a. Cherry Garcia ice cream: heterogeneous mixture b. mayonnaise: colloid c, d, e. seltzer water, nail polish remover, and brass: solutions

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "a. Cherry Garcia ice cream: heterogeneous mixture b. mayonnaise: colloid c, d, e. seltzer water, nail polish remover, and brass: solutions"

Transcription

1 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.

2 Solutions 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 g bismuth subsalicylate. (w/v)% 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

3 Chapter 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 mg vitamin C 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 mg guaifenisin 100 ml cough syrup

4 Solutions Use the formula, ppm (g of solute)/(g of solution) 10 6 to calculate parts per million as in Eample mg DDT g DDT 1000 mg g DDT 1,400 g plankton ppm DDT 1.0 kg tissue 1000 g 1,000 g tissue 1.0 kg g DDT 1,000 g tissue ppm DDT 2.0 mg DDT g DDT 1000 mg g DDT 1,000 g tissue ppm DDT d. 225 µg DDT g DDT 1,000,000 µg g DDT g breast milk ppm DDT 8.16 Calculate the of each aqueous solution as in Eample mol NaCl 0.50 L solution ml 0.25 L 2.0 mol NaCl 0.25 L solution ml L mol NaCl L solution 10. d g NaCl 1 mol NaCl g NaCl mol NaCl mol NaCl L solution e g NaCl 350 ml 1 mol NaCl g NaCl mol NaCl 0.35 L mol NaCl 0.35 L solution 1.2 f g NaCl 750 ml 1 mol NaCl g NaCl 1.03 mol NaCl 0.75 L 1.03 mol NaCl 0.75 L solution 1.4

5 Chapter Calculate the of each solution to determine which has the higher concentration g NaOH 150 ml 1 mol NaOH g NaOH mol NaOH 0.15 L mol NaOH 0.15 L solution 1.7 more concentrated 15.0 g NaOH 250 ml 1 mol NaOH g NaOH mol NaOH 0.25 L mol NaOH 0.25 L solution Use the as a conversion factor to determine the number of milliliters as in Sample Problem 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 ml glucose solution mol glucose 1.5 mol/l L solution 13 ml glucose solution d mol glucose 1.5 mol/l L solution 2.0 ml glucose solution 3.0 mol glucose 1.5 mol/l 2.0 L solution 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

6 Solutions 8 6 d. mol (0.025 L)(2.0 mol/l) mol NaCl mol (0.25 L)(0.25 mol/l) 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 mol NaCl g NaCl 1 mol NaCl 7.3 g NaCl 1.25 mol NaCl 2.0 L solution 2.5 mol NaCl g NaCl 1 mol NaCl 150 g NaCl 1.25 mol NaCl 0.55 L solution 0.69 mol NaCl g NaCl 1 mol NaCl 40. g NaCl d mol NaCl 50. ml solution 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 g sucrose 1 mol sucrose g sucrose 0.25 mol sucrose 5.8 ml solution 2.0 g sucrose 1 mol sucrose g sucrose 0.25 mol sucrose 23 ml solution 1.25 g sucrose 1 mol sucrose g sucrose 0.25 mol sucrose 15 ml solution d mg sucrose 1 mol sucrose 1000 mg g sucrose 0.25 mol sucrose 0.58 ml solution

7 Chapter 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) ml V1 2 V V 2 V1 1 (4.0 )(750 ml) 6.0 (0.10 )(450 ml) ml 7.5 ml d. 2 V 2 V1 1 (3.5 )(25 ml) 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 mol of sucrose molecules 0.51 o C 2.0 mol 1.0 o C Boiling point C C C 2.0 mol of KNO o C 2.0 mol KNO o C mol KNO 3 Boiling point C C C

8 Solutions mol of CaCl o C 2.0 mol CaCl o C mol CaCl 2 Boiling point C C C d g of NaCl 20.0 g NaCl 1 mol NaCl g NaCl mol NaCl 0.51 o C mol NaCl 0.35 o C mol NaCl Boiling point C C C rounded to 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 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 o C 2.0 mol KNO o C elting point 7.4 o C mol KNO mol of CaCl o C 2.0 mol CaCl elting point o C 11 o C mol CaCl 2 d g of NaCl 20.0 g NaCl 1 mol NaCl g NaCl mol NaCl 1.86 o C 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 g ethylene glycol 1 mol g 4.0 mol ethylene glycol temperature decrease 1.86 o C 4.0 mol C 2 H 6 O C

9 Chapter 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 NaCl has greater osmotic pressure than 1.0 glucose solution because NaCl has two particles for each NaCl, whereas glucose has only one 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 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 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 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.

10 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

11 Chapter 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 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 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 Glycine is soluble in water because the nitrogen and oygen atoms can form hydrogen bonds to water Cholesterol is not water soluble because it is a large nonpolar molecule with a single OH group 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 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.

12 Solutions 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 A decrease in pressure (a) decreases the solubility of a gas and (b) has no affect on the solubility of a solid 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 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.

