Ions in Aqueous Solutions and Colligative Properties

CHAPTER 13 Ions in Aqueous Solutions and Colligative Properties

Practice, p. 436

⎯H2O + − 1a. Given: dissolution of am- NH4Cl(s) U NH4 (aq) + Cl (aq) monium chloride, + 1 mol NH 1 mol ammonium 4 − chloride 1 mol Cl Unknown: Chemical 2 mol total ions equation, mol of each ion mol of total ions

b. Given: dissolution of H2O + − Na S(s) ⎯U 2 Na (aq) + S2 (aq) sodium sulfide, 2 + 1 mol sodium 2 mol Na sulfide − 1 mol S2 Unknown: chemical equation, 3 mol total ions mol of each ion mol of total ions

c. Given: dissolution of H2O + − Ba(NO ) (s) ⎯U Ba2 (aq) + 2NO (aq) barium nitrate, 3 2 3 + 0.50 mol barium 0.50 mol Ba2 nitrate − 1.0 mol NO Unknown: chemical 3 equation, 1.50 mol total ions mol of each ion mol of total ions

ATE, Additional Sample Problems, p. 436

⎯H2O 2+ − A-1. Given: amount of solute Mg(ClO3)2(s) U Mg (aq) + 2 ClO3(aq) = 1 mol Mg(ClO3)2 2+ − 1 mol Mg + 2 mol ClO3 = 3 mol ions produced solvent = H2O Unknown: number of moles of ions produced

MODERN CHEMISTRY CHAPTER 13 SOLUTIONS MANUAL Copyright © by Holt, Rinehart and Winston. All rights reserved. 149 ⎯H2O + − A-2. Given: amount of solute NH4NO3(s) U NH4 (aq) + NO3(aq) = 3.5 mol + − 3.5 mol NH4 + 3.5 mol NO3 = 7.0 mol ions produced NH4NO3 solvent = H2O Unknown: number of moles of ions produced

Section Review, p. 443

⎯H2O 2+ − 1. Given: dissolution of Sr(NO3)2(s) U Sr (aq) + 2NO3(aq) 0.5 mol strontium + 0.5 mol Sr2 nitrate − Unknown: Chemical 1.0 mol NO3 equation + moles Sr2 − moles NO3

ATE, Additional Sample Problems, p. 449

C-1. Given: mass of solute = ∆tf = Kfm 58.0 g C H O 6 12 6 mol C H O mass of solvent = ⎯6⎯12 6 = (58.0 g C6H12O6) 0.322 mol C6H12O6 180. g C6H12O6 0.185 kg H2O moles of solute Unknown: freezing point m = ⎯⎯⎯ of solution mass of solvent (kg) 0.322 mol m = ⎯⎯ = 1.74 m 0.185 kg

∆tf = (−1.86°C/m)(1.74 m) =−3.24°C

f.p. solution = f.p. solvent +∆tf = 0.000°C + (−3.24°C) =−3.24°C

C-2. Given: mass of solute = mol H NCONH 39.2 g (39.2 g H NCONH ) ⎯⎯⎯2 2 = 0.6527 mol H NCONH 2 2 60.06 g H NCONH 2 2 H2NCONH2 2 2 mass of solvent = moles of solute m = ⎯⎯⎯ 0.485 kg acetic mass of solvent (kg) acid 0.6527 mol m = ⎯⎯ = 1.346 m Unknowns: molality (m) 0.485 kg of solution, freezing ∆tf = Kfm point of = (−3.90°C/m)(1.346 m) =−5.248°C solution f.p. solution = f.p. solvent +∆tf f.p. = 16.6°C + (−5.246°C) = 11.4°C

MODERN CHEMISTRY CHAPTER 13 SOLUTIONS MANUAL Copyright © by Holt, Rinehart and Winston. All rights reserved. 150 D-1. Given: f.p. of solution = f.p. solution = f.p. solvent +∆tf −6.40°C ∆t = f.p. solution − f.p. solvent mass of solvent = f =− ° − ° =− ° (500. gH2O) 6.40 C 0.000 C 6.40 C

Unknowns: molality ∆tf = Kfm (m) of m =∆t /K solution of f f = (−6.40°C)/(−1.86°C/m) = 3.44 m HOCH2CH2OH, mass of solute moles of solute needed m = ⎯⎯⎯ mass of solvent (kg) moles of solute = (m)(mass of solvent) = (3.44)(0.500) = 1.72 mol

