Version 001 Calculating Concentrations WKST vanden bout (51165 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. ChemPrin3e G 13 001 (part 1 of 2 10.0 points Determine the mass of anhydrous copper(ii sulfate that must be used to prepare 250 ml of 1.9 M CuSO 4 (aq. Correct answer: 75.8148 g. V 250 ml 0.25 L FW CuSO4 63.546 g/mol + 32.066 g/mol + 4(15.9994 g/mol 159.61 g/mol M 1.9 M m CuSO4 (1.9 M(0.25 L(159.61 g/mol 75.8148 g 002 (part 2 of 2 10.0 points Determine the mass of CuSO 4 5H 2 O that must be used to prepare 250 ml of 1.9 M CuSO 4 (aq. Correct answer: 118.603 g. FW CuSO4 5H 2O FW CuSO4 +5(FW H2O 159.61 g/mol + 10(1.0079 g/mol + 5(15.9994 g/mol 249.69 g/mol m CuSO4 5H 2O (1.9 M(0.25 L(249.69 g/mol 118.603 g 003 10.0 points A chemist studying the properties of photographic emulsions needed to prepare 500 ml of 0.178 M AgNO 3 (aq. What mass of silver nitratemustbeplacedintoa500mlvolumetric flask, dissolved, and diluted to the mark with water? Correct answer: 15.1187 g. v 500 ml 0.5 L M 0.178 M FW AgNO3 107.8682 g/mol+14.0067 g/mol + 3(15.9994 g/mol 169.873 g/mol m AgNO3 (0.178 M(0.5 L (169.873 g/mol 15.1187 g Brodbelt 013 015 004 10.0 points How much NaNO 3 is needed to prepare 225 ml of a 1.55 M solution of NaNO 3? 1. 4.10 g 2. 29.6 g correct 3. 12.3 g 4. 0.132 g 5. 0.244 g V 225 ml M 1.55 M? g NaNO 3 225 ml 1 L soln 1000 ml 1.55 mol NaNO 3 1 L soln 85 g NaNO 3 1 mol NaNO 3 29.6 g NaNO 3 ChemPrin3e G 05 Brodbelt 03 13
Version 001 Calculating Concentrations WKST vanden bout (51165 2 005 10.0 points How many moles of HCl are present in 40.0 ml of a 0.035 M solution? 1. 0.012 mol 2. 0.0060 mol 3. 0.25 mol 4. 0.0012 mol 5. 0.0014 mol correct V 40.0 ml M 0.035 M?molHCl 40.0mLsoln 1 L 0.035 mol HCl 1000 ml 1 L soln 0.0014 mol HCl Brodbelt 013 515 006 10.0 points What is the final concentration of Ca(OH 2 when 255 ml of 0.250 M Ca(OH 2 is mixed with 55.0 ml of 0.65 M Ca(OH 2? 1. 0.390 M 2. 0.642 M 3. 0.780 M 4. 2.90 M 5. 0.900 M 6. 0.321 M correct V 1 Ca(OH2 255 ml [Ca(OH 2 ] 1 0.250 M V 2 Ca(OH2 55 ml [Ca(OH 2 ] 2 0.65 M The total moles of Ca(OH 2 in the final solution will be the sum of the moles present in the two individual solutions:? mol Ca(OH 2 (soln 1 0.255 L soln 0.250 mol Ca(OH 2 1 L soln 0.06375 molca(oh 2? mol Ca(OH 2 (soln 2 0.055 L soln 0.65 mol Ca(OH 2 1 L soln 0.03575 molca(oh 2 Total mol Ca(OH 2 0.06375 mol + 0.03575 mol 0.0995 molca(oh 2 The total volume of the final solution is the sum of the volumes of the individual solutions. Total L 0.255 L + 0.055 L 0.31 L soln Molarity is moles solute per L of solution.? M Ca(OH 2 0.0995 mol Ca(OH 2 0.31 L soln 0.321 M Ca(OH 2 Msci 14 0813 007 10.0 points What is the effective molality of a solution containing 12.0 g of KI in 550 g water? Assume 100 percent ionic dissociation. 1. 0.072 molal 2. 0.26 molal correct 3. 0.42 molal 4. 0.066 molal 5. 0.13 molal 6. 0.59 molal m KI 12.0 g m H2O 550 g Because KI is a 1:1 salt, you get one cation and one anion for every single formula unit
Version 001 Calculating Concentrations WKST vanden bout (51165 3 that dissolves. Therefore, you ll get a DOU- BLING of the stated molality for the effective molality. The formula weight of KI is 166 g/mol, so the number of moles of KI is ( 1 mol KI 12.0 g KI 0.0723 mol KI, 166 g KI and the (stated molality of the solution would then be 0.0723 mol KI 0.131 m. 0.550 kg H 2 O But recall that the effective molality would be twice the stated molality here, so the effective molality is 0.262 m. Nlib 11 0020 008 10.0 points If you mix 3 moles of ethylene glycol (antifreeze in 4165 grams of water, what is the molality of the solution? Correct answer: 0.720288 m. n ethylene glycol 3 mol m water 4165 g Molality (m is moles solute per kilogram of solvent. The solute is ethylene glycol. The solvent is water and 4165 g 4.165 kg.? m 3 mol ethylene glycol 4.165 kg H 2 O 0.720288 m Nsci 14 0808 009 10.0 points 1.9 g of NaCl and 6.1 g of KBr were dissolved in48gofwater. What isthemolalityofnacl