13 Chapter 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 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 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 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 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 ml solution 100% 17% (w/v) NaNO 3 (w/v)% 40.0 g NaOH 500. ml solution 100% 8.00% (w/v) NaOH

14 Solutions 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 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 mol KCl 1.50 L solution ml solution L solution 0.44 mol NaNO L solution g NaCl 1 mol NaCl g 650 ml solution mol NaCl 0.65 L solution mol NaCl 0.65 L solution 0.66 d g NaHCO 3 1 mol NaHCO mol NaHCO mol NaHCO L solution 0.036

15 Chapter Calculate the of each aqueous solution as in Eample mol NaOH 1.50 L solution ml solution 0.75 L solution 0.48 mol KNO L solution g KCl 1 mol KCl 74.55g 650 ml solution mol KCl 0.65 L solution mol KCl 0.65 L solution 0.52 d g Na 2 CO 3 1 mol Na 2 CO g mol Na 2 CO mol Na 2 CO L solution 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 g 1 mol NaCl 37 g NaCl

16 Solutions 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 g 1 mol NaCl 58 g NaCl 8.67 Calculate the moles of solute in each solution. ()(V) (0.25 ) (0.15 L solution) mol NaNO 3 ()(V) (2.0 ) (0.045 L solution) 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 mol NaNO g NaNO g NaNO 3 1 mol NaNO 3

17 Chapter mol HNO mol HCl g HNO g HNO 3 1 mol HNO g HCl 140 g HCl 1 mol HCl 8.70 Calculate the number of grams of each solute mol NaNO g NaNO 3 12 g NaNO 3 1 mol NaNO mol HNO g HNO 3 37 g HNO 3 1 mol HNO 3 16 mol HCl 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 ml ethanol 100 ml g ethanol mol ethanol ml ethanol 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 g acetic acid 1.6 mol acetic acid 1.6 mol acetic acid L solution 8.74 Calculate the amount of glucose and convert to. 15 g glucose 100 ml mol glucose g glucose

18 Solutions Use the formula, ppm (g of solute)/(g of solution) 10 6, to calculate parts per million as in Eample g CHCl 3 80 µg CHCl g CHCl ppm CHCl 3 1,000,000 µg 3 1,000 g 700 µg glyphosate g glyphosate 1,000,000 µg g glyphosate 1,000 g ppm glyphosate 8.76 Use the formula ppm (g of solute)/(g of solution) 10 6, to calculate parts per million as in Eample ,300 µg Cu g Cu 1,000,000 µg g Cu 1,000 g ppm Cu 10 µg As As 1,000,000 µg As 1,000 g ppm As 100 µg Cr Cr 1,000,000 µg Cr 1,000 g 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

19 Chapter Add H 2 O. B has half as much NaCl, and is therefore half the : 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 Add H 2 O C D 1 V 1 2 V 2 (2.0 )(100. ml) (1.2 )( ml) ml Since the original volume was 100 ml, 70 ml of water is added Concentration is the amount of solute per unit volume in a solution. Dilution is the addition of solvent to decrease the concentration of solute 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 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)

20 Solutions 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 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 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) ml 2 V 2 V1 1 (0.75 )(1500 ml) ml 2 V 2 V1 1 (0.25 )(15 ml) ml d. 2 V 2 V1 1 (0.025 )(250 ml) 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) ml

21 Chapter V 2 V1 1 (0.5 )(150 ml) ml 2 V 2 V1 1 (0.025 )(1,200 ml) ml d. 2 V 2 V1 1 (1.0 )(750 ml) ml 8.89 Ocean water contains nonvolatile dissolved salts, which increase the boiling point 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 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 mol of fructose molecules 0.51 o C 3.0 mol 1.5 o C C C C 1.2 mol of KI 0.51 o C mol KI 1.2 o C mol KI C C C 1.5 mol Na 3 PO o C mol Na 3 PO o C C C C mol Na 3 PO 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 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 mol KI mol KI 4.5 o C 0.0 C 4.5 C 4.5 C

22 Solutions mol Na 3 PO o C mol Na 3 PO 4 11 o C 0.0 C 11 C 11 C mol Na 3 PO 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 g ethylene glycol (C 2 H 6 O 2 ) 1 mol ethylene glycol 2.4 mol ethylene glycol 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 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 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.

23 Chapter 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 o C 1.86 o C 0.20 mol NaCl o C elting point 0.74 o C mol NaCl higher melting point 0.15 mol CaCl 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 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 o C 0.51 o C 0.20 mol NaCl o C Boiling point o C mol NaCl 0.15 mol CaCl o C Boiling point 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.

24 Solutions 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) 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 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 Sugar is more soluble at higher temperatures, so more sugar dissolves in hot coffee than iced coffee 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 g glucose mol glucose mol glucose 0. blood Convert L to ml and mg to g. 5.0 L 90. mg glucose 100 ml blood 1000 mg 4.5 g glucose 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.

25 Chapter 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 g glucose 0.42 mol glucose 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 When a raisin is placed in water, water moves into the raisin (hypotonic solution), so the raisin swells 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 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 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 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 g g ppm g g 240 g solution ppm 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 g HCl 7.3 g HCl 1 mol HCl

26 Solutions When a red blood cell is placed in pure water, water will move into the blood cell, causing it to swell and eventually rupture L ml blood (w/v)% g ethanol ml blood 100% 0.08% (w/v) ethanol 4 g ethanol 1000 mg 4 g ethanol 4,000 mg ethanol ml ethanol 100 ml solution 250 ml solution ml ethanol g acetaminophen 1 ml ppm acetaminophen g acetaminophen g acetaminophen 10 6 µg 15 µg acetaminophen, which is in the therapeutic range g acetaminophen 1 ml blood 1 mol 5000 ml blood mol acetaminophen g The two values are equivalent because is the same as 200. mg/dl to 2 significant figures mol L g mole 1000 mg mg/dl 10 dl

Chemistry 51 Chapter 8 TYPES OF SOLUTIONS. A solution is a homogeneous mixture of two substances: a solute and a solvent.

