Solutions and Their Properties

11.1 Toluene is nonpolar and is insoluble in water.

Br₂ is nonpolar but because of its size is polarizable and is soluble in water.

KBr is an ionic compound and is very soluble in water.

toluene < Br₂ < KBr (solubility in H₂O)

11.2 (a) Na⁺ has the larger (more negative) hydration energy because the Na⁺ ion is smaller than the Cs⁺ ion and water molecules can approach more closely and bind more tightly to the Na⁺ ion.

(b) Ba²⁺ has the larger (more negative) hydration energy because of its higher charge.

11.3 NaCl, 58.44 amu; 1.00 mol NaCl = 58.44 g

1.00 L H₂O = 1000 mL = 1000 g (assuming a density of 1.00 g/mL)

mass % NaCl = = 5.52 mass %

11.4 ppm = x 10⁶ ppm

total mass of solution = density x volume = (1.3 g/L)(1.0 L) = 1.3 g

35 ppm = x 10⁶ ppm

mass of CO₂ = = 4.6 x 10^-5 g CO₂

11.5 Assume 1.00 L of sea water.

mass of 1.00 L = (1000 mL)(1.025 g/mL) = 1025 g

= 3.50 mass %; mass NaCl = = 35.88 g

There are 35.88 g NaCl per 1.00 L of solution.

M = = 0.614 M

11.6 C₂₇H₄₆O, 386.7 amu; CHCl₃, 119.4 amu; 40.0 g x = 0.0400 kg

molality = = 0.0249 m

= = 2.96 x 10^-3

11.7 CH₃CO₂Na, 82.03 amu

kg H₂O = = 0.300 kg H₂O

(0.150 mol CH₃CO₂Na) = 12.3 g CH₃CO₂Na

mass of solution needed = 300 g + 12.3 g = 312 g

11.8 Assume you have a solution with 1.000 kg (1000 g) of H₂O. If this solution is 0.258 m, then it must also contain 0.258 mol glucose.

mass of glucose = 0.258 mol x = 46.5 g glucose

mass of solution = 1000 g + 46.5 g = 1046.5 g

density = 1.0173 g/mL

volume of solution = = 1028.7 mL

volume = 1028.7 mL x = 1.029 L; molarity = = 0.251 M

11.9 Assume 1.00 L of solution.

mass of 1.00 L = (1.0042 g/mL)(1000 mL) = 1004.2 g of solution

0.500 mol CH₃CO₂H x = 30.02 g CH₃CO₂H

1004.2 g - 30.02 g = 974.2 g = 0.9742 kg of H₂O; molality = = 0.513 m

11.10 Assume you have 100.0 g of seawater.

mass NaCl = (0.0350)(100.0 g) = 3.50 g NaCl

mass H₂O = 100.0 g - 3.50 g = 96.5 g H₂O

NaCl, 58.44 amu; mol NaCl = 3.50 g x = 0.0599 mol NaCl

mass H₂O = 96.5 g x = 0.0965 kg H₂O; molality = = 0.621 m

11.11 M = k P; k = = 3.2 x 10^-2 mol/(L atm)

11.12 (a) M = k P = [3.2 x 10^-2 mol/(L atm)](2.5 atm) = 0.080 M

(b) M = k P = [3.2 x 10^-2 mol/(L atm)](4.0 x 10^-4 atm) = 1.3 x 10^-5 M

11.13 C₇H₆O₂, 122.1 amu; C₂H₆O, 46.07 amu

= 0.981

P_soln = P_solv X_solv = (100.5 mm Hg)(0.981) = 98.6 mm Hg

11.14 P_soln = P_solv X_solv; = = 0.976

NaBr dissociates into two ions in aqueous solution.

X_solv =

X_solv = 0.976 =

0.976 = ; solve for x.

x = 0.171 mol Na⁺ = 0.171 mol Br^- = 0.171 mol NaBr

NaBr, 102.9 amu; mass NaBr = 0.171 mol x = 17.6 g NaBr

11.15 At any given temperature, the vapor pressure of a solution is lower than the vapor pressure of the pure solvent. The upper curve represents the vapor pressure of the pure solvent. The lower curve represents the vapor pressure of the solution.

