Chapter 8 1 Chapter 8 Solutions Solutions to In-Chapter Problems 8.1 A heterogeneous miture does not have a uniform composition throughout a sample. A solution is a homogeneous miture that contains small particles. Liquid solutions are transparent. A colloid is a homogeneous miture with larger particles, often having an opaque appearance. Cherry Garcia ice cream: heterogeneous miture mayonnaise: colloid c, d, e. seltzer water, nail polish remover, and brass: solutions 8.2 A substance that conducts an electric current in water is called an electrolyte. If a solution contains ions, it will conduct electricity. A substance that does not conduct an electric current in water is called a nonelectrolyte. If a solution contains neutral molecules, it will not conduct electricity. KCl in H 2 O: electrolyte KI in H 2 O: electrolyte sucrose (C 12 H 22 O 11 ) in H 2 O: nonelectrolyte 8.3 Use the general solubility rule: Like dissolves like. Generally, ionic and polar compounds are soluble in water. Nonpolar compounds are soluble in nonpolar solvents. H H NaNO 3 HO C C OH e. ionic compound water soluble H H polar compound water soluble NH 2 OH polar compound water soluble CH 4 nonpolar compound NOT water soluble d. KBr ionic compound water soluble 8.4 Benzene (C 6 H 6 ) and heane (C 6 H 14 ): both nonpolar compounds would form a solution. Na 2 SO 4 and H 2 O: one ionic compound and one polar compound would form a solution. NaCl and heane (C 6 H 14 ): one ionic and one nonpolar compound cannot form a solution. d. H 2 O and CCl 4 : one polar and one nonpolar compound cannot form a solution. 8.5 Identify the cation and anion and use the solubility rules to predict if the ionic compound is water soluble. Li 2 CO 3 : cation Li + with anion CO 3 2 water soluble. gco 3 : cation g 2+ with anion CO 3 2 water insoluble. KBr: cation K + with anion Br water soluble. d. PbSO 4 : cation Pb 2+ with anion SO 4 2 water insoluble. e. CaCl 2 : cation Ca 2+ with anion Cl water soluble. f. gcl 2 : cation g 2+ with anion Cl water soluble.
Solutions 8 2 8.6 agnesium salts are not water soluble unless the anion is a halide, nitrate, acetate, or sulfate. The interionic forces between g 2+ and OH must be stronger than the forces of attraction between the ions and water. Therefore, milk of magnesia is a heterogeneous miture, rather than a solution. 8.7 A soft drink becomes flat when CO 2 escapes from the solution. CO 2 comes out of solution faster at room temperature than at the cooler temperature of the refrigerator because the solubility of gases in liquids decreases as the temperature increases. 8.8 For most ionic and molecular solids, solubility generally increases as temperature increases. The solubility of gases decreases with increasing temperature. Pressure changes do not affect the solubility of liquids and solids. The higher the pressure, the higher the solubility of a gas in a solvent. Na 2 CO 3 (s) N 2 (g) increasing temperature increased solubility decreased solubility decreasing temperature decreased solubility increased solubility increasing pressure no change increased solubility d. decreasing pressure no change decreased solubility 8.9 Use the formula in Eample 8.2 to solve the problem; 525 mg 0.525 g bismuth subsalicylate. (w/v)% 0.525 g bismuth 15 ml solution 100% 3.5% (w/v) bismuth subsalicylate 8.10 Use the formula in Eample 8.2 to solve the problems. (w/v)% (w/v)% 4.3 g ethanol 30. ml solution 0.02 antiseptic 30. ml solution 100% 14% (w/v) ethanol 100% 0.070% (w/v) antiseptic 8.11 Use the formula in Eample 8.3 to solve the problem. (v/v)% 21 ml ethanol 250 ml mouthwash 100% 8.4% (v/v) ethanol