Chemistry 51 Chapter 8 TYPES OF SOLUTIONS. A solution is a homogeneous mixture of two substances: a solute and a solvent. TYPES OF SOLUTIONS A solution is a homogeneous mixture of two substances: a solute and a solvent. Solute: substance being dissolved; present in lesser amount. Solvent: substance doing the dissolving; present

More information

Solute and Solvent 7.1. Solutions. Examples of Solutions. Nature of Solutes in Solutions. Learning Check. Solution. Solutions

Solute and Solvent 7.1. Solutions. Examples of Solutions. Nature of Solutes in Solutions. Learning Check. Solution. Solutions Chapter 7 s 7.1 s Solute and Solvent s are homogeneous mixtures of two or more substances. consist of a solvent and one or more solutes. 1 2 Nature of Solutes in s Examples of s Solutes spread evenly throughout

More information

Chapter 7 Solutions 1

Chapter 7 Solutions 1 1 Chapter 7 Solutions Solutions: Solute and Solvent Solutions are homogeneous mixtures of two or more substances form when there is sufficient attraction between solute and solvent molecules have two components:

More information

Solutions & Colloids

Solutions & Colloids Chemistry 100 Bettelheim, Brown, Campbell & Farrell Ninth Edition Introduction to General, Organic and Biochemistry Chapter 6 Solutions & Colloids Solutions Components of a Solution Solvent: The substance

More information

Answers and Solutions to Text Problems

Answers and Solutions to Text Problems 9 Answers and Solutions to Text Problems 9.1 a. δ O δ + δ + H H In a water molecule, the oxygen has a partial negative charge and the hydrogens have partial positive charges. b. δ δ + O H δ + δ + δ H H

More information

Chemistry B11 Chapter 6 Solutions and Colloids

Chemistry B11 Chapter 6 Solutions and Colloids Chemistry B11 Chapter 6 Solutions and Colloids Solutions: solutions have some properties: 1. The distribution of particles in a solution is uniform. Every part of the solution has exactly the same composition

More information

Chapter 14. Mixtures

Chapter 14. Mixtures Chapter 14 Mixtures Warm Up What is the difference between a heterogeneous and homogeneous mixture? Give 1 example of a heterogeneous mixture and 1 example of a homogeneous mixture. Today s Agenda QOTD:

More information

Chapter 14 Solutions

Chapter 14 Solutions Chapter 14 Solutions 1 14.1 General properties of solutions solution a system in which one or more substances are homogeneously mixed or dissolved in another substance two components in a solution: solute

More information

From the book (10, 12, 16, 18, 22, 24 52, 54, 56, 58, 62, 64, 66, 68, 74, 76, 78, 80, 82, 86, 88, 90, 92, 106 and 116)

From the book (10, 12, 16, 18, 22, 24 52, 54, 56, 58, 62, 64, 66, 68, 74, 76, 78, 80, 82, 86, 88, 90, 92, 106 and 116) Chem 112 Solutions From the book (10, 12, 16, 18, 22, 24 52, 54, 56, 58, 62, 64, 66, 68, 74, 76, 78, 80, 82, 86, 88, 90, 92, 106 and 116) 1. Which of the following compounds are nonelectrolytes? A. NaF

More information

Chapter 7, Reactions and Solutions

Chapter 7, Reactions and Solutions 1. Classify the following reaction as precipitation, acid-base or oxidation-reduction: Ce4+(aq) + Fe2+(aq) Ce3+(aq) + Fe3+(aq) Ans. oxidation-reduction 2. Classify the following reaction as precipitation,

More information

Solutions CHAPTER Specific answers depend on student choices.

Solutions CHAPTER Specific answers depend on student choices. CHAPTER 15 1. 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

More information

Sample Exercise 13.1 Predicting Solubility Patterns

Sample Exercise 13.1 Predicting Solubility Patterns Sample Exercise 13.1 Predicting Solubility Patterns Predict whether each of the following substances is more likely to dissolve in the nonpolar solvent carbon tetrachloride (CCl 4 ) or in water: C 7 H

More information

A) HCl C) 52 g KCl in 100 g water at 80ºC A) temperature of the solution increases B) supersaturated D) low temperature and high pressure D) KClO3

A) HCl C) 52 g KCl in 100 g water at 80ºC A) temperature of the solution increases B) supersaturated D) low temperature and high pressure D) KClO3 1. Which compound becomes less soluble in water as the temperature of the solution is increased? A) HCl B) 2. The solubility of O3(s) in water increases as the A) temperature of the solution increases

More information

REVIEW QUESTIONS Chapter 8

REVIEW QUESTIONS Chapter 8 Chemistry 51 ANSWER KEY REVIEW QUESTIONS Chapter 8 1. Identify each of the diagrams below as strong electrolyte, weak electrolyte or non-electrolyte: (a) Non-electrolyte (no ions present) (b) Weak electrolyte

More information

Chapter 13: Properties of Solutions

Chapter 13: Properties of Solutions Chapter 13: Properties of Solutions Problems: 9-10, 13-17, 21-42, 44, 49-60, 71-72, 73 (a,c), 77-79, 84(a-c), 91 solution: homogeneous mixture of a solute dissolved in a solvent solute: solvent: component(s)

More information

Two Ways to Form Solutions. Role of Disorder in Solutions 2/27/2012. Types of Reactions

Two Ways to Form Solutions. Role of Disorder in Solutions 2/27/2012. Types of Reactions Role of Disorder in Solutions Disorder (Entropy) is a factor Solutions mix to form maximum disorder Two Ways to Form Solutions 1. Physical Dissolving (Solvation) NaCl(s) Na + (aq) + Cl - (aq) C 12 H 22

More information

Chapter 12: Solutions

Chapter 12: Solutions Chapter 12: Solutions Problems: 3, 5, 8, 12, 14, 16, 22, 29, 32, 41-58, 61-68, 71-74 solution: homogeneous mixture of a solute dissolved in a solvent solute: solvent: component present in smaller amount

More information

Types of Solutions. Chapter 17 Properties of Solutions. Types of Solutions. Types of Solutions. Types of Solutions. Types of Solutions