11.16 C₂H₅OH, 46.07 amu; H₂O, 18.02 amu

(a) = 0.5426 mol C₂H₅OH

100.0 g H₂O x = 5.549 mol H₂O

= 0.08907

= 0.9109

P_soln =

P_soln = (0.08907)(61.2 mm Hg) + (0.9109)(23.8 mm Hg) = 27.1 mm Hg

(b) = 2.171 mol C₂H₆O

25.0 g H₂O x = 1.387 mol H₂O

= 0.6102

= 0.3898

P_soln =

P_soln = (0.6102)(61.2 mm Hg) + (0.3898)(23.8 mm Hg) = 46.6 mm Hg

11.17 At any given temperature, the vapor pressure of a mixture of two pure liquids falls between the individual vapor pressures of the two pure liquids themselves. The upper and lower curves represent the vapor pressures of the two pure liquids. The middle curve represents the vapor pressure of the mixture.

11.18 C₉H₈O₄, 180.2 amu; CHCl₃ is the solvent. For CHCl₃, K_b = 3.63

75.00 g x = 0.075 00 kg

T_b = K_b m = = 0.40^oC

solution boiling point = 61.7^oC + T_b = 61.7^oC + 0.40^oC = 62.1^oC

11.19 K₂SO₄, 174.3 amu; there are 3 ions (solute particles)/K₂SO₄

T_f = K_f (3 m) = = 2.15^oC

Solution freezing point = 0.00^oC - T_f = 0.00^oC - 2.15^oC = -2.15^oC

11.20 There are 2 ions/KBr. T_f = K_f 2 m

freezing point = -2.95^oC = 0.00^oC - T_f; T_f = 2.95^oC

m = = 0.793 m

11.21 The red curve represents the vapor pressure of pure chloroform.

(a) The normal boiling point for a liquid is the temperature where the vapor pressure of the liquid equals 1 atm (760 mm Hg). The approximate boiling point of pure chloroform is 62^oC.

(b) The approximate boiling point of the solution is 69^oC.

T_b = 69^oC - 62^oC = 7^oC

T_b = K_b m; m = = = 2 m

11.22 For CaCl₂ there are 3 ions (solute particles)/CaCl₂

= MRT; For CaCl₂, = 3MRT

= (3)(0.125 mol/L)(310 K) = 9.54 atm

11.23 = MRT; M = = 0.156 M

11.24 T_f = K_f m; m = = 0.0557 m

0.0557 ; solve for the molar mass.

molar mass = 128 g/mol

11.25 = MRT; M = = 8.02 x 10^-3

molarity = 8.02 x 10^-3

Solve for the molar mass. molar mass = 342 g/mol

11.26 (a) and (c)

(b) The mixture will begin to boil at ~50^oC.

(d) After two cycles of boiling and condensing, the approximate composition of the liquid would 90% dichloromethane and 10% chloroform.

11.27 Both solvent molecules and small solute particles can pass through a semipermeable dialysis membrane. Only large colloidal particles such as proteins can't pass through. Only solvent molecules can pass through a semipermeable membrane used for osmosis.

Understanding Key Concepts

11.28 (a) < (b) < (c)

11.29 At any given temperature, the vapor pressure of a mixture of two pure liquids falls between the individual vapor pressures of the two pure liquids themselves. Because the vapor pressure of the mixture is greater than the vapor pressure of the solvent, the second liquid is more volatile (has a higher vapor pressure) than the solvent.

11.30 Assume that only the blue (open) spheres (solvent) can pass through the semipermeable membrane. There will be a net transfer of solvent from the right compartment (pure solvent) to the left compartment (solution) to achieve equilibrium.

11.31 NaCl is a nonvolatile solute. Methyl alcohol is a volatile solute. When NaCl is added to water, the vapor pressure of the solution is decreased, which means that the boiling point of the solution will increase. When methyl alcohol is added to water, the vapor pressure of the solution is increased which means that the boiling point of the solution will decrease.

11.32 When 100 mL of 9 M H₂SO₄ at 0^oC is added to 100 mL of liquid water at 0^oC, the temperature rises because H_soln for H₂SO₄ is exothermic.