62.0⎯⎯⎯ g HOCH2CH2OH = (1.72 mol HOCH2CH2OH)107 g HOCH2CH2OH mol HOCH2CH2OH

Practice, p. 450

1. Given: mass of solute = mol C H O (10.3 g C H O ) ⎯⎯6 12 6 = 0.0572 mol C H O 10.3 g C H O 6 12 6 6 12 6 6 12 6 180. g C6H12O6 mass of solvent = moles of solute 0.250 kg H2O m = ⎯⎯⎯ mass of solvent (kg) Unknown: freezing-point 0.0572 mol depression of m = ⎯⎯ = 0.22 m solution 0.250 kg

∆tf = Kfm

∆tf = (−1.86°C/m)(0.2288 m) =−0.426°C

2. Given: f.p. of solution = f.p. solution = f.p. solvent +∆tf −0.325°C ∆tf = f.p. solution − f.p. solvent solute = C6H12O6 =− ° − ° =− ° solvent = H2O 0.325 C 0.000 C 0.325 C

Unknown: molality (m) of ∆tf = Kfm solution m =∆tf/Kf

(−0.325°C) = ⎯⎯ = 0.175 m (−1.86°C/m)

3. Given: amount of solute moles of solute = 0.500 mol m = ⎯⎯⎯ amount of solvent mass of solvent (kg) 0.5000 kg ether 0.500 mol = m = ⎯⎯ = 1.00 m f.p. ether 0.500 kg −116.3°C ∆tf = Kfm Unknown: freezing point ∆ = − ° =− ° of solution tf ( 1.79 C/m)(1.00 m) 1.79 C

f.p. solution = f.p. ether +∆tf =−116.3°C + (−1.79°C) =−118.1°C

a. Unknown: freezing- f.p. solution = f.p. solvent +∆tp point de- ∆t = f.p. solution − f.p. solvent pression of p solution =−9.0°C − 0.0°C =−9.0°C

b. Unknown: molality (m) ∆tf = Kfm of solution m =∆tf/Kf

(−9.0°C) = ⎯⎯ = 4.8 m (−1.86°C/m)

Practice, p. 451

1. Given: mass of solute = mol C H O (50.0 g C H O ) ⎯⎯12 22 11 = 0.146 mol C H O 50.0 g C H O 12 22 11 12 22 11 12 22 11 342 g C12H22O11 (sucrose) = moles of solute mass of solvent m = ⎯⎯⎯ 0.500 kg H2O mass of solvent (kg) 0.146 mol Unknown: boiling-point m = ⎯⎯ = 0.292 m elevation 0.500 kg

∆tb = Kbm

∆tb = (0.51°C/m)(0.292 m) = 0.15°C

2. Given: mass of solute = mol C H O (450.0 g C H O ) ⎯⎯12 22 11 = 1.32 mol C H O 450.0 g C H O 12 22 11 12 22 11 12 22 11 342 g C12H22O11 (sucrose) = moles of solute mass of solvent m = ⎯⎯⎯ 0.250 kg H2O mass of solvent (kg) b.p. H O = 100.0°C 2 1.32 mol m = ⎯⎯ = 5.28 m Unknown: boiling point of 0.250 kg solution ∆tb = Kbm

∆tb = (0.51°C/m)(5.28 m) = 2.7°C

b.p. solution = b.p. solvent +∆tb = 100.0°C + 2.7°C = 102.7°C

3. Given: boiling-point ele- ∆tb = Kbm vation = 1.02°C m =∆tb/Kb solvent = H2O = 1.02°C/(0.51°C/m) = 2.0 m Unknown: molality (m) of solution

a. Unknown: boiling-point b.p. solution = b.p. solvent +∆tb elevation ∆tb = b.p. solution − b.p. solvent = 100.75°C − 100.0°C = 0.75°C

b. Unknown: molality (m) of ∆tb = Kbm solution m =∆tb/Kb

0.75°C = ⎯⎯ = 1.5 m (0.51°C/m)

ATE, Additional Sample Problems, p. 451

E-1. Given: mass of ⎯⎯⎯mol HOCH2CH2OC4H9 = (25.0 g HOCH2CH2OC4H9) solute 25.0 g 118 g HOCH2CH2OC4H9 HOCH CH OC H 2 2 4 9 = (butyl cellosolve) 0.212 mol HOCH2CH2OC4H9 = mass of solvent moles of solute 0.0687 kg ether m = ⎯⎯⎯ mass of solvent (kg) Unknown: boiling point of 0.212 mol solution m = ⎯⎯ = 3.09 m 0.0687 kg