in the solution? Holt da 13 1 sample 1 010 10.0 points What is the molarity of a 3.047 L solution that is made from 11.29 g of NaCl? Correct answer: 0.0634032 M. V solution 3.047 L M? m NaCl 11.29 g M L soln? mol NaCl 11.29 g NaCl ( 1 mol NaCl 58.44 g NaCl 0.19319 mol NaCl mol NaCl? M L soln 0.0634032 M 0.19319 mol 3.047 L Molality 08 44a 011 10.0 points Calculate the molality of sucrose in a solution composed of 11.31 g of sucrose (C 12 H 22 O 11 dissolved in 606 ml of water. Correct answer: 0.054524 m. m C12H 22O 11 11.31 g V H2O 606 ml 0.606 L MW C12H 22O 11 342.296 g/mol Correct answer: 0.6773 m. m NaCl 1.9 g m water 48 g m KBr 6.1 g ( 1 kg m H2O (0.606 L 0.606 kg 1 L Thus the molality is mol NaCl m NaCl kg water 1.9 g 58.4 g mol 1 NaCl 0.048 kg water 0.6773 m moles solute m C12H 22O 11 ( 11.31 g sucrose 342.296 g/mol sucrose 0.606 kg H 2 O 0.054524 m
Version 001 Calculating Concentrations WKST vanden bout (51165 4 Msci 14 0809 012 10.0 points What additional information, if any, would enable you to calculate the molality of a 7.35 molar solution of a nonelectrolyte solid dissolved in water? 1. Both the density of the solution and the molecular weight of the solute would be needed. correct 2. Only the density of the solution would be needed. 3. Only the molecular weight of the solute would be needed. 4. Only the density of water would be needed. 5. None is needed. molarity molality L solution The density of the solution can be used to convert volume (1 L of solution into mass of solution. Then the molecular weight of the solute (given or calculated from the formula can be used to convert the number of moles solute in 1 L solution into mass of solute in grams. The mass of the solvent is the difference between the mass of the solution and the mass of the solute (both of which have been calculated. Substitute the values into the molality formula and calculate. Molality 08 46c 013 10.0 points Calculate the molality of 10.5 M NH 3 (aq with a density of 0.9344 g/cm 3. d 0.9344 g/cm 3 Assume 1 L of 10.5 M NH 3 (aq; it will contain 10.5 mol NH 3 with a mass of (10.5 mol(17.0305 g/mol 178.82 g. The density of the 1 L of solution is 0.9344 g/cm 3 1000 cm3 1 L 934.4 g/l, so the total mass of the solution is 934.4 g, which leaves 934.4 g 178.82 g 755.58 g of water. Therefore, the molality is m 10.5 mol NH 3 0.75558 13.8966 m. Nsci 14 0803 014 10.0 points Formalin is a solution of 40.0% formaldehyde (H 2 CO,10.0%methylalcohol(CH 3 OH,and 50.0% water by mass. Calculate the mole fraction of methyl alcohol in formalin. Correct answer: 0.0706436. In a 100 g formalin, solution the masses are formaldehyde 40.0 g, methyl alcohol 10.0 g, and water 50.0 g. n CH3OH +n H2CO +n water 10 g 32 g/mol + 40 g 30 g/mol + 50 g 18 g/mol 4.42361 mol The mole fraction of methyl alcohol is X n CH 3OH 10.0 g 32.0 g/mol 4.42361 mol 0.0706436 Correct answer: 13.8966 m. MW 17.0305 g/mol M 10.5 M Nsci 14 0801 015 10.0 points Toluene (C 6 H 5 CH 3 is a liquid compound
Version 001 Calculating Concentrations WKST vanden bout (51165 5 similar to benzene (C 6 H 6. Calculate the mole fraction of toluene in the solution that contains 112 g toluene and 72.0 g benzene. Correct answer: 0.568. m toluene 112 g m benzene 72.0 g ( 1 mol n toulene (112 g toluene 92.14 g 1.21 mol ( 1 mol n benzene (72.0 g benzene 78.11 g 0.922 mol The total number of moles of all species present is 1.21 mol+0.922 mol 2.13 mol The mole fraction of toluene is then X toluene n toluene 1.21 mol 2.13 mol 0.568 Msci 14 0802 016 10.0 points The mole fraction of a certain nonelectrolyte compound in a solution containing only that substance and water is 0.100. The molecular weight of water is 18.0 g/mol. What additional information is needed to determine the molality of the solution? 1. The density of the solute. 2. The density of the solution. 3. The molecular weight of the compound. 4. The mole fraction of water in the solution. 