Chapter 8 3 8.12 Use the stepwise analysis in Eample 8.4 to solve the problem. 8.0% (v/v) mouthwash 30. ml known quantities? ml ethanol desired quantity 8.0 ml ethanol 100 ml mouthwash or 100 ml mouthwash 8.0 ml ethanol 8.0 ml ethanol 30. ml mouthwash 2.4 ml ethanol 100 ml mouthwash 8.13 Use the stepwise analysis in Eample 8.5 to solve the problem. 0.50% (w/v) vitamin C 1,000. mg vitamin C known quantities? ml solution desired quantity mg g conversion factors g ml solution conversion factors 1000 mg or 1000 mg 0.50 g vitamin C 100 ml solution or 100 ml solution 0.50 g vitamin C Choose the conversion factors with the unwanted units mg and g in the denominator. 100 ml solution 1000. mg vitamin C 2.0 10 2 ml solution 1000 mg 0.50 g vitamin C 8.14 Use the stepwise analysis in Eample 8.4 to solve the problem. 0.20% (w/v) detromethorphan 2.0% (w/v) guaifenisin 3.0 tsp 15 ml cough syrup known quantities? mg of each drug desired quantity 15 ml cough syrup 0.20 g detromethorphan 1000 mg 30. mg detromethorphan 100 ml cough syrup 15 ml cough syrup 2.0 g guaifenisin 1000 mg 3.0 10 2 mg guaifenisin 100 ml cough syrup
Solutions 8 4 8.15 Use the formula, ppm (g of solute)/(g of solution) 10 6 to calculate parts per million as in Eample 8.6. 0.042 mg DDT 0.000 042 g DDT 1000 mg 0.000 042 g DDT 1,400 g plankton 10 6 0.030 ppm DDT 1.0 kg tissue 1000 g 1,000 g tissue 1.0 kg 5 10 4 g DDT 1,000 g tissue 10 6 0.5 ppm DDT 2.0 mg DDT 0.0020 g DDT 1000 mg 0.0020 g DDT 1,000 g tissue 10 6 2.0 ppm DDT d. 225 µg DDT 0.000 225 g DDT 1,000,000 µg 0.000 225 g DDT 1.0 10 3 g breast milk 10 6 0.23 ppm DDT 8.16 Calculate the of each aqueous solution as in Eample 8.7. 1.0 mol NaCl 0.50 L solution 2.0 250 ml 0.25 L 2.0 mol NaCl 0.25 L solution 8.0 5.0 ml 0.0050 L 0.050 mol NaCl 0.0050 L solution 10. d. 12.0 g NaCl 1 mol NaCl 58.44 g NaCl 0.205 mol NaCl 0.205 mol NaCl 0.10 2.0 L solution e. 24.4 g NaCl 350 ml 1 mol NaCl 58.44 g NaCl 0.418 mol NaCl 0.35 L 0.418 mol NaCl 0.35 L solution 1.2 f. 60.0 g NaCl 750 ml 1 mol NaCl 58.44 g NaCl 1.03 mol NaCl 0.75 L 1.03 mol NaCl 0.75 L solution 1.4
Chapter 8 5 8.17 Calculate the of each solution to determine which has the higher concentration. 10.0 g NaOH 150 ml 1 mol NaOH 40.00 g NaOH 0.250 mol NaOH 0.15 L 0.250 mol NaOH 0.15 L solution 1.7 more concentrated 15.0 g NaOH 250 ml 1 mol NaOH 40.00 g NaOH 0.375 mol NaOH 0.25 L 0.375 mol NaOH 0.25 L solution 1.5 8.18 Use the as a conversion factor to determine the number of milliliters as in Sample Problem 8.9. 1.5 solution 0.15 mol glucose? solution known quantities desired quantity 0.15 mol glucose 1.5 mol/l 0.10 L solution 0.10 L solution 1.0 10 2 ml glucose solution 0.020 mol glucose 1.5 mol/l 0.013 L solution 13 ml glucose solution d. 0.0030 mol glucose 1.5 mol/l 0.0020 L solution 2.0 ml glucose solution 3.0 mol glucose 1.5 mol/l 2.0 L solution 2.0 10 3 ml glucose solution 8.19 Use the as a conversion factor to determine the number of moles. mol (2.0 L)(2.0 mol/l) 4.0 mol NaCl mol (2.5 L)(0.25 mol/l) 0.63 mol NaCl
Solutions 8 6 d. mol (0.025 L)(2.0 mol/l) 0.050 mol NaCl mol (0.25 L)(0.25 mol/l) 0.063 mol NaCl 8.20 Use the to convert the volume of the solution to moles of solute. Then use the molar mass to convert moles to grams as in Eample 8.8. volume molar mass conversion factor 0.10 L solution 1.25 mol NaCl 0.125 mol NaCl 58.44 g NaCl 1 mol NaCl 7.3 g NaCl 1.25 mol NaCl 2.0 L solution 2.5 mol NaCl 58.44 g NaCl 1 mol NaCl 150 g NaCl 1.25 mol NaCl 0.55 L solution 0.69 mol NaCl 58.44 g NaCl 1 mol NaCl 40. g NaCl d. 1.25 mol NaCl 50. ml solution 58.44 g NaCl 1 mol NaCl 3.7 g NaCl 8.21 Use the molar mass to convert grams to moles. Then use the to convert moles of solute to volume of solution. Sample Problem 8.10 is similar, but the order of steps is different. molar mass conversion factor conversion factor 0.500 g sucrose 1 mol sucrose 342.3 g sucrose 0.25 mol sucrose 5.8 ml solution 2.0 g sucrose 1 mol sucrose 342.3 g sucrose 0.25 mol sucrose 23 ml solution 1.25 g sucrose 1 mol sucrose 342.3 g sucrose 0.25 mol sucrose 15 ml solution d. 50.0 mg sucrose 1 mol sucrose 1000 mg 342.3 g sucrose 0.25 mol sucrose 0.58 ml solution
