Solutions CHAPTER Specific answers depend on student choices.

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1 CHAPTER Specific answers depend on student choices.. A heterogeneous mixture does not have a uniform composition: the composition varies in different places within the mixture. Examples of non homogeneous mixtures include salad dressing (mixture of oil, vinegar, water, herbs, and spices) and granite (combination of minerals).. Water is the solvent, sugar the solute. 4. solid 5. When an ionic solute dissolves in water, a given ion is pulled into solution by the attractive ion dipole force exerted by several water molecules. For example, in dissolving a positive ion, the ion is approached by the negatively charged end of several water molecules: if the attraction of the water molecules for the positive ion is stronger than the attraction of the negative ions near it in the crystal, the ion leaves the crystal and enters solution. After entering solution, the dissolved ion is surrounded completely by water molecules, which tends to prevent the ion from reentering the crystal. 6. Like dissolves like. The hydrocarbons in oil have intermolecular forces that are very different from those in water, and so the oil spreads out rather than dissolving in the water. 7. Salts are composed of positive and negative ions packed closely together in a crystal lattice. The crystal is held together by the ionic forces among oppositely charged particles in the solid. When a crystal of a salt is placed in water, the polar water molecules arrange their dipoles around the ions of the crystal in such a way as to attract the ions. For example, a positive ion at a corner of the crystal will be surrounded by a group of water molecules aiming the negative end of their dipoles at the positive ion. If the forces of attraction of the water molecules for the positive ion are larger than the ionic forces holding the ion in the crystal, then the ion will leave the crystal and enter solution. See Figure Carbon dioxide is somewhat soluble in water, especially if pressurized (otherwise, the soda you may be drinking while studying Chemistry would be flat ). Carbon dioxide s solubility in water is approximately 1.5 g/l at 5 C under a pressure of approximately 1 atm. Although the carbon dioxide molecule overall is non-polar, that is because the two individual C O bond dipoles cancel each other due to the linearity of the molecule. However these bond dipoles are able to interact with water, making CO more soluble in water than non-polar molecules such as O or N that do not possess individual bond dipoles. 9. A saturated solution is one that contains the maximum amount of solute possible at a particular temperature. A saturated solution is one that is in equilibrium with undissolved solute. 10. unsaturated 11. variable 07

2 1. large 1. The mass percent of a particular component in a solution is the mass of that component, divided by the total mass of the solution, multiplied by 100% a. b. c. d..14 g KCl 100 = 14.6% KCl (.14 g KCl g water).14 g KCl 100 = 7.89% KCl (.14 g KCl g water).14 g KCl 100 = 5.40% KCl (.14 g KCl g water).14 g KCl 100 = 4.10% KCl (.14 g KCl g water) mg = g g CaCl a. 100 = ( g CaCl g water) 0.116% CaCl g CaCl b. 100 = ( g CaCl g water) % CaCl 6.11 g CaCl c. 100 = (6.11 g CaCl g water) 10.4% CaCl 6.11 kg CaCl d. 100 = (6.11 kg CaCl kg water) 10.4% CaCl 1.51 g NH4Cl 17. a. 75 g 100. g solution = 5.66 g NH 4Cl b. 15 g.91 g NaCl 100. mg solution =.64 g NaCl c. 1.1 kg = g g KNO g 100. g solution = 64.5 g KNO d. 478 mg = g g g NH NO 100. g solution = g NH 4NO 08

3 .91 g FeCl 18. a. 55 g solution 100.g solution = 0.5 g FeCl 55 g solution 0.5 g FeCl = g water (505 g water) 11.9 g sucrose b. 5 g solution = 6.8 g sucrose 100. g solution 5 g solution 6.8 g sucrose = 198. g water (198 g water) c kg = g g NaCl g solution = 181. g NaCl (181 g NaCl) 100. g solution g solution 181. g NaCl = g water ( g water) 15.1 g KNO d. 65 g solution 100. g solution = 95.9 g KNO 65 g solution 95.9 g KNO = 59.1 g water (59 g water) 19. Total mass = 9.1 g +.59 g g = 96.8 g 9.1 g Fe %Fe = 100 = 95.7% Fe 96.8 g alloy.59 g C %C = g alloy =.69% C %Cr = 1.59 g Cr 100 = 1.65% Cr 96.8 g alloy 0. Percent means per hundred. So the percentages in Question 19 represent the number of grams of the particular element present in 100. g of the alloy. Since 1.00 kg represents ten times the mass of 100. g, we would need 957 g Fe, 6.9 g C, and 16.5 g Cr to prepare 1.00 kg of the alloy. 1. To say that the solution contains 7.51% of ammonium nitrate means that every 100. grams of the solution will contain 7.51 g of ammonium nitrate. 1.5 kg = g g NH4NO g solution = 9.9 g NH 4 NO 100 g solution g solution 9.9 g NH 4 NO = g water ( g water) 67.1 g CaCl. (67.1 g CaCl + 75 g water) 100 = 19.6% CaCl 09