Types of Solutions. Chapter 17 Properties of Solutions. Types of Solutions. Types of Solutions. Types of Solutions. Types of Solutions Big Idea: Liquids will mix together if both liquids are polar or both are nonpolar. The presence of a solute changes the physical properties of the system. For nonvolatile solutes the vapor pressure, boiling

More information

Name Date Class. SECTION 16.1 PROPERTIES OF SOLUTIONS (pages 471 477)

Name Date Class. SECTION 16.1 PROPERTIES OF SOLUTIONS (pages 471 477) 16 SOLUTIONS SECTION 16.1 PROPERTIES OF SOLUTIONS (pages 471 477) This section identifies the factors that affect the solubility of a substance and determine the rate at which a solute dissolves. Solution

More information

Chapter 13 Properties of Solutions. Classification of Matter

Chapter 13 Properties of Solutions. Classification of Matter Chapter 13 Properties of Solutions Learning goals and key skills: Describe how enthalpy and entropy changes affect solution formation Describe the relationship between intermolecular forces and solubility,

More information

Intermolecular forces, acids, bases, electrolytes, net ionic equations, solubility, and molarity of Ions in solution:

Intermolecular forces, acids, bases, electrolytes, net ionic equations, solubility, and molarity of Ions in solution: Intermolecular forces, acids, bases, electrolytes, net ionic equations, solubility, and molarity of Ions in solution: 1. What are the different types of Intermolecular forces? Define the following terms:

More information

Lecture 6: Lec4a Chemical Reactions in solutions

Lecture 6: Lec4a Chemical Reactions in solutions Lecture 6: Lec4a Chemical Reactions in solutions Zumdahl 6 th Ed, Chapter 4 Sections 1-6. 4.1 Water, the Common Solvent 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes 4.3 The Composition

More information

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) All of the following statements describing solutions are true except A) Solutions are homogeneous.

More information

Unit 13 Practice Test

Unit 13 Practice Test Name: Class: Date: Unit 13 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1) The dissolution of water in octane (C 8 H 18 ) is prevented by.

More information

2. Why does the solubility of alcohols decrease with increased carbon chain length?

2. Why does the solubility of alcohols decrease with increased carbon chain length? Colligative properties 1 1. What does the phrase like dissolves like mean. 2. Why does the solubility of alcohols decrease with increased carbon chain length? Alcohol in water (mol/100g water) Methanol

More information

Solutions. Chapter 13. Properties of Solutions. Lecture Presentation

Solutions. Chapter 13. Properties of Solutions. Lecture Presentation Lecture Presentation Chapter 13 Properties of Yonsei University homogeneous mixtures of two or more pure substances: may be gases, liquids, or solids In a solution, the solute is dispersed uniformly throughout

More information

Chapter 13 Properties of Solutions

Chapter 13 Properties of Solutions Chapter 13 Properties of Solutions 13.1 The Solution Process - Solutions are homogeneous mixtures of two or more pure substances. - In a solution, the solute is dispersed uniformly throughout the solvent.

More information

Chapter 14 The Chemistry of Solutes and Solutions. Solute-Solvent Interactions. Solute-Solvent Interactions. Solute-Solvent Interactions

Chapter 14 The Chemistry of Solutes and Solutions. Solute-Solvent Interactions. Solute-Solvent Interactions. Solute-Solvent Interactions John W. Moore Conrad L. Stanitski Peter C. Jurs Solubility & Intermolecular Forces Solution = homogeneous mixture of substances. It consists of: http://academic.cengage.com/chemistry/moore solvent - component

More information

Chapter 4: Solution Stoichiometry Cont. Aqueous Solutions

Chapter 4: Solution Stoichiometry Cont. Aqueous Solutions Chapter 4: Solution Stoichiometry Cont. 1 Aqueous Solutions Molarity (dilution calculations, solution stoichiometry); Solubility and Solubility Rules Molecular, Ionic and Net Ionic Equations Precipitation

More information

Chapter 13 Properties of Solutions

Chapter 13 Properties of Solutions Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 13 Properties of are homogeneous mixtures of two or more pure substances. In a solution,

More information

SAMPLE PROBLEM 8.1. Solutions of Electrolytes and Nonelectrolytes SOLUTION STUDY CHECK

SAMPLE PROBLEM 8.1. Solutions of Electrolytes and Nonelectrolytes SOLUTION STUDY CHECK Solutions of Electrolytes and Nonelectrolytes SAMPLE PROBLEM 8.1 Indicate whether solutions of each of the following contain only ions, only molecules, or mostly molecules and a few ions: a. Na 2 SO 4,

More information

Chemistry Ch 15 (Solutions) Study Guide Introduction

Chemistry Ch 15 (Solutions) Study Guide Introduction Chemistry Ch 15 (Solutions) Study Guide Introduction Name: Note: a word marked (?) is a vocabulary word you should know the meaning of. A homogeneous (?) mixture, or, is a mixture in which the individual

More information

Unit 3: Solubility Equilibrium

Unit 3: Solubility Equilibrium Unit 3: Chem 11 Review Preparation for Chem 11 Review Preparation for It is expected that the student understands the concept of: 1. Strong electrolytes, 2. Weak electrolytes and 3. Nonelectrolytes. CHEM

More information

Honors Unit 10 Notes Solutions

Honors Unit 10 Notes Solutions Name: Honors Unit 10 Notes Solutions [Chapter 10] Objectives: 1. Students will be able to calculate solution concentration using molarity, molality, and mass percent. 2. Students will be able to interpret

More information

12.3 Colligative Properties

12.3 Colligative Properties 12.3 Colligative Properties Changes in solvent properties due to impurities Colloidal suspensions or dispersions scatter light, a phenomenon known as the Tyndall effect. (a) Dust in the air scatters the