Chapter 8 7 8.22 Calculate a new ( 2 ) using the equation, 1 V 1 2 V 2. 1 V 1 2 V 2 (3.8 )(25.0 ml) (275 ml) 0.35 glucose solution 8.23 Calculate the volume needed (V 1 ) using the equation, 1 V 1 2 V 2. 2 V 2 V1 1 (2.5 )(525 ml) 6.0 220 ml V1 2 V 2 1 2 V 2 V1 1 (4.0 )(750 ml) 6.0 (0.10 )(450 ml) 6.0 5.0 10 2 ml 7.5 ml d. 2 V 2 V1 1 (3.5 )(25 ml) 6.0 15 ml 8.24 Calculate the new concentration of ketamine, and determine the volume needed to supply 75 mg. C 1 V 1 C2 V 2 (100. mg/ml)(2.0 ml) 10.0 ml 20. mg/ml 75 mg 1 ml 20. mg 3.8 ml 8.25 Determine the number of particles contained in the solute. Use 0.51 o C/mol as a conversion factor to relate the temperature change to the number of moles of solute particles, as in Eample 8.10. 2.0 mol of sucrose molecules 0.51 o C 2.0 mol 1.0 o C Boiling point 100.0 C + 1.0 C 101.0 C 2.0 mol of KNO 3 0.51 o C 2.0 mol KNO 3 2 2.0 o C mol KNO 3 Boiling point 100.0 C + 2.0 C 102.0 C
Solutions 8 8 2.0 mol of CaCl 2 0.51 o C 2.0 mol CaCl 2 3 3.1 o C mol CaCl 2 Boiling point 100.0 C + 3.1 C 103.1 C d. 20.0 g of NaCl 20.0 g NaCl 1 mol NaCl 58.44 g NaCl 0.342 mol NaCl 0.51 o C 2 0.342 mol NaCl 0.35 o C mol NaCl Boiling point 100.0 C + 0.35 C 100.35 C rounded to 100.4 C 8.26 Determine the number of particles contained in the solute. Use 1.86 o C/mol as a conversion factor to relate temperature change to the number of moles of solute particles, as in Sample Problem 8.14. 2.0 mol of sucrose molecules 1.86 o C 2.0 mol 3.7 o C elting point 3.7 C 2.0 mol of KNO 3 1.86 o C 2.0 mol KNO 3 2 7.4 o C elting point 7.4 o C mol KNO 3 2.0 mol of CaCl 2 1.86 o C 2.0 mol CaCl 2 3 11 elting point o C 11 o C mol CaCl 2 d. 20.0 g of NaCl 20.0 g NaCl 1 mol NaCl 58.44 g NaCl 0.342 mol NaCl 1.86 o C 2 0.342 mol NaCl 1.27 o C elting point 1.27 o C mol NaCl 8.27 Determine the number of moles of ethylene glycol. Use 1.86 o C/mol as a conversion factor to relate temperature change to the number of moles of solute particles, as in Sample Problem 8.14. 250 g ethylene glycol 1 mol 62.07 g 4.0 mol ethylene glycol temperature decrease 1.86 o C 4.0 mol C 2 H 6 O 2 7.4 C
Chapter 8 9 8.28 The solvent (water) flows from the less concentrated solution to the more concentrated solution. A 5.0% sugar solution has greater osmotic pressure than a 1.0% sugar solution. 4.0 NaCl has greater osmotic pressure than 3.0 NaCl. 0.75 NaCl has greater osmotic pressure than 1.0 glucose solution because NaCl has two particles for each NaCl, whereas glucose has only one. 8.29 Water flows across the membrane. ore water initially flows from the 1.0 side to the more concentrated 1.5 side. When equilibrium is reached there is equal flow of water in both directions. The height of the 1.5 side will be higher and the height of the 1.0 NaCl will be lower. 8.30 A hypotonic solution has a lower osmotic pressure than body fluids. A hypertonic solution has a higher osmotic pressure than body fluids. A 3% (w/v) glucose solution is hypotonic, so it will cause hemolysis. A 0.15 KCl solution is isotonic, so no change results. A 0.15 Na 2 CO 3 solution is hypertonic, so it will cause crenation. Solutions to End-of-Chapter Problems 8.31 A shows KI dissolved in water, since the K + and I ions are separated when KI is dissolved. The solution contains ions, so it conducts an electric current. B shows KI as a molecule, without being separated into ions, a situation that does not occur. 8.32 B shows NaF dissolved in water, since the Na + and F ions are separated when NaF is dissolved. The solution contains ions, so it conducts an electric current. A shows NaF as a molecule, without being separated into ions, a situation that does not occur. 8.33 Use the definitions from 8.1 to classify each substance as a heterogeneous miture, a solution, or a colloid. bronze (alloy of Sn and Cu): solution d. household ammonia: solution diet soda: solution e. gasoline : solution orange juice with pulp: heterogeneous miture f. fog: colloid 8.34 Use the definitions from 8.1 to classify each substance as a heterogeneous miture, a solution, or a colloid. soft drink: solution d. lava rock: heterogeneous miture cream: colloid e. bleach: solution wine: solution f. apple juice: solution 8.35 A solution that has less than the maimum number of grams of solute is said to be unsaturated. A solution that has the maimum number of grams of solute that can dissolve is said to be saturated. A solution that has more than the maimum number of grams of solute is said to be supersaturated.