4 . To say that the solution to be prepared is 4.50% by mass CaCl means that 4.50 g of CaCl will be contained in every 100. g of the solution g CaCl 175 g solution 100. g solution = 7.88 g CaCl 4. To say that the solution is 6.5% KBr by mass, means that 100. g of the solution will contain 6.5 g KBr. 6.5 g KBr 15 g solution = 7.81 g KBr 100. g solution g solution 5.00 g NaCl g solution = 14. g NaCl 7.50 g Na CO 85 g solution g solution = 1.4 g Na CO 6. molar mass O =.00 g 1.00 g O = mol O.00 g from the balanced chemical equation, it will take (0.015) = mol H O to produce this quantity of oxygen. molar mass H O = 4.0 g 4.0 g HO mol H O H O =.1 g H O.1 g H O 100. g solution g H O = approximately 71 g 7. First find the mass of the solution, because the percentage is by mass. = ml g solution ml solution 1 ml solution = 1840 g solution = g solution g HSO4 g solution 100. g solution = g H SO g 0.95 g stablizer 100. g = 9.5 g 9. Each liter of the solution contains mol of hydrogen chloride mol; 0.0 mol 10

5 1. A standard solution is one whose concentration is known very accurately and with high precision. A standard solution is typically prepared by weighing out a precise amount of solute, and then dissolving the solute in a precise amount of solvent, or to a precise final volume using a volumetric flask.. The molarity represents the number of moles of solute per liter of solution: choice b is the only scenario that fulfills this definition. moles of solute. Molarity = liters of solution a. 15 ml = 0.15 L 0.5 = 4.17 M 0.15 L b. 50. ml = 0.50 L L =.08 M c ml = L L = 1.04 M d. 0.5 = 0.51 M moles of solute 4. Molarity = liters of solution a. 5 ml = 0.5 L mol KNO 0.5 L b. 10. ml = L mol CaCl L c..15 mol NaCl 5.00 L d ml = L mol NaBr L 5. Molarity =.51 g NaCl =.5 M = 1.0 M = 0.60 M = 4.99 M moles of solute ; molar mass of NaCl = g liters of solution NaCl g NaCl = mol NaCl 11

6 a. 5 ml = 0.05 L mol NaCl =.4 M 0.05 L b. 50. ml = L mol = 1. M L c. 75 ml = L mol = 0.80 M L d. 6. Molarity = mol = M moles of solute liters of solution ; molar mass of CaCl = g; 15 ml = 0.15 L CaCl a g CaCl g CaCl = mol CaCl mol = 0.40 M 0.15 L CaCl b..4 g CaCl g CaCl mol = M 0.15 L CaCl c. 8.7 g CaCl g CaCl mol = 0.69 M 0.15 L CaCl d g CaCl g CaCl = mol CaCl = mol CaCl = mol CaCl mol 0.15 L = 0.89 M 7. molar mass of CaCl = g; 5 ml = 0.5 L mol g 0.5 L =.75 g CaCl 1 L 8. molar mass of KBr = g; 15 ml = 0.15 L 0.15 L g = 4.0 g KBr 1 L 1

7 9. molar mass CaCO = g; 50.0 ml = L g CaCO g = mol CaCO One mole of CaCO contains one mole of Ca + ion; therefore, mol Ca L solution = M 40. molar mass of I = 5.8 g; 5 ml = 0.5 L 5.15 g I 5.8 g = 0.00 mol I 0.00 mol I 0.5 L solution = M 41. molar mass of NaOH = g; 5 ml = 0.5 L NaOH 4.5 g NaOH = 1.06 mol NaOH g 1.06 mol NaOH 0.5 L solution = 4.7 M 4. molar mass of KNO = g; 5.0 ml = L KNO 1.1 g KNO = mol KNO g mol KNO L solution = M moles of solute 4. Molarity = liters of solution a. 4.5 ml = L mol CaCl L solution solution = mol CaCl b. 11. ml = L mol NaOH L solution solution = mol NaOH c. 1. HCl 1.5 L solution = 15. HCl solution d. 7.5 ml = L 1.98 mol NaCl L solution = mol NaCl solution 1

8 44. a. 1.5 ml = L mol HCl L solution = mol HCl solution b. 7. ml = 0.07 L 0. mol NaOH 0.07 L solution solution = mol NaOH c. 6.8 ml = L 0.50 HNO L solution solution = mol HNO d ml = L mol KOH L solution solution = mol KOH 45. a. 1. HCl.50 L solution =.75 mol HCl solution molar mass HCl = 6.46 g 6.46 g HCl.75 mol HCl = 1194 g HCl = g HCl HCl b ml = L mol NaOH L solution solution = mol NaOH molar mass NaOH = g g NaOH mol NaOH = g NaOH c. 15 ml = 0.15 L.0 HNO 0.15 L solution solution = mol HNO molar mass HNO = 6.0 g 6.0 g HNO mol HNO = 17.1 g HNO d mol CaCl 4.1 L solution solution =.168 mol CaCl molar mass CaCl = g g CaCl.168 mol CaCl = 41 g CaCl 14