More information

Solutions Review Questions

Solutions Review Questions Name: Thursday, March 06, 2008 Solutions Review Questions 1. Compared to pure water, an aqueous solution of calcium chloride has a 1. higher boiling point and higher freezing point 3. lower boiling point

More information

CHAPTER 13: SOLUTIONS

CHAPTER 13: SOLUTIONS CHAPTER 13: SOLUTIONS Problems: 1-8, 11-15, 20-30, 37-88, 107-110, 131-132 13.2 SOLUTIONS: HOMOGENEOUS MIXTURES solution: homogeneous mixture of substances present as atoms, ions, and/or molecules solute:

More information

Chapter 12. Solutions. Lecture Presentation

Chapter 12. Solutions. Lecture Presentation 12.1 Thirsty Solutions: Why You Shouldn t Drink Seawater 544 12.2 Types of Solutions and Solubility 546 12.3 Energetics of Solution Formation 551 12.4 Solution Equilibrium and Factors Affecting Solubility

More information

Solutions. How Solutions Form

Solutions. How Solutions Form Solutions How Solutions Form Solvent substance doing the dissolving, present in greater amount Definitions Solution - homogeneous mixture Solute substance being dissolved Definitions Solute - KMnO 4 Solvent

More information

Solution concentration = how much solute dissolved in solvent

Solution concentration = how much solute dissolved in solvent Solutions 1 Solutions Concentration Solution concentration = how much solute dissolved in solvent Coffee crystal = solute Water = solvent Liquid Coffee = solution so a solute is dissolved in solvent to

More information

David A. Katz Department of Chemistry Pima Community College

David A. Katz Department of Chemistry Pima Community College Solutions David A. Katz Department of Chemistry Pima Community College A solution is a HOMOGENEOUS mixture of 2 or more substances in a single phase. One constituent t is usually regarded as the SOLVENT

More information

The component present in larger proportion is known as solvent.

The component present in larger proportion is known as solvent. 40 Engineering Chemistry and Environmental Studies 2 SOLUTIONS 2. DEFINITION OF SOLUTION, SOLVENT AND SOLUTE When a small amount of sugar (solute) is mixed with water, sugar uniformally dissolves in water

More information

Chapter 13. Properties of Solutions

Chapter 13. Properties of Solutions 13.4 Ways of Expressing Concentration All methods involve quantifying the amount of solute per amount of solvent (or solution). Concentration may be expressed qualitatively or quantitatively. The terms

More information

The Solution Process CHEMISTRY. Properties of Solutions. The Central Science. Prof. Demi Levendis Room GH807 Gate House

The Solution Process CHEMISTRY. Properties of Solutions. The Central Science. Prof. Demi Levendis Room GH807 Gate House CHEMISTRY The Central Science Properties of Solutions The Solution Process Solutions: Air; brass; body fluids; sea water When a solution forms some questions we can ask are: What happens on a molecular

More information

SCH 3UI Unit 9 Outline: Solutions, Acids and Bases

SCH 3UI Unit 9 Outline: Solutions, Acids and Bases SCH 3UI Unit 9 Outline: Solutions, Acids and Bases Lesson Topics Covered Homework Questions and Assignments 1 Note: Introduction to Solutions review of the organization of matter define: solution, solute,

More information

Guide to Chapter 11. Solutions and their properties

Guide to Chapter 11. Solutions and their properties Guide to Chapter 11. Solutions and their properties We will spend three lecture days on this chapter. You may want to start by reviewing the concepts of heterogeneous solutions (Chapter 2) Read the introductory

More information

Chapter 13. Properties of Solutions

Chapter 13. Properties of Solutions Sample Exercise 13.1 (p. 534) By the process illustrated below, water vapor reacts with excess solid sodium sulfate to form the hydrated form of the salt. The chemical reaction is Na 2 SO 4(s) + 10 H 2

More information

Solution a homogeneous mixture = A solvent + solute(s) Aqueous solution water is the solvent

Solution a homogeneous mixture = A solvent + solute(s) Aqueous solution water is the solvent Solution a homogeneous mixture = A solvent + solute(s) Aqueous solution water is the solvent Water a polar solvent: dissolves most ionic compounds as well as many molecular compounds Aqueous solution:

More information

Aqueous Ions and Reactions

Aqueous Ions and Reactions Aqueous Ions and Reactions (ions, acids, and bases) Demo NaCl(aq) + AgNO 3 (aq) AgCl (s) Two clear and colorless solutions turn to a cloudy white when mixed Demo Special Light bulb in water can test for

More information

Sample Test 1 SAMPLE TEST 1. CHAPTER 12

Sample Test 1 SAMPLE TEST 1. CHAPTER 12 13 Sample Test 1 SAMPLE TEST 1. CHAPTER 12 1. The molality of a solution is defined as a. moles of solute per liter of solution. b. grams of solute per liter of solution. c. moles of solute per kilogram

More information

Chapter Thirteen. Physical Properties Of Solutions

Chapter Thirteen. Physical Properties Of Solutions Chapter Thirteen Physical Properties Of Solutions 1 Solvent: Solute: Solution: Solubility: Types of Solutions Larger portion of a solution Smaller portion of a solution A homogeneous mixture of 2 or more

More information

Ch 8.5 Solution Concentration Units % (m/m or w/w) = mass of solute x 100 total mass of solution mass of solution = mass solute + mass solvent

Ch 8.5 Solution Concentration Units % (m/m or w/w) = mass of solute x 100 total mass of solution mass of solution = mass solute + mass solvent 1 Ch 8.5 Solution Concentration Units % (m/m or w/w) = mass of solute x 100 total mass of solution mass of solution = mass solute + mass solvent % (v/v) = volume of solute x 100 volume of solution filled

More information

CHM1 Review for Exam 9

CHM1 Review for Exam 9 Topics 1. Reaction Types a. Combustion b. Synthesis c. Decomposition d. Single replacement i. Metal activity series ii. Nonmetal activity series e. Double replacement i. Precipitates and solubility rules