Solutions 8 10 Adding 30 g to 100 ml of H 2 O at 20 o C: unsaturated Adding 65 g to 100 ml of H 2 O at 50 o C: saturated Adding 20 g to 50 ml of H 2 O at 20 o C: saturated d. Adding 42 g to 100 ml of H 2 O at 50 o C and slowly cooling to 20 o C to give a clear solution with no precipitate: supersaturated 8.36 A solution that has less than the maimum number of grams of solute is said to be unsaturated. A solution that has the maimum number of grams of solute that can dissolve is said to be saturated. A solution that has more than the maimum number of grams of solute is said to be supersaturated. Adding 200 g to 100 ml of H 2 O at 20 C: unsaturated Adding 245 g to 100 ml of H 2 O at 50 C: unsaturated Adding 110 to 50 ml of H 2 O at 20 C: saturated d. Adding 220 g to 100 ml of H 2 O at 50 C and slowly cooling to 20 C to give a clear solution with no precipitate: supersaturated 8.37 Use the general solubility rule: Like dissolves like. Generally, ionic and polar compounds are soluble in water. Nonpolar compounds are soluble in nonpolar solvents. H H C C H O H LiCl H C C CH 3 H C C C H d. C C ionic H H water soluble H H polar water soluble nonpolar NOT water soluble Na 3 PO 4 ionic water soluble 8.38 Use the general solubility rule: Like dissolves like. Generally, ionic and polar compounds are soluble in water, whereas nonpolar compounds are soluble in nonpolar solvents. C 5 H 12 CaCl 2 H C N H d. nonpolar NOT water soluble ionic water soluble 8.39 ethanol will hydrogen bond to water. H H H polar water soluble CH 3 Br polar water soluble hydrogen bond
Chapter 8 11 8.40 Dimethyl ether can hydrogen bond with water. H O H hydrogen bond 8.41 Water-soluble compounds are ionic or are small polar molecules that can hydrogen bond with the water solvent, but nonpolar compounds, such as oil, are soluble in nonpolar solvents. 8.42 A bottle of salad dressing that contains oil and vinegar has two layers because the oil is nonpolar and the vinegar is polar. A heterogenous miture is formed because the oil and vinegar are insoluble in each other. 8.43 Iodine would not be soluble in water but is soluble in CCl 4 since I 2 is nonpolar and CCl 4 is a nonpolar solvent. 8.44 Glycine is soluble in water because the nitrogen and oygen atoms can form hydrogen bonds to water. 8.45 Cholesterol is not water soluble because it is a large nonpolar molecule with a single OH group. 8.46 KCl and CCl 4 do not form a solution. 1-Propanol (C 3 H 8 O) and H 2 O do form a solution. Cyclodecanone (C 10 H 18 O) and H 2 O do not form a solution. d. Pentane (C 5 H 12 ) and heane (C 6 H 14 ) do form a solution. 8.47 The solubility of gases decreases with increasing temperature. The H 2 O molecules are omitted in each representation. Less O 2 is dissolved at 50 C. ore O 2 is dissolved at 10 C.
Solutions 8 12 8.48 The solubility of most ionic solids increases with increasing temperature and decreases with decreasing temperature. The H 2 O molecules are omitted in each representation. Na + Cl ore NaCl is dissolved at 50 o C Less NaCl is dissolved at 10 o C 8.49 For most ionic and molecular solids, solubility generally increases as temperature increases. The solubility of gases decreases as temperature increases. Pressure changes do not affect the solubility of liquids and solids. The higher the pressure, the higher the solubility of a gas in a solvent. NaCl(s) increasing temperature increased solubility decreasing temperature decreased solubility increasing pressure no change d. decreasing pressure no change 8.50 For most ionic and molecular solids, solubility generally increases as temperature increases. The solubility of gases decreases as temperature increases. Pressure changes do not affect the solubility of liquids and solids. The higher the pressure, the higher the solubility of a gas in a solvent. He(g) increasing temperature decreased solubility decreasing temperature increased solubility increasing pressure increased solubility d. decreasing pressure decreased solubility 8.51 A decrease in temperature (a) increases the solubility of a gas and (b) decreases the solubility of a solid. 8.52 A decrease in pressure (a) decreases the solubility of a gas and (b) has no affect on the solubility of a solid. 8.53 any ionic compounds are soluble in water because the ion dipole interactions between ions and water provide the energy needed to break apart the ions from the crystal lattice. The water molecules form a loose shell of solvent around each ion. 8.54 Some ionic compounds are insoluble in water because the attractions between the ions in the crystalline solid are stronger than the forces of attraction between the ions and water.