9 46. a. molar mass of CaCl = g; 17.8 ml = L mol CaCl g CaCl L solution = 0.5 g CaCl 1 L solution CaCl b. molar mass of KCl = g; 7.6 ml = L 0.88 mol KCl g KCl L solution = 0.59 g KCl 1 L solution KCl c. molar mass of FeCl = 16. g; 5.4 ml = L 0.99 mol FeCl 16. g FeCl L solution =.9 g FeCl 1 L solution FeCl d. molar mass KNO = g; 46.1 ml = L mol KNO g KNO L solution =.61 g KNO 1 L solution KNO 47. molar mass of NaOH = g 0.50 mol NaOH g NaOH.5 L solution = 70. g NaOH 1 L solution NaOH 48. molar mass of KBr = g; 5 ml = 0.5 L 0.55 mol KBr g KBr 0.5 L solution = 9.51 g KBr 1 L solution KBr 0.5 NaSO4 49. a. solution = 0.5 Na SO Na SO 4 + mol Na Na SO 4 = 0.50 mol Na+ 0. FeCl b L solution = mol FeCl mol Cl mol FeCl FeCl = 1.65 mol = 1.7 mol Cl c ml = L 0.55 mol Ba(NO ) L solution = mol Ba(NO ) mol NO mol Ba(NO ) Ba(NO ) = 0.1 NO d. 50. ml = 0.50 L 0.50 mol (NH 4) SO L solution = mol (NH 4 ) SO 4 15

10 + mol NH mol (NH 4 ) SO 4 (NH ) SO 4 4 = mol NH a. 10. ml = L L AlCl Al = mol Al + AlCl 0.45 AlCl mol Cl L = mol Cl AlCl b L mol NaPO4 mol Na = 1.70 mol Na + Na PO L 0.10 mol NaPO4 PO4 = mol PO 4 Na PO 4 c ml = L mol CuCl Cu L = mol Cu + CuCl 1.5 mol CuCl mol Cl L = mol Cl CuCl d. 5. ml = 0.05 L mol Ca(OH) Ca 0.05 L = mol Ca + Ca(OH) mol Ca(OH) mol OH 0.05 L = mol OH Ca(OH) 51. molar mass NaCl = g; 15 ml = 0.15 L mol NaCl g NaCl 0.15 L = g NaCl for 15 ml of the solution mol NaCl g NaCl = 6.14 g NaCl for of the solution 5. Molar mass of Na CO = g; 50 ml = 0.50 L mol Na CO 106.g 0.50 L = 1. g Na CO 5. moles of solute 54. half 16

11 55. M1 V1 = M V a. M 1 = M M =? V 1 = 5.0 ml = L V = ( ) ml = 80.0 ml = L (0.119 M )(0.050 L) M = ( L) = 0.07 M b. M 1 = M M =? V 1 = 45. ml = L V = ( ) = 170. ml = L (0.701 M )(0.045 L) M = (0.170 L) = M c. M 1 =.01 M M =? V 1 = 15 ml = 0.15 L V = ( ) ml = 675 ml = L (.01 M )(0.15 L) M = (0.675 L) = M d. M 1 =.07 M M =? V 1 = 75. ml = L V = ( ) ml = 410. ml = L (.07 M )(0.075 L) M = (0.410 L) = 0.80 M 56. M1 V1 = M V a. M 1 = 0.51 M M =? V 1 = 10.0 ml V = ( ) = 5.0 ml M = (0.51 M )(10.0 ml) (5.0 ml) = M b. M 1 =.00 M M =? V 1 = 15 ml V = ( ) =.5 ml M = (.00 M )(15 ml) (.5 ml) = 1.69 M c. M 1 = M M =? V 1 = 5.0 ml V = 500. ml M = (0.851 M )(5 ml) (500. ml) = M d. M 1 = 1.5 M M =? V 1 = 5.0 ml V = 50.0 ml 17

12 M = (1.5 M )(5 ml) (50.0 ml) = 0.65 M 57. M1 V1 = M V HCl: (.0 M )(5 ml) (1.1 M ) = 55.8 ml = 56 ml HNO : (.0 M )(5 ml) (15.9 M ) H SO 4 : (.0 M )(5 ml) (18.0 M ) = 4.45 ml = 4 ml = 7.5 ml = 8 ml HC H O : (.0 M )(5 ml) (17.5 M ) = 8.6 ml = 9 ml H PO 4 : (.0 M )(5 ml) (14.9 M ) = 45. ml = 45 ml 58. M 1 = 19.4 M M =.00 M V 1 =? ml M 1 = (.00 M )(.50 ml) = L (541 ml) (19.4 M ) 59. M1 V1 = M V V =.50 L M 1 =.00 M V 1 =? (0.50 M )(0.75 L) V 1 = (.00 M ) = L = 48.1 ml M = 0.50 M V = 75 ml = 0.75 L Dilute 48.1 ml of the.00 M solution to a final volume of 75 ml. 60. M1 V1 = M V M 1 = 1.01 M V 1 =? ml M = (0.150 M )(5 ml) (1.01 M ) = 48. ml M = M V = 5 ml Dilute 48. ml of the 1.01 M solution to a final volume of 5 ml. 61. M1 V1 = M V M 1 = 0.00 M V 1 = 500. ml = L V =? M = M 18