More information

EXPERIMENT # 3 ELECTROLYTES AND NON-ELECTROLYTES

EXPERIMENT # 3 ELECTROLYTES AND NON-ELECTROLYTES EXPERIMENT # 3 ELECTROLYTES AND NON-ELECTROLYTES Purpose: 1. To investigate the phenomenon of solution conductance. 2. To distinguish between compounds that form conducting solutions and compounds that

More information

Chapter mol H2S 34.08g H2S L 1 mol H S. = mg O 2 / 100 ml. = 1.48 mg N 2 / 100 ml

Chapter mol H2S 34.08g H2S L 1 mol H S. = mg O 2 / 100 ml. = 1.48 mg N 2 / 100 ml Practice Exercises 1.1 C H S = k H P H S P H S = 1.0 atm C H S = 0.11 mol HS 4.08g HS L 1 mol H S =.7 g L 1.7 g L 1 H S = k H (1.0 atm H s S) k H =.7 g L 1 Hydrogen sulfide is more soluble in water than

More information

Name: Class: Date: 2 4 (aq)

Name: Class: Date: 2 4 (aq) Name: Class: Date: Unit 4 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1) The balanced molecular equation for complete neutralization of

More information

Name Solutions Extra Credit

Name Solutions Extra Credit Name Solutions Extra Credit Page 1 1. Which compound is most soluble in water? A) silver acetate B) silver chloride C) silver nitrate D) silver sulfate 2. According to Reference Table F, which compound

More information

SOLUBILITY CURVES WORKSHEET

SOLUBILITY CURVES WORKSHEET SOLUBILITY CURVES WORKSHEET 1.) Which compound is least soluble at 20 o C? At 80 o C? 2.) Which substance is the most soluble at 10 o C? At 50 o C? At 90 o C? 3.) The solubility of which substance is most

More information

Chapter 6. Solution, Acids and Bases

Chapter 6. Solution, Acids and Bases Chapter 6 Solution, Acids and Bases Mixtures Two or more substances Heterogeneous- different from place to place Types of heterogeneous mixtures Suspensions- Large particles that eventually settle out

More information

Introducing Driving Force #3 - Formation of a Solid

Introducing Driving Force #3 - Formation of a Solid Introducing Driving Force #3 - Formation of a Solid In each of the next two types of chemical reactions, the reactants are aqueous solutions Reactants are ionic substances (solutes) dissolved in water

More information

Review of Basic Concepts, Molarity, Solutions, Dilutions and Beer s Law

Review of Basic Concepts, Molarity, Solutions, Dilutions and Beer s Law Review of Basic Concepts, Molarity, Solutions, Dilutions and Beer s Law Aqueous Solutions In Chemistry, many reactions take place in water. This is also true for Biological processes. Reactions that take

More information

9. The molar mass of hydrazine is 32 g/mol and its empirical formula is NH 2. What is its molecular formula?

9. The molar mass of hydrazine is 32 g/mol and its empirical formula is NH 2. What is its molecular formula? Dr. Rogers Solutions Homework 1. Give the name for the following compounds and state whether they are ionic or not A. Ba(NO 3 ) 2 B. NaH C. PCl 5 D. CO E. NH 4 OH F. Ca(MnO 4 ) 2 G. N 2 O 4 H. NaHSO 4

More information

CHAPTER 4. AQUEOUS REACTION CHEMISTRY

CHAPTER 4. AQUEOUS REACTION CHEMISTRY CAPTER. AQUEOUS REACTION CEMISTRY solution - homogeneous mixture of or more substances; uniform distribution of particles and same properties throughout. A solution is composed of a solute dissolved in

More information

SOLUTIONS UNIT. What composition and properties characterize a true solution(five properties)?

SOLUTIONS UNIT. What composition and properties characterize a true solution(five properties)? SOLUTIONS UNIT Assignment #1 How do heterogeneous mixtures differ from homogeneous mixtures? Of the following, which are homogeneous mixtures and which are heterogeneous mixtures: black coffee, household

More information

Colligative Properties

Colligative Properties Colligative Properties Vapor pressures have been defined as the pressure over a liquid in dynamic equilibrium between the liquid and gas phase in a closed system. The vapor pressure of a solution is different

More information

Chemistry 12 Review Sheet on Unit 3 Solubility of Ionic Substances

Chemistry 12 Review Sheet on Unit 3 Solubility of Ionic Substances Chemistry 12 Review Sheet on Unit 3 Solubility of Ionic Substances 1. Identify each of the following as ionic or molecular substances: a) NaCl (aq)... b) CH 3 COOH (aq)... c) CCl 4(l)... d) HNO 3(aq)...

More information

1. Balance the following equation. What is the sum of the coefficients of the reactants and products?

1. Balance the following equation. What is the sum of the coefficients of the reactants and products? 1. Balance the following equation. What is the sum of the coefficients of the reactants and products? 1 Fe 2 O 3 (s) + _3 C(s) 2 Fe(s) + _3 CO(g) a) 5 b) 6 c) 7 d) 8 e) 9 2. Which of the following equations

More information

CHAPTER 13: ANSWERS TO ASSIGNED PROBLEMS Hauser- General Chemistry I revised 8/03/08

CHAPTER 13: ANSWERS TO ASSIGNED PROBLEMS Hauser- General Chemistry I revised 8/03/08 CHAPTER 13: ANSWERS TO ASSIGNED PROBLEMS Hauser- General Chemistry I revised 8/03/08 13.21 The solubility of Cr(NO 3 ) 3 9 H 2 O in water is 208 g per 100 g of water at 15 C. A solution of Cr(NO 3 ) 3

More information

Chapter 14 Solutes and Solvents

Chapter 14 Solutes and Solvents Chapter 14 Solutes and Solvents A solution is a homogeneous mixture of two or more substances. The relative abundance of the substances in a solution determines which is the solute and which is the solvent.