Chapter 8 13 8.55 Identify the cation and anion and use the solubility rules to predict if the ionic compound is water soluble. K 2 SO 4 : cation K + with anion SO 4 2 water soluble. gso 4 : cation g 2+ with anion SO 4 2 water soluble. ZnCO 3 : cation Zn 2+ with anion CO 3 2 not water soluble. d. KI: cation K + with anion I water soluble. e. Fe(NO 3 ) 3 : cation Fe 3+ with anion NO 3 water soluble. f. PbCl 2 : cation Pb 2+ with anion Cl not water soluble. g. CsCl : cation Cs + with anion Cl water soluble. h. Ni(HCO 3 ) 2 : cation Ni 2+ with anion HCO 3 not water soluble. 8.56 Identify the cation and anion and use the solubility rules to predict if the ionic compound is water soluble. Al(NO 3 ) 3 : cation Al 3+ with anion NO 3 water soluble. NaHCO 3 : cation Na + with anion HCO 3 water soluble. Cr(OH) 2 : cation Cr 2+ with anion OH not water soluble. d. LiOH: cation Li + with anion OH water soluble. e. CuCO 3 : cation Cu 2+ with anion CO 3 2 not water soluble. f. (NH 4 ) 2 SO 4 : cation NH 4 + with anion SO 4 2 water soluble. g. Fe(OH) 3 : cation Fe 3+ with anion OH not water soluble. h. (NH 4 ) 3 PO 4 : cation NH 4 + with anion PO 4 3 water soluble. 8.57 Write two conversion factors for each concentration. 5% (w/v) 5 g 100 ml 100 ml 5 g 10 ppm 10 g 10 6 g 10 6 g 10 g 6.0 6.0 mol 1.0 L 1.0 L 6.0 mol 8.58 Write two conversion factors for each concentration. 15% (v/v) 15 ml 100 ml 100 ml 15 ml 15 ppm 15 g 10 6 g 10 6 g 15 g 12.0 12.0 mol 1.0 L 1.0 L 12.0 mol 8.59 Use the formula in Eample 8.2 to solve the problems. (w/v)% 10.0 g LiCl 750 ml solution 100% 1.3% (w/v) LiCl (w/v)% 25 g NaNO 3 150 ml solution 100% 17% (w/v) NaNO 3 (w/v)% 40.0 g NaOH 500. ml solution 100% 8.00% (w/v) NaOH
Solutions 8 14 8.60 Use the formula in Eample 8.2 to solve the problems. (w/v)% 5.5 g LiCl 550 ml solution 100% 1.0% (w/v) LiCl (w/v)% 12.5 g NaNO 3 250 ml solution 100% 5.0% (w/v) NaNO 3 (w/v)% 20.0 g NaOH 400. ml solution 100% 5.00% (w/v) NaOH 8.61 Use the formula in Eample 8.3 to solve the problem. (v/v)% 25 ml ethyl acetate 150 ml solution 100% 17% (v/v) ethyl acetate 8.62 Use the formula in Eample 8.3 to solve the problem. (v/v)% 75 ml acetone 250 ml solution 100% 30. % (v/v) acetone 8.63 Calculate the of each aqueous solution as in Eample 8.7. 3.5 mol KCl 1.50 L solution 2.3 855 ml solution 0.855 L solution 0.44 mol NaNO 3 0.855 L solution 0.51 25.0 g NaCl 1 mol NaCl 58.44 g 650 ml solution 0.428 mol NaCl 0.65 L solution 0.428 mol NaCl 0.65 L solution 0.66 d. 10.0 g NaHCO 3 1 mol NaHCO 3 84.0 0.119 mol NaHCO 3 0.119 mol NaHCO 3 3.3 L solution 0.036
Chapter 8 15 8.64 Calculate the of each aqueous solution as in Eample 8.7. 2.4 mol NaOH 1.50 L solution 1.6 750 ml solution 0.75 L solution 0.48 mol KNO 3 0.75 L solution 0.64 25.0 g KCl 1 mol KCl 74.55g 650 ml solution 0.335 mol KCl 0.65 L solution 0.335 mol KCl 0.65 L solution 0.52 d. 10.0 g Na 2 CO 3 1 mol Na 2 CO 3 105.99 g 0.0943 mol Na 2 CO 3 0.0943 mol Na 2 CO 3 3.8 L solution 0.025 8.65 Add 12 g of acetic acid to the flask and then water to bring the volume to 250 ml. (w/v)% g acetic acid 250 ml solution 100% 4.8% (w/v) acetic acid 12 g acetic acid Add 55 ml of ethyl acetate to the flask and then water to bring the volume to 250 ml. (v/v)% ml ethyl acetate 250 ml solution 100% 22% (v/v) ethyl acetate 55 ml ethyl acetate Add 37 g of NaCl to the flask and then water to bring the volume to 250 ml. ()(V) (2.5 ) (0.25 L solution) 0.63 mol NaCl 0.63 mol NaCl 58.44 g 1 mol NaCl 37 g NaCl
Solutions 8 16 8.66 Add 5.0 g of KCl to the flask and then water to bring the volume to 250 ml. (w/v)% g KCl 250 ml solution 100% 2.0% (w/v) KCl 5.0 g KCl Add 85 ml of ethanol to the flask and then water to bring the volume to 250 ml. (v/v)% ml ethanol 250 ml solution 100% 34% (v/v) ethanol 85 ml ethanol Add 58 g of NaCl to the flask and then water to bring the volume to 250 ml. ()(V) (4.0 ) (0.25 L solution) 1.0 mol NaCl 1.0 mol NaCl 58.44 g 1 mol NaCl 58 g NaCl 8.67 Calculate the moles of solute in each solution. ()(V) (0.25 ) (0.15 L solution) 0.038 mol NaNO 3 ()(V) (2.0 ) (0.045 L solution) 0.090 mol HNO 3 ()(V) (1.5 ) (2.5 L solution) 3.8 mol HCl 8.68 Calculate the moles of solute in each solution. ()(V) (0.55 ) (0.25 L solution) 0.14 mol NaNO 3 ()(V) (4.0 ) (0.145 L solution) 0.58 mol HNO 3 ()(V) (2.5 ) (6.5 L solution) 16 mol HCl 8.69 Calculate the number of grams of each solute. 0.038 mol NaNO 3 85.00 g NaNO 3 3.2 g NaNO 3 1 mol NaNO 3