13 V = (0.00 M )(0.500 L) (0.150 M ) = L = 667 ml Therefore = 167 ml of water must be added. 6. (100. ml)(1.5 M ) = 10. ml (1.1 M ) ml = L The number of moles of Ni + ion present is given by L mol Ni 1 L = mol Ni + Since Ni + and DMG react on a 1: mole basis, then = mol of DMG are needed for the reaction mol DMG 1 L mol DMG = 0.09 L = 9. ml of the DMG solution is needed. 64. Na CO (aq) + CaCl (aq) CaCO (s) + NaCl(s) mmol Ca + ion: 7. ml mmol Ca 1.00 ml =.91 mmol Ca + From the balanced chemical equation,.91 mmol CO will be needed to precipitate this quantity of Ca + ion..91 mmol CO 1.00 ml = 1. ml 0.15 mmol 65. The balanced net ionic equation for the precipitation reaction is Cu + (aq) + S (aq) CuS(s) copper ion present: 7.5 ml mmol Cu 1.00 ml =.8 mmol Cu + As the coefficients of the balanced equation are both 1, then.8 mmol of S ion will be required to precipitate the Cu + ion. volume sulfide solution:.8 mmol S 1.00 ml mmol S = 1.7 ml Na S 66. molar mass Na C O 4 = 14.0 g 7.5 ml = L moles Ca + ion = L mol Ca = mol Ca + ion Ca + (aq) + C O 4 (aq) CaC O 4 (s) As the precipitation reaction is of 1:1 stoichiometry, then mol of C O 4 ion is needed. Moreover, each formula unit of Na C O 4 contains one C O 4 ion, so mol of Na C O 4 is required. 19

14 mol Na C O g = 0.5 g Na C O 4 required 67. Pb(NO ) (aq) + K CrO 4 (aq) PbCrO 4 (s) + KNO (aq) molar masses: Pb(NO ), 1. g; PbCrO 4,. g Pb(NO ) 1.00 g Pb(NO ) = mol Pb(NO ) 1. g Pb(NO ) 5.0 ml = L 1.00 mol KCrO L solution = mol K CrO 4 solution Pb(NO ) is the limiting reactant: mol PbCrO 4 will form.. g PbCrO mol PbCrO 4 PbCrO = g PbCrO ml = L 0.50 mol AlCl L = mol AlCl AlCl (aq) + NaOH(s) Al(OH) (s) + NaCl(aq) mol NaOH mol AlCl = mol NaOH AlCl molar mass NaOH = g mol NaOH g NaOH 4 = 0.00 g NaOH 69. NaOH(aq) + HNO (aq) NaNO (aq) + H O(l) 7. ml = 0.07 L 0.49 HNO 0.07 L = mol HNO 1 L NaOH mol HNO = mol NaOH HNO mol NaOH 1 L solution 0.50 mol NaOH 70. NaOH(aq) + HCl(aq) H O(l) + NaCl(aq) 15 ml = 0.15 L 0.15 L.0 NaOH 1 L = 0.76 mol NaOH = 0.06 L = 6.6 ml 0

15 0.76 mol NaOH 0.76 mol HCl HCl NaOH 1 L solution mol HCl = 0.76 mol HCl = 0.78 L = 78 ml 71. HCl(aq) + NaHCO (s) NaCl(aq) + H O(l) + CO (g) molar mass of NaHCO = g; 47.1 ml = L g NaHCO g = mol NaHCO The balanced chemical equation indicates that HCl and NaHCO react on a 1:1 basis, so mol of HCl must be present. Therefore, the molarity of the HCl is given by mol L = M ml = L mol NaOH L H + (aq) + OH (aq) H O(l) mol OH 100 ml = L = mol NaOH + H 1mol OH = mol H mol H L 5 + = M H + (aq) 7. a. NaOH(aq) + HC H O (aq) NaC H O (aq) + H O(l) 5.0 ml = L L mol acetic acid mol acetic acid = mol acetic acid NaOH acetic acid = mol NaOH mol NaOH = L =.85 ml NaOH 1.00 mol NaOH b. HF(aq) + NaOH(aq) NaF(aq) + H O(l) 5.0 ml = L L mol HF 0.10 mol HF HF NaOH = mol HF = mol NaOH 1