∆tb = Kbm = (2.02°C/m)(4.37 m) = 6.24°C

b.p. solution = b.p. solvent +∆tb = 34.6°C + 6.24°C = 40.8°C

E-2. Given: mass of solvent = b.p. solution = b.p. solvent +∆tb 1.00 kg H2O ∆ = − boiling point of so- tb b.p. solution b.p. solvent lution = 104.5°C = 104.5°C − 100.0°C = 4.5°C solute = CH2OH- ∆tb = Kbm CHOHCH2OH (glycerol) m =∆tb/Kb = ° ° = Unknown: mass of solute 4.5 C/(0.51 C/M) 8.8 m moles of solute m = ⎯⎯⎯ mass of solvent (kg) moles of solute = (m)(mass of solvent) = (8.8 mol/kg)(1.00 kg) = 8.8 mol 92.08 g glycerol (8.8 mol glycerol) ⎯⎯ = 810 g glycerol mol glycerol

MODERN CHEMISTRY CHAPTER 13 SOLUTIONS MANUAL Copyright © by Holt, Rinehart and Winston. All rights reserved. 153 Practice, p. 455 moles of solute 2.0 mol 1. Given: solute = 2.0 mol m ==⎯⎯⎯ ⎯ mass of solvent (kg) 1.0 kg MgSO4 = 2+ 2− mass of solvent MgSO4 (s) U Mg (aq) + SO4 (aq) 1.0 kg H2O Each formula unit of MgSO4 yields two ions in solution. Unknown: expected ∆ = freezing-point tf Kfm depression of −1.86°C • kg H2O 2.0 mol MgSO4 2 mol ions solution = ⎯⎯⎯ ⎯⎯ ⎯⎯ mol ions kg H2O mol MgSO4 =−7.4°C

2. Given: mass of solute = mol NaCl (150 g NaCl) ⎯⎯ = 2.57 mol NaCl 150 g NaCl 58.44 g NaCl mass of solvent = moles of solute 2.57 mol 1.0 kg H2O m ==⎯⎯⎯ ⎯⎯ mass of solvent (kg) 1.0 kg Unknown: expected + − boiling-point NaCl(s) U Na (aq) + Cl (aq) elevation Each formula unit of NaCl yields two ions in solution. of solution ∆tb = Kbm ° • 2.57 mol NaCl = ⎯⎯0.51 C kg H2O ⎯⎯2⎯⎯ mol ions mol ions kg H2O mol NaCl = 2.6°C

3. Given: solute = NaCl f.p. solution = f.p. solvent +∆tf solvent = H O 2 ∆t = f.p. solution − f.p. solvent freezing point of f solution =−0.20°C =−0.20°C − 0.0°C =−0.20°C + − freezing point of NaCl(s) U Na (aq) + Cl (aq) H2O = 0.0°C Each formula unit of NaCl yields two ions in solution. Unknown: molality (m) ∆tf = Kfm

m =∆tf/Kf −0.20°C • mol ions 1 mol NaCl = ⎯⎯⎯ ⎯⎯ = 0.054 m NaCl −1.86°C 2 mol ions

Section Review, p. 456

2. Given: amount of solute = moles of solute m = ⎯⎯⎯ 2 mol mass of solvent (kg) amount of solvent = 2 mol 1 kg ==⎯ 2 m freezing point of 1 kg = ° solution 7.8 C ∆t = K m below its normal f f freezing point Kf =∆tf/m =− ° =− ° Unknowns: molal freezing- 7.8 C/2 m 3.9 C/m point constant Identity of solvent: acetic acid of unknown solvent (Kf), identity of solvent

MODERN CHEMISTRY CHAPTER 13 SOLUTIONS MANUAL Copyright © by Holt, Rinehart and Winston. All rights reserved. 154 4. a. Given: molality (m) of + − KNO3(s) U K (aq) + NO3(aq) solution = 0.2 m Each formula unit of KNO yields two ions in solution. solute = KNO3 3 ∆ = Unknown: expected tf Kfm freezing- −1.86°C • kg H O 0.2 mol KNO 2 mol/ions point = ⎯⎯2 ⎯⎯3 ⎯⎯ depression mol/ions kg H2O mol/KNO3 =−0.744°C

Chapter Review

H O + − 1. a. Given: solute = KCl KCl(s) ⎯2U K (aq) + Cl (aq) + − Unknown: number of Ions produced = 1 mol K + 1 mol Cl = 2 mol moles of ions present ⎯H2O 2+ − Mg(NO3)2 U Mg (aq) + 2NO3(aq) b. Given: amount of + − Ions produced = 1 mol Mg2 + = solvent = 1 L 2 mol NO3 3 mol M of solution = 1 M solute = Mg(NO3)2 Unknown: number of moles of ions present

9. Given: dissolution of H2O + − KI(s) ⎯U K (aq) + I (aq) KI, NaNO3, MgCl2 + and Na2SO4 in 1 mol K water, − 1 mol of each 1 mol I compound 2 mol total ions Unknown: mol of ⎯H2O + − each ion, NaNO3(s) U Na (aq) + 2NO3(aq) mol of + total ions 1 mol Na − 1 mol NO3 2 mol total ions