Chapter 8 17 0.090 mol HNO 3 3.8 mol HCl 63.02 g HNO 3 5.7 g HNO 3 1 mol HNO 3 36.46 g HCl 140 g HCl 1 mol HCl 8.70 Calculate the number of grams of each solute. 0.14 mol NaNO 3 85.00 g NaNO 3 12 g NaNO 3 1 mol NaNO 3 0.58 mol HNO 3 63.02 g HNO 3 37 g HNO 3 1 mol HNO 3 16 mol HCl 36.46 g HCl 580 g HCl 1 mol HCl 8.71 Calculate the number of milliliters of ethanol in the bottle of wine. (v/v)% ml ethanol 750 ml solution 100% 11.0% (v/v) ethanol 83 ml ethanol 8.72 Use the density and molar mass to calculate the. 20.0 ml ethanol 100 ml 0.790 g ethanol mol ethanol 3.43 1 ml ethanol 46.07 g ethanol 8.73 Calculate the amount of acetic acid (number of grams and number of moles) in the solution and convert to. (w/v)% g acetic acid 1890 ml solution 100% 5.0% (w/v) acetic acid 95 g acetic acid 95 g acetic acid 1 mol acetic acid 60.05 g acetic acid 1.6 mol acetic acid 1.6 mol acetic acid 0.85 1.89 L solution 8.74 Calculate the amount of glucose and convert to. 15 g glucose 100 ml mol glucose 0.83 180.18 g glucose
Solutions 8 18 8.75 Use the formula, ppm (g of solute)/(g of solution) 10 6, to calculate parts per million as in Eample 8.6. 0.000 08 g CHCl 3 80 µg CHCl 3 0.000 08 g CHCl 10 6 0.08 ppm CHCl 3 1,000,000 µg 3 1,000 g 700 µg glyphosate 0.0007 g glyphosate 1,000,000 µg 0.0007 g glyphosate 1,000 g 10 6 0.7 ppm glyphosate 8.76 Use the formula ppm (g of solute)/(g of solution) 10 6, to calculate parts per million as in Eample 8.6. 1,300 µg Cu 0.001 3 g Cu 1,000,000 µg 0.001 3 g Cu 1,000 g 10 6 1.3 ppm Cu 10 µg As 0.000 0 As 1,000,000 µg 0.000 0 As 1,000 g 10 6 0.01 ppm As 100 µg Cr 0.000 Cr 1,000,000 µg 0.000 Cr 1,000 g 10 6 0.1 ppm Cr 8.77 When solution X is diluted, the volume will increase, but the amount of solute will stay the same (B). X A B C Increased volume decreased solute Increased volume same solute 8.78 C is the most concentrated and A is the least concentrated. Same volume decreased solute
Chapter 8 19 8.79 Add H 2 O. B has half as much NaCl, and is therefore half the : 0.05. A B 1 V 1 2 V 2 (0.1 )(50.0 ml) (0.05 )( ml) 100 ml Since the original volume was 50 ml, 50 ml of water is added. 8.80 Add H 2 O. 2.0 3 5 1.2 C D 1 V 1 2 V 2 (2.0 )(100. ml) (1.2 )( ml) 1.7 10 2 ml Since the original volume was 100 ml, 70 ml of water is added. 8.81 Concentration is the amount of solute per unit volume in a solution. Dilution is the addition of solvent to decrease the concentration of solute. 8.82 Dilution is the addition of solvent to decrease the concentration of solute. The amount of solute is a constant. The addition of solvent would decrease the molar concentration of the 2.5 NaOH solution, not increase it, so it is impossible to prepare 200 ml of a 5.0 NaOH solution by diluting a 2.5 NaOH solution. 8.83 Use the equation, (C 1 )(V 1 ) (C 2 )(V 2 ), to calculate the new concentration after dilution. C 1 V 1 C2 V 2 (30.0%)(100. ml) 200. ml 15.0% (w/v) C 1 V 1 C2 V 2 (30.0%)(100. ml) 500. ml 6.00% (w/v) C 1 V 1 C2 V 2 (30.0%)(250 ml) 1500 ml 5.0% (w/v) d. C 1 V 1 C2 V 2 (30.0%)(350 ml) 750 ml 14% (w/v)
Solutions 8 20 8.84 Use the equation (C 1 )(V 1 ) (C 2 )(V 2 ) to calculate the new concentration after dilution. (w/v)% 1.00 g 10.0 ml solution 100% 10.0% (w/v) [1] C 1 V 1 C2 V 2 (10.0%)(1.0 ml) 10.0 ml 1.0% (w/v) [2] C 1 V 1 C2 V 2 (10.0%)(1.0 ml) 2.5 ml 4.0% (w/v) [3] C 1 V 1 C2 V 2 (10.0%)(1.0 ml) 50.0 ml 0.20% (w/v) [4] C 1 V 1 C2 V 2 (10.0%)(1.0 ml) 120 ml 0.083% (w/v) 8.85 Use the equation, ( 1 )(V 1 ) ( 2 )(V 2 ), to calculate the new after dilution. 1 V 1 2 V 2 (12.0 )(125 ml) 850 ml 1.8 8.86 Use the equation ( 1 )(V 1 ) ( 2 )(V 2 ) to calculate the new after dilution. 1 V 1 2 V 2 (6.0 )(250 ml) 450 ml 3.3 8.87 Use the equation, ( 1 )(V 1 ) ( 2 )(V 2 ), to calculate the volume needed to prepare each solution. 2 V 2 V1 1 (1.0 )(25 ml) 2.5 10. ml 2 V 2 V1 1 (0.75 )(1500 ml) 2.5 450 ml 2 V 2 V1 1 (0.25 )(15 ml) 2.5 1.5 ml d. 2 V 2 V1 1 (0.025 )(250 ml) 2.5 2.5 ml 8.88 Use the equation ( 1 )(V 1 ) ( 2 )(V 2 ) to calculate the volume needed to prepare each solution. 2 V 2 V1 1 (4.0 )(45 ml) 5.0 36. ml
Chapter 8 21 2 V 2 V1 1 (0.5 )(150 ml) 5.0 20 ml 2 V 2 V1 1 (0.025 )(1,200 ml) 5.0 6.0 ml d. 2 V 2 V1 1 (1.0 )(750 ml) 5.0 150 ml 8.89 Ocean water contains nonvolatile dissolved salts, which increase the boiling point. 8.90 Osmotic pressure is the pressure that prevents the flow of additional solvent into a solution on one side of a semipermeable membrane. Pure water will not have osmotic pressure since water is on both sides of the semipermeable membrane and there will be no pressure difference. 8.91 Determine the number of particles contained in the solute. Use 0.51 o C/mol as a conversion factor to relate temperature change to the number of moles of solute particles, as in Eample 8.10. 3.0 mol of fructose molecules 0.51 o C 3.0 mol 1.5 o C 100.0 C + 1.5 C 101.5 C 1.2 mol of KI 0.51 o C 2 1.2 mol KI 1.2 o C mol KI 100.0 C + 1.2 C 101.2 C 1.5 mol Na 3 PO 4 0.51 o C 4 1.5 mol Na 3 PO 4 3.1 o C 100.0 C + 3.1 C 103.1 C mol Na 3 PO 4 8.92 Determine the number of particles contained in the solute. Use 1.86 C/mol as a conversion factor to relate temperature change to the number of moles of solute particles, as in Eample 8.10. 3.0 mol of fructose molecules 1.86 o C 3.0 mol 5.6 o C 0.0 C 5.6 C 5.6 C 1.2 mol of KI 1.86 o C 2 1.2 mol KI mol KI 4.5 o C 0.0 C 4.5 C 4.5 C