16 mol NaOH = L =.57 ml 1.00 mol NaOH c. H PO 4 (aq) + NaOH(aq) Na PO 4 (aq) + H O(l) 10.0 ml = L 0.14 mol HPO L = mol H PO 4 mol NaOH mol H PO 4 = mol NaOH H PO mol NaOH = L = 4.9 ml 1.00 mol NaOH d. H SO 4 (aq) + NaOH(aq) Na SO 4 (aq) + H O(l) 5.0 ml = L 0.0 mol HSO L = mol H SO 4 mol NaOH mol H SO 4 = mol NaOH H SO mol NaOH 1.00 mol NaOH = L = 15.4 ml 74. Experimentally, neutralization reactions are usually performed with volumetric glassware that is calibrated in milliliters rather than liters. For convenience in calculations for such reactions, the arithmetic is often performed in terms of milliliters and millimoles, rather than in liters and moles: 1 mmol = Note that the number of moles of solute per liter of solution, the molarity, is numerically equivalent to the number of millimoles of solute per milliliter of solution. a. HNO (aq) + NaOH(aq) NaNO (aq) + H O(l) 1.7 ml mmol 1.00 ml = 6.6 mmol NaOH present in the sample 1 mmol HNO 6.6 mmol NaOH 1 mmol NaOH = 6.6 mmol HNO required to react 6.6 mmol HNO 1.00 ml mmol HNO = 6.0 ml HNO required b. HNO (aq) + Ba(OH) Ba(NO ) + H O(l) mmol 4.9 ml = 0.1 mmol Ba(OH) present in sample 1.00 ml mmol HNO 0.1 mmol Ba(OH) = 0.44 mmol HNO required 1 mmol Ba(OH)

17 0.44 mmol HNO c. HNO (aq) + NH (aq) NH 4 NO (aq) 1.00 ml mmol HNO =.4 ml HNO is required 0.10 mmol 49.1 ml = 5.06 mmol NH present in the sample 1.00 ml 1 mmol HNO 5.06 mmol NH 1 mmol NH = 5.06 mmol HNO required 5.06 mmol HNO 1.00 ml mmol HNO = 50.1 ml HNO required d. KOH(aq) + HNO (aq) KNO (aq) + H O(l) 1.1 L 0.10 mol = 0. KOH present in the sample HNO 0. KOH KOH = 0. HNO required 0. HNO 75. one mole of protons (hydrogen ions) normal 0.10 HNO = 1. L HNO required 77. When H SO 4 reacts with OH, the reaction is H SO 4 (aq) + OH (aq) H O(l) + SO 4 (aq). As each mol of H SO 4 provides two moles of H + ion, it is only necessary to take half a mole of H SO 4 to provide one mole of H + ion. The equivalent weight of H SO 4 is thus half the molar mass equivalents OH ion are needed to react with 1.5 equivalents of H + ion. By definition, one equivalent of OH ion exactly neutralizes one equivalent of H + ion. number of equivalents of solute 79. N = number of liters of solution a. equivalent weight HCl = molar mass HCl = 6.46 g MV 1 1 (5. ml)(0.105 M ) M = = = N V (75. ml) molar mass b. equivalent weight H PO 4 = 0.5 mol HPO4 equivalents HPO4 = N e H PO 4

18 molar mass c. equivalent weight Ca(OH) = mol Ca(OH) equivalents Ca(OH) = N Ca(OH) number of equivalents of solute 80. N = number of liters of solution a. equivalent weight NaOH = molar mass NaOH = g 0.11 g NaOH 1 equiv NaOH g 10. ml = L.8 10 equiv N = = 0.77 N L b. equivalent weight Ca(OH) =.8 10 equiv NaOH molar mass g = = 7.05 g 1 g 1 equiv 1.5 mg = equiv Ca(OH) 10 mg 7.05 g 100. ml = L equiv N = =.7 10 N L c. equivalent weight H SO 4 = 1.4 g 1 equiv g = 0.5 equiv H SO ml = L 0.5 equiv N = L = 1.6 N molar mass g = 1 equiv HCl 81. a M HCl = 0.50 N HCl HCl equiv HSO4 b M H SO 4 H SO = 0.10 N 4 = g c equiv HPO4 M H PO 4 H PO 4 = N = 0.16 N 8. a M NaOH 1 equiv NaOH NaOH = 0.14 N NaOH 4

19 equiv Ca(OH) b M Ca(OH) Ca(OH) = N Ca(OH) equiv HPO4 c. 4.4 M H PO 4 H PO = 1. N H PO molar mass H PO 4 = 98.0 g HPO4 5. g H PO g H PO = 0.59 mol H PO mol HPO4 = M equiv HPO M H PO 4 H PO = 1.08 N H PO Molar mass of Ca(OH) = g 1 g 5.1 mg Ca(OH) = mol Ca(OH) 10 mg g ml = L (volumetric flask volume: 4 significant figures) mol L = M Ca(OH) N = equiv Ca(OH) M Ca(OH) Ca(OH) = N Ca(OH) 85. H SO 4 (aq) + NaOH(aq) Na SO 4 (aq) + H O(l) N acid V acid = N base V base (0.5 N) (15.0 ml) = (0.50 N) (V base ) V base = 10.5 ml = 11 ml 86. H SO 4 (aq) + NaOH(aq) Na SO 4 (aq) + H O(l) N acid V acid = N base V base (0.104 N)(V acid ) = (0.15 N)(15. ml) V acid =. ml The M sulfuric acid solution is twice as concentrated as the N sulfuric acid solution (e = equivalents), so half as much will be required to neutralize the same quantity of NaOH = 11.1 ml 87. NaOH(aq) + H SO 4 (aq) Na SO 4 (aq) + H O(l) For the 0.15 N H SO 4 : 5