⎯H2O 2+ − MgCl2(s) U Mg (aq) + 2Cl (aq) + 1 mol Mg2 − 2 mol Cl 3 mol total ions

⎯H2O + 2− Na2SO4(s) U 2Na (aq) + SO4 (aq) + 2 mol Na 2− 1 mol SO4 3 mol total ions

MODERN CHEMISTRY CHAPTER 13 SOLUTIONS MANUAL Copyright © by Holt, Rinehart and Winston. All rights reserved. 155 ⎯H2O 2+ − 10. a. Given: amount of Sr(NO3)2(s) U Sr (aq) + 2NO3(aq) 2+ solute = 1 mol Sr + Ions produced: 0.5 mol Sr(NO ) × ⎯⎯ = 0.5 mol Sr2 0.50 mol 3 2 1molSr(NO3)2 Sr(NO3)2 − 2 mol NO3 − solvent = H2O 0.5 mol Sr(NO3)2 × ⎯⎯ = 1.00 mol NO3 1mol Sr(NO3)2 Unknown: total num- 2+ − ber of moles 0.50 mol Sr + 1.00 mol NO3 = 1.50 mol ions of solute ions formed

H O + − b. Given: amount of Na PO (s) ⎯2U 3Na (aq) + PO3 (aq) 3 4 4 + solute = 3 mol Na + Ions produced: 0.5 mol Na PO × ⎯⎯ = 1.5 mol Na 0.50 mol 3 4 1mol Na3PO4 Na3PO4 3− ⎯1 molP⎯O4 3− 0.5 mol Na3PO4 × = 0.5 mol PO4 Unknown: total num- 1 mol Na3PO4 ber of moles + 3− of solute 1.50 mol Na + 0.50 mol PO4 = 2.00 mol ions ions formed

13. Given: mass of CuCl = mol CuCl 2 (13.45 g CuCl ) ⎯⎯2 = 0.10004 mol CuCl 13.45 g 2 2 134.45 g CuCl2 Unknown: maximum ⎯1 mol ⎯PbCl2 amount of 0.10004 mol CuCl2 × = 0.10004 mol PbCl2 precipitate 1 mol CuCl2 formed ⎯278.11 g⎯PbCl2 = (0.10004 mol PbCl2) 27.82 g PbCl2 mol PbCl2

19. a. Given: molality = ∆t = K m 1.50 m f f = − ° =− ° C12H22O11 ( 1.86 C/m)(1.50 m) 2.79 C (sucrose) solvent = H2O Unknown: freezing- point depression

b. Given: mass of solute mol sucrose = = (171 g sucrose) ⎯⎯ 0.500 mol sucrose 171 g 342 g sucrose C12H22O11 moles of solute (sucrose) m = ⎯⎯⎯ mass of solvent mass of solvent (kg) = 1.00 kg H O 2 0.500 mol m ==⎯⎯ 0.5 m Unknown: freezing- 1.00 kg point ∆ = K m depression tf f ∆tf = (−1.86°C/m)(0.500 m) =−0.93°C

c. Given: mass of mol sucrose = = (77.0 g sucrose) ⎯⎯ 0.225 mol sucrose solute 77.0 g 342 g sucrose C12H22O11 moles of solute (sucrose) m = ⎯⎯⎯ mass of solvent mass of solvent (kg) = 0.400 kg H O 2 0.225 mol Unknown: freezing- m ==⎯⎯ 0.563 m point 0.400 kg depression ∆tf = Kfm

∆tf = (−1.86°C/m)(0.563 m) =−1.05°C

MODERN CHEMISTRY CHAPTER 13 SOLUTIONS MANUAL Copyright © by Holt, Rinehart and Winston. All rights reserved. 156 20. Given: solvent = H2O; ∆tf = Kfm solute = m =∆t /K nonelectrolyte f f a. freezing-point a. m = (−0.930°C)/(−1.86°C/m) = 0.500 m depression =−0.93°C b. freezing-point b. m = (−3.72°C)/(−1.86°C/m) = 2.00 m depression =−3.72°C c. freezing-point c. m = (−8.37°C)/(−1.86°C/m) = 4.50 m depression =−8.37°C Unknown: molality (m) of each solution

21. Given: mass of solute = 20.0 g C6H12O6 (glucose) mass of solvent = 0.250 kg H2O

a. Unknown: freezing- ∆tf = Kfm point depression mol glucose (20.0 g glucose) ⎯⎯ = 0.111 mol glucose of solvent 180 g glucose moles of solute m = ⎯⎯⎯ mass of solvent (kg) 0.111 mol m ==⎯⎯ 0.444 m 0.250 kg