Solutions 8 22 1.5 mol Na 3 PO 4 1.86 o C 4 1.5 mol Na 3 PO 4 11 o C 0.0 C 11 C 11 C mol Na 3 PO 4 8.93 Determine the number of particles contained in the solute. Use 1.86 o C/mol as a conversion factor to relate temperature change to the number of moles of solute particles, as in Sample Problem 8.14. 150 g ethylene glycol (C 2 H 6 O 2 ) 1 mol ethylene glycol 2.4 mol ethylene glycol 62.07 g ethylene glycol 1.86 o C 2.4 mol 4.5 o C elting point 4.5 o C 8.94 Determine the number of particles contained in the solute. Use 1.86 C/mol as a conversion factor to relate temperature change to the number of moles of solute particles, as in Sample Problem 8.14.!T 0.0 o C 10. o C 10. o C 1.86 o C mol 10. o C mol 5.4 mol 5.4 mol ethylene glycol 62.07 g ethylene glycol 330 g ethylene glycol 1 mol ethylene glycol 8.95 The NaCl solution has a higher boiling point because it contains two particles per mole, whereas the glucose solution contains one per mole. The glucose solution has the higher melting point because it contains one particle per mole, whereas the NaCl solution contains two particles per mole. The NaCl solution has the higher osmotic pressure because it contains two particles per mole, whereas the glucose solution contains one per mole. d. The glucose solution has a higher vapor pressure at a given temperature because it contains one particle per mole, whereas the NaCl solution contains two particles per mole. 8.96 The CaCl 2 solution has a higher boiling point because it contains three particles per mole, whereas the NaCl solution contains two particles per mole. The NaCl solution has the higher melting point because it contains two particles per mole, whereas the CaCl 2 solution contains three particles per mole. The CaCl 2 solution has the higher osmotic pressure because it contains three particles per mole, whereas the NaCl solution contains two particles per mole. d. The NaCl solution has a higher vapor pressure at a given temperature because it contains two particles per mole, whereas the CaCl 2 solution contains three particles per mole.
Chapter 8 23 8.97 Determine the number of particles contained in the solute. If necessary, use 1.86 o C/mol as a conversion factor to relate the temperature change to the number of moles of solute particles. A 0.10 glucose solution has a higher melting point than 0.10 NaOH even though the solutions have the same molar concentration, because the NaOH solution contains twice as many particles. Therefore the NaOH solution will have its melting point reduced by twice as much as the glucose solution. A 0.20 NaCl solution has a higher melting point than a 0.15 CaCl 2 solution. 1.86 o C 1.86 o C 0.20 mol NaCl 2 0.74 o C elting point 0.74 o C mol NaCl higher melting point 0.15 mol CaCl 2 3 0.84 o C elting point 0.84 o C mol CaCl 2 A 0.10 Na 2 SO 4 solution has a higher melting point than 0.10 Na 3 PO 4 even though the solutions have the same molar concentration, because the Na 2 SO 4 solution contains fewer particles (three vs. four). d. A 0.10 glucose solution has a higher melting point than a 0.20 glucose solution since it has a lower molar concentration, which causes less melting point depression. 8.98 Determine the number of particles contained in the solute. If necessary, use 0.51 C/mol as a conversion factor to relate the temperature change to the number of moles of solute particles. A 0.10 NaOH solution has a higher boiling point than 0.10 glucose, even though the solutions have the same molar concentration, because the NaOH solution contains twice as many particles. Therefore, the NaOH solution will have its boling point increased by twice as much as the glucose solution. A 0.15 CaCl 2 solution has a higher boiling point than a 0.20 NaCl solution. 0.51 o C 0.51 o C 0.20 mol NaCl 2 0.20 o C Boiling point 100.20 o C mol NaCl 0.15 mol CaCl 2 3 0.23 o C Boiling point 100.23 o C mol CaCl 2 higher boiling point A 0.10 Na 3 PO 4 solution has a higher boiling point than 0.10 Na 2 SO 4, even though the solutions have the same molar concentration, because the Na 2 SO 4 solution contains fewer particles (three vs. four). d. A 0.20 glucose solution has a higher boiling point than a 0.10 glucose solution since it has a higher molar concentration, which causes an increased boiling point elevation.