20 N acid V acid = N base V base (0.15 N) (4. ml) = (0.151 N) (V base ) V base = 0.0 ml of the N NaOH solution needed For the 0.15 M H SO 4 : As each H SO 4 formula unit produces two H + ions, the normality of this solution will be twice its molarity M H SO 4 = 0.50 N H SO 4 N acid V acid = N base V base (0.50 N) (4.1 ml) = (0.151 N) (V base ) V base = 9.9 ml of the N NaOH solution needed 88. NaOH(aq) + H SO 4 (aq) Na SO 4 (aq) + H O(l) 7.4 ml NaOH mmol 1.00 ml =.791 mmol NaOH 1 mmol HSO4.791 mmol NaOH mmol NaOH = 1.96 mmol H SO mmol HSO4 = M H SO 4 = N H SO ml 89. total mass of solution = 50.0 g g g = g 50.0 g ethanol % ethanol = 100 = 47.6% ethanol g total % water = % sugar = 50.0 g water g total 5.0 g sugar g total 100 = 47.6% water 100 = 4.8% sugar 1.5 g sugar g solution 4.8 g sugar = 1 g solution 10.0 g ethanol g solution 47.6 g ethanol = 1.0 g solution 90. Molarity is defined as the number of moles of solute contained in 1 liter of total solution volume (solute plus solvent after mixing). In the first case, where 50. g of NaCl is dissolved in 1.0 L of water, the total volume after mixing is not known and the molarity cannot be calculated. In the second example, the final volume after mixing is known and the molarity can be calculated simply. 91. mmol CoCl = 50.0 ml 0.50 M CoCl = 1.5 mmol CoCl This contains 1.5 mmol Co + and 5.0 mmol Cl 6

21 mmol NiCl = 5.0 ml 0.50 M NiCl = 8.75 mmol NiCl This contains 8.75 mmol Ni + and 17.5 mmol Cl Total mmol Cl after mixing = = 4.5 mmol Cl Total volume after mixing = 50.0 ml ml = 75.0 ml M cobalt(ii) ion = M nickel(ii) ion = M chloride ion = g solution new % = mmol Co 75.0 ml mmol Ni 75.0 ml 4.5 mmol Cl 75.0 ml 5 g NaCl 100. g solution = M = M = M = g NaCl g NaCl 100 =.6 =. % 575 g solution 9. AgNO (s) + NaCl(aq) AgCl(s) + NaNO (aq) molar masses: AgNO, g; AgCl, 14.4 g AgNO 10.0 g AgNO g AgNO = mol AgNO 50. ml = L mol NaCl L = mol NaCl NaCl is the limiting reactant mol AgCl form g AgCl mol AgCl = 0.07 g AgCl (7 mg) As AgNO contains Ag +, the mol Ag + remaining in solution = = mol AgNO mol AgNO = mol Ag + M silver ion = mol Ag L = M 94. molar mass NaHCO = g; 5. ml = 0.05 L NaHCO (s) + HCl(aq) NaCl(aq) + H O(l) 6.0 HCl mol HCl = mol NaHCO required = 0.05 L =

22 g = 1.7 g NaHCO required 95. NiCl (aq) + H S(aq) NiS(s) + HCl(aq) 10. ml = L mol NiCl L = mol NiCl From the balanced equation, mol H S will be required mol H S.4 L = L = 11 ml H S 96. molar mass H O = 18.0 g 1.0 L water = ml water g water HO g H O 18.0 g H O = 56 mol H O ml = L 14.5 mol NH L = 1.45 mol NH 1.45 mol NH.4 L =.5 L ml HCl solution = L HCl solution mol HCl L solution = mol HCl solution mol HCl.4 L = 1.1 L HCl gas at STP 99. we say like dissolves like we mean that two substances will be miscible if they have similar intermolecular forces, so that the forces existing in the mixture will be similar to the forces existing in each separate substance. Molecules do not have to be identical to be miscible, but a similarity in structure (e.g., an OH group) will aid solubility g HCl 100.g solution.1 g HCl = 0.1 g solution 0.1 g solution 1.00 ml solution g solution = 6. ml solution g AgNO g solution 0.50 g AgNO = 00 g solution (.0 10 ml solution) 8