∆tf = Kfm

∆tf = (−1.86°C/m)(0.444 m) =−0.826°C

b. Unknown: freezing f.p. solution = f.p. solvent +∆tf point of = 0.000°C + (−0.826°C) =−0.826°C solution

22. Given: solute = C2H4(OH)2 f.p. solution = f.p. solvent +∆tf (antifreeze) ∆t = f.p. solution − f.p. solvent mass of solvent = f =− ° − ° =− ° 0.500 kg H2O 20.0 C 0.000 C 20.0 C freezing point of ∆t = K m solution =−20.0°C f f m =∆t /K Unknown: mass of solute f f = − ° − ° = in g ( 20.0 C)/( 1.86 C/m) 10.8 m moles of solute m = ⎯⎯⎯ mass of solvent (kg) moles of solute = (m)(mass of solvent) = (10.8 mol/kg)(0.500 kg) = 5.40 mol

62.06⎯⎯⎯ g C2H4(OH)2 = (5.40 mol C2H4(OH)2)334 g C2H4(OH)2 mol C2H4(OH)2

MODERN CHEMISTRY CHAPTER 13 SOLUTIONS MANUAL Copyright © by Holt, Rinehart and Winston. All rights reserved. 157 23. Given: freezing point mol C Cl H (7.24 g C Cl H ) ⎯⎯2 4 2 = 0.0431 mol C Cl H C H (benzene) = 2 4 2 2 4 2 6 6 168 g C2Cl4H2 5.45°C = moles of solute mass of solute m = ⎯⎯⎯ 7.24 g C2Cl4H2 mass of solvent (kg) = mass of solvent 0.0431 mol 0.115 kg benzene m = ⎯⎯ = 0.375 m specific gravity 0.115 kg = benzene 0.879 f.p. solution = f.p. solvent +∆tf freezing point of ∆ = − solution = 3.55°C tf f.p. solution f.p. solvent = 3.55°C − 5.45°C =−1.90°C Unknown: molal freezing- point constant ∆tf = Kfm (K ) for f K =∆t /m benzene f f = (−1.90°C)/(0.375 m) =−5.07°C/m

24. Given: mass of solute = mol solute 1.500 g (1.500 g solute)⎯⎯ = 0.01200 mol solute molar mass of 125.0 g solute = solute 125.0 g moles of solute mass of solvent = m = ⎯⎯⎯ 0.03500 kg cam- mass of solvent (kg) phor 0.01200 mol m = ⎯⎯ = 0.3429 m Unknown: freezing point 0.03500 kg of solution ∆tf = Kfm

∆tf = (−39.7°C/m)(0.3429 m) =−13.6°C

f.p. solution = f.p. solvent +∆tf = 178.8°C + (−13.6°C) = 165.2°C

25. a. Given: molality = ∆tb = Kbm 2.5 m C6H12O6 (glucose) = (0.51°C/m)(2.5 m) = 1.3°C solvent = H2O Unknown: boiling- point elevation of H2O

b. Given: mass of mol glucose (3.20 g glucose) ⎯⎯ = 0.0178 mol glucose solute = 3.20 g 180. g glucose C6H12O6 moles of solute (glucose) m = ⎯⎯⎯ mass of solvent mass of solvent (kg) = 1.00 kg H2O 0.0178 mol m = ⎯⎯ = 0.0178 m Unknown: boiling- 1.00 kg point eleva- ∆tb = Kbm tion of H2O ∆tb = (0.51°C/m)(0.0178 m) = 0.0091°C

MODERN CHEMISTRY CHAPTER 13 SOLUTIONS MANUAL Copyright © by Holt, Rinehart and Winston. All rights reserved. 158 mol sucrose c. Given: mass of solute (20.0 g sucrose)⎯⎯ = 0.0585 mol sucrose = 20.0 g 342 g sucrose C12H22O11 moles of solute (sucrose) m = ⎯⎯⎯ mass of solvent mass of solvent (kg) 0.500 kg H O 0.585 mol 2 m ==⎯⎯ 0.117 m Unknown: boiling- 0.500 kg point eleva- ∆tb = Kbm tion of H2O ∆tb = (0.51°C/m)(0.117 m) = 0.060°C

