other two terms can have. RT m RT = V MW V mRT V (0.166 g) (0.08206 L atm/K/mol) MW = (0.016 L)(1.2 Torr) 760 Torr (293 K) 1 atm 5 = 1.57986 10 g/mol . M= ... View Full Document

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