23 10. molar mass CaCl = g CaCl 14. g CaCl g CaCl 50.0 ml = L 0.18 mol CaCl L g CaCl =.56 M = 0.18 mol CaCl 5.0 g KNO 104. a. (5.0 g KNO + 75 g H O) 100 = 6.% KNO b..5 mg = g g KNO (0.005 g KNO g H O) 100 = 0.5% KNO 11 g KNO c. (11 g KNO + 89 g H O) 100 = 11% KNO 11 g KNO d. (11 g KNO + 49 g H O) 100 = 18% KNO 105. To say a solution is 15.0% by mass NaCl means that g of the solution would contain 15.0 g of NaCl 100. g solution a g NaCl = 66.7 g solution 15.0 g NaCl 15.0 g NaCl b. 5.0 g NaCl 100. g solution 15.0 g NaCl = 167 g solution c g NaCl 100. g solution 15.0 g NaCl = 667 g solution d. 1.0 lb = g 100. g solution g NaCl 15.0 g NaCl =.0 10 g solution 106. %C = %Ni = 5.0 g C (5.0 g C g Ni g Fe) 1.5 g Ni (5.0 g C g Ni g Fe) 100 = 4.7% C 100 = 1.4% Ni 9

24 %Fe = 100. g Fe (5.0 g C g Ni g Fe) 100 = 9.9% Fe g dextrose 100. g solution 10. g dextrose = 50 g solution 108. To say that the solution is 5.5% by mass Na CO means that 5.5 g of Na CO are contained in every 100 g of the solution. 5.5 g Na CO 500. g solution 100. g solution = 8 g Na CO 1.5 g KNO g solution 100. g solution = 1.9 g KNO 110. For NaCl: 15 g solution For KBr: 15 g solution 7.5 g NaCl 100. g solution = 9.4 g NaCl.5 g KBr =.1 g KBr 100. g solution 0.10 mol NaPO L = mol Na PO 4 The sample would contain mol of PO 4 and ( ) = mol of Na + ion. 11. a. 5 ml = 0.05 L 0.10 mol CaCl 0.05 L solution = 4.0 M b..5 mol KBr.5 L solution = 1.0 M c. 755 ml = L 0.55 mol NaNO L solution = 0.7 M 4.5 mol Na SO4 d. 1.5 L solution =.6 M 11. a. molar mass BaCl = 08. g 5.0 g BaCl 08. g BaCl = mol BaCl 0.40 mol BaCl.5 L solution = M 0

25 b. molar mass KBr = g; 75 ml = L.5 g KBr = mol KBr g KBr mol KBr L solution = 0.9 M c. molar mass Na CO = g; 175 ml = L 1.5 g Na CO g Na CO = 0.08 mol Na CO 0.08 mol Na CO = 1.16 M L d. molar mass CaCl = g 55 g CaCl = mol CaCl g CaCl mol CaCl 1. L solution = 0.41 M 114. molar mass C 1 H O 11 = 4. g; 450. ml = L 15 g C 1 H O g = 0.65 mol C 1H O mol L solution = 0.81 M 115. molar mass HCl = 6.46 g 49 g HCl = 1.04 mol HCl 6.46 g 1.04 mol HCl solution = 1.0 M 116. molar mass NaCl = g 1.5 g NaCl = mol NaCl g mol NaCl solution = 0.06 M 117. a. 1.5 L solution 4.0 mol H SO solution = 4.5 mol H SO 4 1

26 b. 5 ml = 0.05 L 0.05 L solution 5.4 mol NaCl solution = 0.19 mol NaCl 18 mol HSO4 c. 5. L solution solution = 94 mol H SO 4 d L mol NaF solution = mol NaF 118. a mol KCl 4.5 L solution = mol KCl solution molar mass KCl = 74.6 g 74.6 g KCl mol KCl KCl =. g KCl b ml = L 0.5 mol NaNO L solution solution = mol NaNO molar mass NaNO = g g NaNO mol NaNO = 0.89 g NaNO c. 5 ml = 0.05 L.0 mol HCl 0.05 L solution = mol HCl solution molar mass HCl = 6.46 g 6.46 g HCl mol HCl =.7 g HCl HCl d ml = L mol HSO L solution = mol H SO 4 solution molar mass H SO 4 = g g HSO mol H SO 4 H SO = 4.95 g H SO molar mass AgNO = g AgNO 10. g AgNO g AgNO = mol AgNO mol AgNO solution 0.5 mol AgNO = 0.4 L solution

27 0.50 mol NaPO4 10. a. 1.5 L = 0.15 mol Na PO mol Na PO 4 + mol Na Na PO 4 = 0.98 mol Na+ PO mol Na PO 4 Na PO b..5 ml = L 6.0 mol HSO L + mol H 0.0 H SO 4 H SO 4 = 0. PO 4 = 0.0 H SO = 0.04 mol H+ SO4 0.0 H SO 4 = 0.0 SO 4 H SO c. 5 ml = 0.05 L 0.15 mol AlCl 0.05 L = mol AlCl + Al mol AlCl AlCl = mol Al mol AlCl Cl AlCl = 0.01 Cl 1.5 mol BaCl d L = mol BaCl mol BaCl mol BaCl + Ba BaCl mol Cl BaCl = 1.88 mol Ba + =.75 mol Cl 11. molar mass CaCO = g ; 500. ml = L mol CaCO L = mol CaCO needed g CaCO mol CaCO CaCO = 1.00 g CaCO 1. M1 V1 = M V a. M 1 = 0.00 M M =? V 1 = 15 ml V = = 75 ml