When 100 mL of 9 M H₂SO₄ at 0^oC is added to 100 g of solid ice at 0^oC, some of the ice will melt (an endothermic process) and the temperature will fall because the H₂SO₄ (solute) lowers the freezing point of the ice/water mixture.

11.33 H_soln for HBr is exothermic. The HBr solution will be warm to touch.

H_soln for AgNO₃ is endothermic. The AgNO₃ solution will be cool to touch.

11.34 The vapor pressure of the NaCl solution is lower than that of pure H₂O. More H₂O molecules will go into the vapor from the pure H₂O than from the NaCl solution. More H₂O vapor molecules will go into the NaCl solution than into pure H₂O. The result is represented by (b).

11.35 (b) ~95^oC

Additional Problems

Solutions and Energy Changes

11.36 The surface area of a solid plays an important role in determining how rapidly a solid dissolves. The larger the surface area, the more solid-solvent interactions, and the more rapidly the solid will dissolve. Powdered NaCl has a much larger surface area than a large block of NaCl, and it will dissolve more rapidly.

11.37 (a) a gas in a liquid -- carbonated soft drink

(b) a solid in a solid -- metal alloys (14-karat gold)

(c) a liquid in a solid -- dental amalgam (Hg in Ag)

11.38 Substances tend to dissolve when the solute and solvent have the same type and magnitude of intermolecular forces; thus the rule of thumb "like dissolves like."

11.39 Both Br₂ and CCl₄ are nonpolar, and intermolecular forces for both are dispersion forces. H₂O is a polar molecule with dipole-dipole forces and hydrogen bonding. Therefore, Br₂ is more soluble in CCl₄.

11.40 Energy is required to overcome intermolecular forces holding solute particles together in the crystal. For an ionic solid, this is the lattice energy. Substances with higher lattice energies tend to be less soluble than substances with lower lattice energies.

11.41 SO₄^2- has the larger hydration energy because of its higher charge. Both SO₄^2- and ClO₄^- are comparable in size, so size is not a factor.

11.42 Ethyl alcohol and water are both polar with small dispersion forces. They both can hydrogen bond, and are miscible.

Pentyl alcohol is slightly polar and can hydrogen bond. It has, however, a relatively large dispersion force because of its size, which limits its water solubility.

11.43 The intermolecular forces associated with octane are dispersion forces. Both pentyl alcohol and methyl alcohol can hydrogen bond. Pentyl alcohol has relatively large dispersion forces because of its size. Methyl alcohol does not. Pentyl alcohol is soluble in octane; methyl alcohol is not.

11.44 CaCl₂, 110.98 amu

For a 1.00 m solution:

heat released = 81,300 J

mass of solution = 1000 g H₂O + 110.98 g CaCl₂ = 1110.98 g

T = = = 17.5 K = 17.5^oC

Final temperature = 25.0^oC + 17.5^oC = 42.5^oC

11.45 NH₄ClO₄, 117.48 amu

For a 1.00 m solution:

heat absorbed = 33,500 J

mass of solution = 1000 g H₂O + 117.48 g NH₄ClO₄ = 1117.48 g

T = = = -7.2 K = -7.2^oC

Final temperature = 25.0^oC - 7.2^oC = 17.8^oC

Units of Concentration

11.46 molarity = ; molality =

11.47 A saturated solution contains enough solute so that there is an equilibrium between dissolved solute and undissolved solid.

A supersaturated solution contains a greater-than-equilibrium amount of solute.

11.48 (a) Dissolve 0.150 mol of glucose in water; dilute to 1.00 L.

(b) Dissolve 1.135 mol of KBr in 1.00 kg of H₂O.

(c) Mix together 0.15 mol of CH₃OH with 0.85 mol of H₂O.

11.49 (a) Dissolve 15.5 mg urea in 100 mL water

(b) Choose a K⁺ salt, say KCl, and dissolve 0.0075 mol (0.559 g) in water; dilute to

100 mL.