More information

CHE 107 Exam 1 Spring 2016

CHE 107 Exam 1 Spring 2016 CHE 107 Exam 1 Spring 2016 Your Name: Your ID: Question #: 1 Molecular View State Density Shape Volume Strength of Intermolecu lar Forces solid high definite definite 1 [strong, weak] liquid high indefinite

More information

Unit 6 Water and Its Properties

Unit 6 Water and Its Properties Unit 6 Water and Its Properties 15.1 Water and Its Properties I. Liquid Water A. Surface Tension 1. Surface Tension a. A force that tends to pull adjacent parts of a liquid's surface together, thereby

More information

Chemistry: The Central Science. Chapter 13: Properties of Solutions

Chemistry: The Central Science. Chapter 13: Properties of Solutions Chemistry: The Central Science Chapter 13: Properties of Solutions Homogeneous mixture is called a solution o Can be solid, liquid, or gas Each of the substances in a solution is called a component of

More information

Chemistry Continuing Practice Packet

Chemistry Continuing Practice Packet Chemistry Continuing Practice Packet Mrs. Andrechek Webberville High School The best way to learn and remember chemistry is to practice. Check out the following websites: http://www.chemistrylecturenotes.com

More information

Dr. Taylor Chapter 8 and 9 Homework Chem 111 Spring 2012

Dr. Taylor Chapter 8 and 9 Homework Chem 111 Spring 2012 Dr. Taylor Chapter 8 and 9 Homework Chem 111 Spring 2012 9.1 (12, 103 and 107) Electrolytes 9.2 (15, 21, 23, 25, 93 and 97) 9.3 (27, 31, 33, 35, 101, 105 and 111) 9.4 (45,47,49 and 51) 8.2 (17, 19, 21

More information

Unit 10: Solutions. Student Name: Key. Class Period: _3, 5, & 10_. Page 1 of 61. Website upload

Unit 10: Solutions. Student Name: Key. Class Period: _3, 5, & 10_. Page 1 of 61. Website upload Unit 10: Solutions Student Name: Class Period: _3, 5, & 10_ Page 1 of 61 Page intentionally blank Page 2 of 61 Unit 10 Vocabulary: 1. Aqueous: A solution in which the solvent is water. 2. Colligative Property:

More information

Calculation of Molar Masses. Molar Mass. Solutions. Solutions

Calculation of Molar Masses. Molar Mass. Solutions. Solutions Molar Mass Molar mass = Mass in grams of one mole of any element, numerically equal to its atomic weight Molar mass of molecules can be determined from the chemical formula and molar masses of elements

More information

Molarity Practice Worksheet

Molarity Practice Worksheet Molarity Practice Worksheet Find the molarity of the following solutions: 1) 0.5 moles of sodium chloride is dissolved to make 0.05 liters of solution. 2) 0.5 grams of sodium chloride is dissolved to make

More information

Chapter 3 mastery check

Chapter 3 mastery check 1. Each water molecules is capable of forming a. one hydrogen bond b. three hydrogen bonds c. four hydrogen bonds d. two covalent bonds and two hydrogen bonds e. four polar covalent bonds 2. The polar

More information

Aqueous Solutions. Water is the dissolving medium, or solvent. Some Properties of Water. A Solute. Types of Chemical Reactions.

Aqueous Solutions. Water is the dissolving medium, or solvent. Some Properties of Water. A Solute. Types of Chemical Reactions. Aqueous Solutions and Solution Stoichiometry Water is the dissolving medium, or solvent. Some Properties of Water Water is bent or V-shaped. The O-H bonds are covalent. Water is a polar molecule. Hydration

More information

Chapter 13: Physical Properties of Solutions

Chapter 13: Physical Properties of Solutions Chapter 13: Physical Properties of Solutions Key topics: Molecular Picture (interactions, enthalpy, entropy) Concentration Units Colligative Properties terminology: Solution: a homogeneous mixture Solute:

More information

1/27/2014. Chapter 12. Solutions. Thirsty Seawater. Seawater. Seawater. Homogeneous Mixtures. Seawater. Lecture Presentation

1/27/2014. Chapter 12. Solutions. Thirsty Seawater. Seawater. Seawater. Homogeneous Mixtures. Seawater. Lecture Presentation Lecture Presentation Chapter 12 Solutions Sherril Soman, Grand Valley State University Thirsty Seawater Drinking seawater can cause dehydration. Seawater Is a homogeneous mixture of salts with water Contains

More information

SOLUTIONS EXPERIMENT 13

SOLUTIONS EXPERIMENT 13 SOLUTIONS EXPERIMENT 13 OBJECTIVE The objective of this experiment is to demonstrate the concepts of concentrations of solutions and the properties of solution. Colloids will be demonstrated. EQUIPMENT

More information

Chemistry Continuing Practice Packet

Chemistry Continuing Practice Packet Chemistry Continuing Practice Packet Mrs. Andrechek Webberville High School The best way to learn and remember chemistry is to practice. Check out the following apps: - Mahjong Chem - Formulas Lite - Atoms

More information

CHEMICAL REACTIONS. When sodium bicarbonate, e.g. baking soda, is combined with an acid, a gas is produced

CHEMICAL REACTIONS. When sodium bicarbonate, e.g. baking soda, is combined with an acid, a gas is produced CHEMICAL REACTIONS OBJECTIVES 1. To study reactions between ions in aqueous solutions 2. To observe exothermic and endothermic reactions 3. To study oxidation-reduction reactions 4. To practice balancing

More information

1. Define the term colligative property and list those physical properties of a solution that can be classified as colligative properties.