28 M = (0.00 M )(15 ml) (75 ml) = M b. M 1 = 0.50 M M =? V 1 = 155 ml V = = 05 ml M = (0.50 M )(155 ml) (05 ml) = 0.17 M c. M 1 = 0.50 M M =? V 1 = L = 500. ml V = = 650. ml M = (0.50 M )(500. ml) = 0.19 M (650 ml) d. M 1 = 18.0 M M =? V 1 = 15 ml V = = 165 ml M = (18.0 M )(15 ml) (165 ml) = 1.6 M 1. M1 V1 = M V M 1 = 18.0 M V 1 =? V 1 = (0.50 M )(5.0 ml) (18.0 ml) 14. M1 V1 = M V = ml M = 0.50 M V = 5.0 ml M 1 = 5.4 M M =? V 1 = 50. ml M = (5.4 M )(50. ml) = 0.90 M (00. ml) 15. M1 V1 = M V M 1 = 6.0 M M =? V 1 =.0 L M = (6.0 M )(.0 L) (1.0 L) = 1.4 M V = 00. ml V = = 1.0 L ml = L 0.0 mol NiCl L NiCl solution NiCl solution = mol NiCl 4

29 Na S mol NiCl NiCl = mol Na S Na S solution mol Na S 0.10 mol Na S = L = 50. ml Na S solution ml = L; molar mass Ba(NO ) = 61. g 0.19 mol HSO L = mol H SO Ba(NO ) mol H SO 4 = mol Ba(NO ) H SO g Ba(NO ) mol Ba(NO ) Ba(NO ) = g Ba(NO ) 18. HNO (aq) + NaOH(aq) NaNO (aq) + H O(l) 5.0 ml = L mol NaOH L = mol NaOH HNO mol NaOH NaOH = mol HNO mol HNO mol HNO = L = 5.0 ml HNO 19. a. HCl(aq) + NaOH(aq) NaCl(aq) + H O(l) 5.0 ml = L 0.10 mol NaOH L = mol NaOH mol NaOH HCl NaOH = mol HCl mol HCl = L HCl = 10. ml HCl 0.50 mol HCl b. HCl(aq) + Ca(OH) (aq) CaCl (aq) + H O(l) 50.0 ml = L Ca(OH) L = mol Ca(OH) mol HCl mol Ca(OH) = mol HCl Ca(OH) 5

30 mol HCl = L =.00 ml 0.50 mol HCl c. HCl(aq) + NH (aq) NH 4 Cl(aq) 0.0 ml = L 0.6 mol NH L = mol NH HCl mol NH NH = mol HCl mol HCl = L = 18.1 ml 0.50 mol HCl d. HCl(aq) + KOH(aq) KCl(aq) + H O(l) 15.0 ml = L KOH L = mol KOH mol KOH mol HCl HCl KOH = mol HCl 0.50 mol HCl = L = 5.95 ml 10. a. equivalent weight HCl = molar mass HCl = 6.46 g; 500. ml = L 1 equiv HCl 15.0 g HCl = equiv HCl 6.46 g HCl N = equiv L = 0.8 N molar mass g b. equivalent weight H SO 4 = = = g; 50. ml = 0.50 L 1 equiv HSO g H SO g H SO = equiv H SO 4 N = equiv 0.50 L = 4.00 N 4 molar mass 98.0 g c. equivalent weight H PO 4 = = =.67 g; 100. ml = L 1 equiv HPO g H PO 4.67 g H PO = equiv H PO 4 4 N = equiv L =.06 N 6

31 1 equiv HCHO 11. a M HC H O HC H O = 0.50 N HC H O equiv HSO4 b M H SO 4 H SO = N H SO 4 4 c M KOH 1 equiv KOH KOH = 0.10 N KOH 1. molar mass NaH PO 4 = 10.0 g; 500. ml = L NaHPO4 5.0 g NaH PO g NaH PO = mol NaH PO mol L 4 = 0.08 M NaH PO 4 = 0.08 M NaH PO 4 equiv NaHPO M NaH PO 4 NaH PO = N NaH PO 4 = 0.17 N NaH PO NaOH(aq) + H PO 4 (aq) Na PO 4 (aq) + H O(l) 14. ml = L 0.14 HPO L = mol H PO mol NaOH mol H PO 4 H PO = mol NaOH mol NaOH 14. N acid V acid = N base V base 4 N acid (10.0 ml) = (.5 10 N)(7.5 ml) N acid = N HNO mol NaOH = L = 57. ml 7

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