Solutions 8 24 8.99 A > B. Since solution A contains a dissolved solute, water will flow from compartment B to compartment A. B > A. Since solution B is more concentrated, water will flow from compartment A to compartment B. No change will occur since the solutions have an equal number of dissolved particles. d. A > B. Solution A contains more dissolved particles than solution B. Therefore, water will flow from compartment B to compartment A. e. No change will occur since the solutions have an equal number of dissolved particles (NaCl has two particles per mole, making it equivalent to the glucose solution). 8.100 Diagram 2 represents the final level of the liquid since solution B is more concentrated. Thus, water will flow from compartment A to compartment B. Diagram 3 represents the final level of the liquid since solution A contains more dissolved particles than solution B. Therefore, water will flow from compartment B to compartment A. Diagram 2 represents the final level of the liquid since solution B contains a dissolved solute. Thus, water will flow from compartment A to compartment B. d. Diagram 3 represents the final level of the liquid since solution A contains a dissolved solute. Thus, water will flow from compartment B to compartment A. e. Diagram 3 represents the final level of the liquid since solution A is more concentrated. Thus, water will flow from compartment B to compartment A. 8.101 At warmer temperatures, CO 2 is less soluble in water and more is in the gas phase and escapes as the can is opened and pressure is reduced. 8.102 Sugar is more soluble at higher temperatures, so more sugar dissolves in hot coffee than iced coffee. 8.103 Calculate the weight/volume percent concentration of glucose, and the. (w/v)% 0.09 g glucose 100 ml blood 100% 0.09% (w/v) glucose 0.09 g glucose 1 mol glucose 180.2 g glucose 0.0005 mol glucose 0.0005 mol glucose 0. blood 0.005 8.104 Convert L to ml and mg to g. 5.0 L 90. mg glucose 100 ml blood 1000 mg 4.5 g glucose 8.105 280 ml of mannitol solution would have to be given. (w/v)% 70. g mannitol ml solution 100% 25% (w/v) mannitol 280 ml The hypertonic mannitol solution draws water out of swollen brain cells and thus reduces the pressure on the brain.
Chapter 8 25 8.106 Calculate the mass of glucose and then convert to moles. 10. g glucose 100 ml solution 750 ml solution 75 g glucose 75 g glucose 1 mol glucose 180.2 g glucose 0.42 mol glucose 8.107 When a cucumber is placed in a concentrated salt solution, water moves out of the cells of the cucumber to the hypertonic salt solution, so the cucumber shrinks and loses its crispness. 8.108 When a raisin is placed in water, water moves into the raisin (hypotonic solution), so the raisin swells. 8.109 NaCl, KCl, and glucose are found in the bloodstream. If they weren t in the dialyzer fluid, they would move out of the bloodstream into the dialyzer, and their concentrations in the bloodstream would fall. 8.110 Pure water is not used in the solution contained in a dialyzer during hemodialysis because water would move out of the dialyzer and into the bloodstream and dilute the blood. 8.111 Convert ounces to milliliters, and then calculate the weight/volume percent concentration. 8.0 oz 29.6 ml 1 oz 240 ml (w/v)% 15 g comple carbohydrates 240 ml solution 100% 6.3% (w/v) comple carbohydrates 8.112 Convert ounces to millimeters, and then calculate the ppm concentration. 8.0 oz 29.6 ml 1 oz 240 ml 1.0 g 240 g solution 1.0 ml 25 mg g 1000 mg 0.025 g g ppm 0.025 g g 240 g solution 10 6 1.0 10 2 ppm 8.113 Use the to determine the number of moles of HCl, and then convert this to grams. mol V 0.10 HCl 2.0 L 0.20 mol HCl 0.20 mol HCl 36.46 g HCl 7.3 g HCl 1 mol HCl
Solutions 8 26 8.114 When a red blood cell is placed in pure water, water will move into the blood cell, causing it to swell and eventually rupture. 8.115 5.0 L 5.0 10 3 ml blood (w/v)% g ethanol 5.0 10 3 ml blood 100% 0.08% (w/v) ethanol 4 g ethanol 1000 mg 4 g ethanol 4,000 mg ethanol 8.116 40. ml ethanol 100 ml solution 250 ml solution 1.0 10 2 ml ethanol 8.117 g acetaminophen 1 ml 10 6 15 ppm acetaminophen 0.000 015 g acetaminophen 0.000 015 g acetaminophen 10 6 µg 15 µg acetaminophen, which is in the therapeutic range 0.000 015 g acetaminophen 1 ml blood 1 mol 5000 ml blood 0.000 50 mol acetaminophen 151.2 g 8.118 The two values are equivalent because 5.2 10 3 is the same as 200. mg/dl to 2 significant figures. 5.2 10 3 mol L 386.74 g mole 1000 mg 2.0 10 2 mg/dl 10 dl