5. No additional information; the molality can be calculated from the information given. correct Here we can assume that we have 1 mol total. (In fact, we can choose any number of moles, but the math is easier if you choose 1 mol. If the mole fraction of the substance is 0.100, you can then assume that you have 0.100 mol of the substance, and the remaining 0.900 mol is H 2 O. The molality of a solution is determined by the following formula: m We ve already assumed that we have 0.100 mol of solute, and we can determine the kg of H 2 O in the usual way: ( 18.0 g 1 mol ( 1 kg 0.900 mol H 2 O 1000 g 0.0162 kg, and we can calculatethe molalityof thissolution: m 0.100 mol 0.0162 kg H 2 O 6.17 m So, it is possible to determine the molality of this solution without any additional information. Msci 14 0818 017 10.0 points The molecular weight ofsugar is342 and that of water is 18.01. What is the mole fraction of sugar in a 2.00 molal solution of sugar dissolved in water? 1. mole fraction 0.0348 correct 2. mole fraction 0.950 3. mole fraction 0.406 4. mole fraction 0.0360 5. mole fraction 0.925 MW sugar 342 MW water 18.01 m sugar 2.00 m Molality Molality of sugar 2.00 mol sugar 1 kg water Mol water in 1 kg water ( 1 mol water (1000 g water 18.01 g water 55.5 mol H 2 O
Version 001 Calculating Concentrations WKST vanden bout (51165 6 The mole fraction of sugar is mol sugar X sugar mol sugar+mol water 2.00 mol 2.00 mol+55.5 mol 3.48 10 2 Nsci 14 0802exam 018 10.0 points 11.6g ofnacl(58.4g/moland 16.9gofKBr (119.8 g/mol were dissolved in 49 g of water (18.0 g/mol. What is the mole fraction of KBr in the solution? Correct answer: 0.0312. m NaCl 11.6 g m KBr 16.9 g m water 49 g FW NaCl 58.4 g/mol FW KBr 119.8 g/mol FW water 18.0 g/mol 11.6 g n KBr 0.096828 mol 119.8 g/mol n NaCl 16.9 g 0.289384 mol 58.4 g/mol 49 g n H2O 2.72222 mol 18.0 g/mol Since n KBr +n NaCl +n H2O, 0.096828 mol+0.289384 mol + 2.72222 mol 3.10843 mol, the mole fraction is n KBr 0.096828 mol X KBr 3.10843 mol 0.0312 mol Correct answer: 814.286 g. m water 100 g X sucrose 0.3 n water X sucrose n 100 g 18.0 g/mol 5.55556 mol n 0.3 n+n water n 0.3(n+n water 0.7n 0.3n water n 0.3n water m 0.7 342 g/mol, so (342 g/mol(0.3(5.55556 mol m 0.7 814.286 g Nsci 14 0807 020 10.0 points A solution is made from 596 ml methanol (CH 3 OH of density 0.800 g/ml and 82 ml of water (H 2 O of density 1.000 g/ml. Assume that the solution behaves ideally and the volumes are additive. Calculate the mole fraction of methanol in this solution. Correct answer: 0.765848. V methanol 596 ml V water 82 ml density methanol 0.800 g/ml density water 1.00 g/ml mole fraction CH 3 OH n CH 3OH Nsci 14 0805 019 10.0 points How many grams of sucrose (C 12 H 22 O 11 must be dissolved in 100 g water to make a solution in which the mole fraction of sucrose is 0.3? n CH3OH 82 ml CH 3 OH 0.800 g CH 3OH 1.0 ml CH 3 OH 1.0 ml CH 3OH 32.0 g CH 3 OH 14.9 mol CH 3 OH
Version 001 Calculating Concentrations WKST vanden bout (51165 7 ( 1.0 ml H2 O n n H2O 82 ml H 2 O total 0.14 mol P total 700 torr 18.0 g H 2 O n H2 0.02 mol 4.55556 mol H 2 O The mole fraction is 14.9 mol X CH3OH 14.9 mol+4.55556 mol 0.765848 Holt da 10 rev 39 021 10.0 points Three of the primary components of air are carbon dioxide, nitrogen, and oxygen. In a sample containing a mixture of only these gases at exactly one atmosphere pressure, the partial pressures of carbon dioxide and nitrogen are given as P CO2 0.285 torr and P N2 578.351 torr. What is the partial pressure of oxygen? Correct answer: 181.364 torr. P CO2 0.285 torr P T 1atm 760torr P N2 578.351 torr P O2? X H2 n H 2 0.02 mol 0.14 mol 0.142857 P H2 X H2 P total (0.142857(700 torr 100 torr P T P CO2 +P N2 +P O2 P O2 P T (P CO2 +P N2 P O2 760 torr (0.285 torr+578.351 torr P O2 181.364 torr Mlib 04 1003 022 10.0 points A 22.4 L vessel contains 0.02 mol H 2 gas, 0.02 mol N gas, and 0.1 mol NH 3 gas. The total pressure is 700 torr. What is the partial pressure of the H 2 gas? 1. 100 torr correct 2. 7 torr 3. None of these 4. 28 torr 5. 14 torr