26. Given: solvent = H2O ∆tb = Kbm a. boiling point of m =∆t /K solution = b b ° 100.25 C b.p. solution = b.p. solvent +∆tb b. boiling point of ∆t = b.p. solution − b.p. solvent solution = b 101.53°C a. ∆tb = 100.25°C − 100.00°C = 0.25°C c. boiling point 0.25°C = m = ⎯⎯ = 0.49 m of solution ° 102.805°C 0.51 C/m Unknown: molality (m) of b. ∆tb = 101.53°C − 100.00°C = 1.53°C each solution 1.53°C m = ⎯⎯ = 3.0 m 0.51°C/m

c. ∆tb = 102.805°C − 100.000°C = 2.805°C 2.805°C m = ⎯⎯ = 5.5 m 0.51°C/m

27. Given: molality = 1.00 m ∆tf = Kfm solvent = H2O + − a. solute = KI a. KI(s) U K (aq) + I (aq) = b. solute CaCl2 Each formula unit of KI yields two ions in solution. c. solute = −1.86°C • kg H O 1.00 mol KI 2 mol ions Ba(NO3)2 ⎯⎯⎯2 ⎯⎯ ⎯⎯ =−3.72°C mol ions kg H O mol KI Unknown: expected 2 freezing-point U 2+ + − depression of b. CaCl2(s) Ca (aq) 2Cl (aq) solutions Each formula unit of KI yields three ions in solution. −1.86°C • kg H O 1.00 mol CaCl 3 mol ions ⎯⎯2 ⎯⎯2 ⎯⎯ =−5.58°C mol ions kg H2O mol CaCl2

2+ − c. Ba(NO3)2(s) U Ba (aq) + 2NO3 (aq)

Each formula unit Ba(NO3)2 yields three ions in solution. −1.86°C • kg H O 1.00 mol Ba(NO ) 3 mol ions ⎯⎯2 ⎯⎯⎯3 2 ⎯⎯ =−5.58°C mol ions kg H2O mol Ba(NO3)2

MODERN CHEMISTRY CHAPTER 13 SOLUTIONS MANUAL Copyright © by Holt, Rinehart and Winston. All rights reserved. 159 28. Given: molality = 0.015 m ∆tf = Kfm AlCl3 3+ − AlCl3(s) U Al (aq) + 3Cl (aq) solvent = H2O Unknown: expected Each formula unit of AlCl3 yields four ions in solution. freezing-point − ° • 1.86 C kg H2O 0.015 mol AlCl3 4 mol ions =− ° depression of ⎯⎯⎯ ⎯⎯ ⎯⎯ 0.11 C mol ions kg H O mol AlCl solution 2 3

29. Given: mass of solute = f.p. solution = f.p. solvent +∆tf 85.0 g NaCl ∆t = K m mass of solvent = f f 0.450 kg H2O moles of solute m = ⎯⎯⎯ Unknown: expected freez- mass of solvent (kg) ing point of mol NaCl solution (85.0 g NaCl) ⎯⎯ = 1.45 mol NaCl 58.44 g NaCl + − NaCl(s) U Na (aq) + Cl (aq) Each formula unit of NaCl yields two ions in solution.

− ° • ⎯⎯1.86 C kg H2O 1.45⎯⎯ mol NaCl ⎯⎯2 mol ions =− ° =∆ 12.0 C tf mol ions 0.450 kg H2O mol NaCl f.p. solution = 0.0°C + (−12.0°C) =−12.0°C

30. Given: mass of solute = b.p. solution = b.p. solvent +∆tb 25.0 g BaCl2 ∆ = mass of solvent = tb Kbm 0.150 kg H2O moles solute m = ⎯⎯⎯ Unknown: expected boil- mass of solvent (kg) ing point of mol BaCl solution ⎯⎯2 = (25.0 g BaCl2)0.120 mol BaCl2 208.23 g BaCl2 2+ − BaCl2(s) U Ba (aq) + 2Cl (aq)

Each formula unit of BaCl2 yields three ions in solution. ° • ⎯⎯0.51 C kg H2O 0.120⎯⎯ mol BaCl2 ⎯⎯3 mol ions = ° =∆ 1.2 C tb mol ions 0.150 kg H2O mol BaCl2 b.p. solution = 100.0°C + 1.2°C = 101.2°C

31. Given: boiling-point ele- ∆tb = Kbm vation of solution + − KI(s) U K (aq) + I (aq) = 0.65°C solvent = H2O Each formula unit of KI yields two ions in solution. solute = KI m =∆tb/Kb Unknown: molality (m) 0.65°C • mol ions 1 mol KI = ⎯⎯ ⎯⎯ = 0.64 m KI • 0.51°C kg H2O 2 mol ions

MODERN CHEMISTRY CHAPTER 13 SOLUTIONS MANUAL Copyright © by Holt, Rinehart and Winston. All rights reserved. 160 34. Given: molality = 1.00 m ∆tf = Kfm MgI2 2+ − MgI2(s) U Mg (aq) + 2I (aq) solvent = H2O freezing-point de- Each formula unit of MgI2 yields three ions in solution. pression =−4.78°C − ° • 1.86 C kg H2O 1.00 mol MgI2 3 mol ions =− ° Unknowns: expected ⎯⎯ ⎯⎯ ⎯⎯ 5.58 C mol ions kg H O mol MgI freezing-point 2 2 depression of Reason for discrepancy: Because of forces of attraction between them, the H2O; reason for discrep- ions cluster somewhat. ancy between experimental and expected values