1. Define the term colligative property and list those physical properties of a solution that can be classified as colligative properties. Solutions Colligative Properties DCI Name Section 1. Define the term colligative property and list those physical properties of a solution that can be classified as colligative properties. Colligative

More information

Solutions. ... the components of a mixture are uniformly intermingled (the mixture is homogeneous). Solution Composition. Mass percentageof solute=

Solutions. ... the components of a mixture are uniformly intermingled (the mixture is homogeneous). Solution Composition. Mass percentageof solute= Solutions Properties of Solutions... the components of a mixture are uniformly intermingled (the mixture is homogeneous). Solution Composition 1. Molarity (M) = 4. Molality (m) = moles of solute liters

More information

A. Types of Mixtures:

A. Types of Mixtures: I. MIXTURES: SOLUTIONS 1) mixture = a blend of two or more kinds of matter, each of which retains its own identity and properties a) homogeneous mixture = a mixture that is uniform in composition throughout

More information

SOLUTE - SOLVENT SYSTEM

SOLUTE - SOLVENT SYSTEM SOLUTIONS: SOLUTE - SOLVENT SYSTEM SCH4U_08 09 Solubility The term solubility is commonly used in two senses qualitatively and quantitatively. Qualitatively, solubility is often used in a relative way

More information

13.3 Factors Affecting Solubility Solute-Solvent Interactions Pressure Effects Temperature Effects

13.3 Factors Affecting Solubility Solute-Solvent Interactions Pressure Effects Temperature Effects Week 3 Sections 13.3-13.5 13.3 Factors Affecting Solubility Solute-Solvent Interactions Pressure Effects Temperature Effects 13.4 Ways of Expressing Concentration Mass Percentage, ppm, and ppb Mole Fraction,

More information

Exercise 3.5 - Naming Binary Covalent Compounds:

Exercise 3.5 - Naming Binary Covalent Compounds: Chapter Exercise Key 1 Chapter Exercise Key Exercise.1 Classifying Compounds: Classify each of the following substances as either a molecular compound or an ionic compound. a. formaldehyde, CH 2 O (used

More information

C) 6 D) thermal energy B) Aluminum reacts with sulfuric acid. C) 1676 kj B) kilojoules C) Magnesium reacts with an acid. D) 100 C and 1 atm

C) 6 D) thermal energy B) Aluminum reacts with sulfuric acid. C) 1676 kj B) kilojoules C) Magnesium reacts with an acid. D) 100 C and 1 atm 1. What is the atomic number of the element whose atoms bond to each other in chains, rings, and networks? A) 10 B) 8 C) 6 D) 4 2. Which statement describes a chemical property of aluminum? A) Aluminum

More information

Solubility Rules First things first, you have to memorize the basic solubility rules in order to (1) know which salts dissociate (break apart) in solu

Solubility Rules First things first, you have to memorize the basic solubility rules in order to (1) know which salts dissociate (break apart) in solu A solution is a homogeneous mixture of a solute distributed through a solvent substance being dissolved substance doing the dissolving Solutions can exist in any phase gas -- air is a good example. It

More information

stoichiometry = the numerical relationships between chemical amounts in a reaction.

stoichiometry = the numerical relationships between chemical amounts in a reaction. 1 REACTIONS AND YIELD ANSWERS stoichiometry = the numerical relationships between chemical amounts in a reaction. 2C 8 H 18 (l) + 25O 2 16CO 2 (g) + 18H 2 O(g) From the equation, 16 moles of CO 2 (a greenhouse

More information

Experiment 9 Chem 110 Lab SOLUTIONS I. INTRODUCTION. Polar or Nonpolar? 1 ethanol (ethyl alcohol) C 2 H 6 O. 2 cyclohexane, C 6 H 12

Experiment 9 Chem 110 Lab SOLUTIONS I. INTRODUCTION. Polar or Nonpolar? 1 ethanol (ethyl alcohol) C 2 H 6 O. 2 cyclohexane, C 6 H 12 Experiment 9 Chem 110 Lab SOLUTIONS I. INTRODUCTION A solution is a homogeneous mixture of two (or more) substances. It is composed of a solvent and a dissolved material called a solute. The solute is

More information

Chapter 11 Properties of Solutions

Chapter 11 Properties of Solutions Chapter 11 Properties of Solutions 11.1 Solution Composition A. Molarity moles solute 1. Molarity ( M ) = liters of solution B. Mass Percent mass of solute 1. Mass percent = 1 mass of solution C. Mole

More information

R = J/mol K R = L atm/mol K

R = J/mol K R = L atm/mol K version: master Exam 1 - VDB/LaB/Spk This MC portion of the exam should have 19 questions. The point values are given with each question. Bubble in your answer choices on the bubblehseet provided. Your

More information

Exam 3 Chemistry 65 Summer Score:

Exam 3 Chemistry 65 Summer Score: Name: Exam 3 Chemistry 65 Summer 2015 Score: Instructions: Clearly circle the one best answer 1. The main interactions between molecules of iodine I2 are examples of A) ionic bonds. B) covalent bonds.

More information

Lab 9. Colligative Properties an Online Lab Activity

Lab 9. Colligative Properties an Online Lab Activity Prelab Assignment Before coming to lab: Lab 9. Colligative Properties an Online Lab Activity Chemistry 162 - K. Marr Revised Winter 2014 This lab exercise does not require a report in your lab notebook.

More information

Molarity is used to convert between moles of substance and liters of solution.

Molarity is used to convert between moles of substance and liters of solution. Appendix C Molarity C.1 MOLARITY AND THE MOLE The molar mass is the mass of a mole of a pure substance while the molarity, M, is the number of moles of a pure substance contained in a liter of a solution.

More information