35. Given: molality = 0.01 m ∆tf = Kfm solvent = H2O Unknown: freezing-point depression of each solution

+ − a. Given: solute = NaI NaI(s) U Na (aq) + I (aq) Each formula unit of NaI yields two ions in solution. −1.86°C • kg H O 0.01 mol NaI 2 mol ions ⎯⎯2 ⎯⎯ ⎯⎯ =−0.04°C mol ions kg H2O mol NaI

2+ − b. Given: solute = CaCl2 CaCl2(s) U Ca (aq) + 2Cl (aq)

Each formula unit of CaCl2 yields three ions in solution. −1.86°C • kg H O 0.01 mol CaCl 3 mol ions ⎯⎯2 ⎯⎯2 ⎯⎯ =−0.06°C mol ions kg H2O mol CaCl2

+ 3− c. Given: solute = K3PO4 K3PO4(s) U 3K (aq) + PO4 (aq)

Each formula unit of K3PO4 yields four ions in solution. −1.86°C • kg H O 0.01 mol K PO 4 mol ions ⎯⎯2 ⎯⎯3 4 ⎯⎯ =−0.07°C mol ions kg H2O mol K3PO4

d. Given: solute = ∆tf = (−1.86°C/m)(0.01 m) =−0.02°C C H O 6 12 6 order of increasing ∆t = − ° − ° − ° − ° (glucose) f d( 0.02 C), a( 0.04 C), b( 0.06 C), c( 0.07 C)

36. Given: solute = CaCl2 ∆tf = Kfm solvent = H O + − 2 CaCl (s) U Ca2 (aq) + 2Cl (aq) freezing point of 2 =− ° solution 2.43 C Each formula unit of CaCl2 yields three ions in solution.

Unknown: molality (m) m =∆tf/Kf

−2.43°C • mol ions 1 mol CaCl2 ⎯⎯⎯ ⎯⎯ = 0.435 m CaCl2 −1.86°C 3 mol/ions

MODERN CHEMISTRY CHAPTER 13 SOLUTIONS MANUAL Copyright © by Holt, Rinehart and Winston. All rights reserved. 161 = 39. a. Given: solute ⎯H2O + 2− K2S(s) U 2K (aq) + S (aq) 0.275 mol K2S + + 2− solvent = H2O Ions produced: 2(0.275 mol K ) = 0.550 mol K ; 0.275 mol S Unknowns: balanced 0.550 + 0.275 = 0.825 mol solute ions formed equation, total number of moles of solute ions formed

b. Given: solute = H2O + − (SO ) (s) ⎯U 2Al3 (aq) + 3SO2 (aq) 0.15 mol Al2 4 3 4 3+ 3+ 2− Al2(SO4)3 Ions produced: 2(0.15 mol Al ) = 0.30 mol Al + 3(0.15 mol SO4 ) = 2− Unknowns: balanced 0.45 mol SO4 equation, 0.30 + 0.45 = 0.75 mol solute ions formed total number of moles of solute ions formed

40. Given: mass of solute = 131.2 g AgNO ⎯⎯mol AgNO3 = 3 (131.2 g AgNO3)0.7724 mol AgNO3 mass of solvent = 169.85 g AgNO3 2.00 kg H O 2 moles of solute 0.7724 mol m ===⎯⎯⎯ ⎯⎯ 0.386 m Unknown: expected mass of solvent (kg) 2.00 kg boiling-point ∆ = elevation tb Kbm of solution + − AgNO3(s) U Ag (aq) + NO3(aq)

Each formula unit AgNO3 yields two ions in solution. 0.51°C • kg H O 0.385 mol AgNO 2 mol ions ⎯⎯2 ⎯⎯⎯3 ⎯⎯ = 0.39°C mol ions kg H2O mol AgNO3

42. Given: solvent = H2O f.p. solution = f.p. solvent +∆tf solute = nonelectrolyte ∆tf = f.p. solution − f.p. solvent freezing point = =−6.51°C − 0°C =−6.51°C −6.51°C ∆tf = Kfm Unknown: boiling point m =∆tf/Kf −6.51°C ==⎯⎯⎯⎯ 3.50 m −1.86°C • kg H2O/mol solute

∆tb = Kbm = (0.51°C/m)(3.50 m) = 1.8°C

∆tb = b.p. solution − b.p. solvent

b.p. solution =∆tb + b.p. solvent = 1.8°C + 100.0°C = 101.8°C

MODERN CHEMISTRY CHAPTER 13 SOLUTIONS MANUAL Copyright © by Holt, Rinehart and Winston. All rights reserved. 162 ⎯H2O + 2− 43. Given: solute = Na2CO3 a. Na2CO3(s) U 2Na (aq) + CO3 (aq) solvent = H2O amount of solute = + + b. 2(0.20 mol Na ) = 0.40 mol Na 0.20 mol Na CO 2 3 − − 1(0.20 mol CO2 ) = 0.20 mol CO2 Unknowns: a. balanced 3 3 equation 0.40 + 0.20 = 0.60 moles of solute ions produced b. number of c. moles of each ion produced c. total number of moles of ions

44. Given: K3PO4(aq) + Ionic equation: Pb(NO3)2(aq) + 3− 2+ − + − 6K (aq) + 2PO4 (aq) + 3Pb (aq) + 6NO3(aq) U 6K (aq) + 6NO3(aq) Unknown: net ionic + Pb3(PO4)2(s) equation Net ionic equation: 2+ 3− 3Pb (aq) + 2PO4 (aq) U Pb3(PO4)2(s)

= 45. Given: solvent H2O moles of solute mass of solute = m = ⎯⎯⎯ mass of solvent (kg) 268 g Al(NO3)3 mass of solvent = ⎯⎯mol Al(NO3)3 = 8.50 kg H2O (268 g Al(NO3)3)1.26 mol Al(NO3)3 212.9 g Al(NO3)3 Unknown: expected freez- + − Al(NO ) (s) U Al3 (aq) + 3NO (aq) ing point of 3 3 3 solution Each formula unit Al(NO3)3 yields four ions in solution.

∆tf = Kfm −1.86°C • kg H O 1.26 mol Al(NO ) 4 mol ions ⎯⎯2 ⎯⎯⎯3 3 ⎯⎯ =−1.10°C mol ions 8.50 kg H2O mol Al(NO3)3

f.p. solution = f.p. solvent +∆tf = 0.00°C + (−1.10°C) =−1.10°C

MODERN CHEMISTRY CHAPTER 13 SOLUTIONS MANUAL Copyright © by Holt, Rinehart and Winston. All rights reserved. 163 46. a. Given: solvent = H2O f.p. solution = f.p. solvent +∆tf = solute KNO3 ∆ = − observed freez- tf f.p. solution f.p. solvent ing point of so- =−1.15°C − 0.25°C =−1.40°C =− ° lution 1.15 C ∆t = K m observed freez- f f + − ing point of KNO3(s) U K (aq) + NO3(aq) pure H O = 2 Each formula unit KNO yields two ions in solution. 0.25°C 3 m =∆t Unknown: molality f/Kf (m) of − ° • = 1.40 C mol ions 1 mol KNO3 KNO ⎯⎯⎯ ⎯⎯ 3 −1.86°C 2 mol ions = 0.376 m

b. Given: mass of solute = moles of solute = m ⎯⎯⎯ 0.415 g KNO3 mass of solvent (kg) volume of solu- = 1.00 g kg tion 0.010 L mass of solvent = 10.00 mL ×=⎯ 10.0 g ×=⎯ 0.0100 kg density of mL 1000 g solution = mol KNO 1.00 g/mL ⎯⎯3 = (0.415 g KNO3) 0.00410 mol KNO3 101.02 g KNO3 Unknowns: actual molality (m) 0.00410 mol ⎯⎯ = 0.410 m KNO3 of KNO3; 0.0100 kg percentage difference Predicted m = 0.376 m between predicted Actual m = 0.410 m concentra- 0.410 m − 0.376 m tion and ⎯⎯ × 100 = 8.29% = percentage difference actual con- 0.410 m centration of KNO3

Math Tutor, p. 462

1. Given: 28.0 g CaCl dis- 28.0 g CaCl × 1000 g × 3 mol ions × (−1.86°C/m) 2 ⎯⎯⎯⎯⎯⎯2 =−4.77°C solved in 110.98 g/mol × 295 g H2O × 1 kg × 1 mol CaCl2 295 g H2O Unknown: freezing point of solution

2. Given: 850 g ethylene 850 g C H O boiling point elevation =×⎯⎯⎯⎯2 6 2 0.51°C/m =+6.3°C glycol dissolved in 62.08 g/mol × 1.100 kg H2O 1100 g water boiling point = 100.°C + 6.3°C = 106.3°C Unknown: boiling point of solution

Standardized Test Prep, p. 463

8. Given: freezing point of −0.58°C ⎯⎯ = 0.31 m nonelectrolyte −1.86°C/m solution =−0.58°C Unknown: molality of solution