11.50 C₇H₆O₂, 122.12 amu, 165 mL = 0.165 L

mol C₇H₆O₂ = 0.165 L x = 0.004 42 mol

mass C₇H₆O₂ = 0.004 42 mol x = 0.540 g

Dissolve 4.42 x 10^-3 mol (0.540 g) of C₇H₆O₂ in enough CHCl₃ to make 165 mL of solution.

11.51 C₇H₆O₂, 122.12 amu

0.0268 mol C₇H₆O₂ x = 3.27 g C₇H₆O₂

Dissolve 3.27 g of C₇H₆O₂ in 1.000 kg of CHCl₃, and take 165 mL of the solution.

11.52 (a) KCl, 74.6 amu

A 0.500 M KCl solution contains 37.3 g of KCl per 1.00 L of solution.

A 0.500 mass % KCl solution contains 5.00 g of KCl per 995 g of water.

The 0.500 M KCl solution is more concentrated (that is, it contains more solute per amount of solvent).

(b) Both solutions contain the same amount of solute. The 1.75 M solution contains less solvent than the 1.75 m solution. The 1.75 M solution is more concentrated.

11.53 (a) KI, 166.00 amu; KBr, 119.00 amu; assume 1.000 L = 1000 mL = 1000 g solution

10 ppm = x 10⁶ ; mass KI = 0.010 g

10,000 ppb = x 10⁹ ; mass KBr = 0.010 g

Both solutions contain the same mass of solute in the same amount of solvent. Because the molar mass of KBr is less than that of KI, the number of moles of KBr is larger than the number of moles of KI. The KBr solution has a higher molarity than the KI solution.

(b) Because the mass % of the two solutions is the same, they both contain the same mass of solute and solution. Because the molar mass of KCl is less than that of citric acid, the number of moles of KCl is larger than the number of moles of citric acid. The KCl solution has a higher molarity than the citric acid solution.

11.54 (a) C₆H₈O₇, 192.12 amu

0.655 mol C₆H₈O₇ x = 126 g C₆H₈O₇

mass % C₆H₈O₇ = = 11.2 mass %

(b) 0.135 mg = 0.135 x 10^-3 g

(5.00 mL H₂O)(1.00 g/mL) = 5.00 g H₂O

mass % KBr = = 0.002 70 mass % KBr

(c) mass % aspirin = = 3.65 mass % aspirin

11.55 (a) molality = = 0.655 m

(b) KBr, 119.00 amu; 5.00 g = 0.005 00 kg

molality = = 2.27 x 10^-4 m

(c) C₉H₈O₄, 180.16 amu; 145 g = 0.145 kg

molality = = 0.211 m

11.56

= 1.2 x 10^-7

Assume one mole of air (29 g/mol)

mol O₃ = n_air = (1 mol)(1.2 x 10^-7) = 1.2 x 10^-7 mol O₃

O₃, 48.00 amu; mass O₃ = 1.2 x 10^-7 mol x = 5.8 x 10^-6 g O₃

ppm O₃ = x 10⁶ = 0.20 ppm

11.57 Assume 1 mL of blood weighs 1 g. 1 dL = 0.1 L = 100 mL = 100 g

ppb = x 10⁹ = 100 ppb

11.58 (a) H₂SO₄, 98.08 amu; molality = = 0.196 m

(b) C₁₀H₁₄N₂, 162.23 amu; CH₂Cl₂, 84.93 amu

2.25 g C₁₀H₁₄N₂ x = 0.0139 mol C₁₀H₁₄N₂

80.0 g CH₂Cl₂ x = 0.942 mol CH₂Cl₂

= 0.0145

= 0.985

11.59 NaOCl, 74.44 amu

A 5.0 mass % aqueous solution of NaOCl contains 5.0 g NaOCl and 95 g H₂O.

molality = = 0.71 m

5.0 g NaOCl x = 0.0672 mol NaOCl

95 g H₂O x = 5.27 mol H₂O

= 0.013

11.60 16.0 mass % =

H₂SO₄, 98.08 amu; density = 1.1094 g/mL

volume of solution = = 90.14 mL = 0.090 14 L

molarity = = 1.81 M

11.61 C₂H₆O₂, 62.07 amu

A 40.0 mass % aqueous solution of C₂H₆O₂ contains 40.0 g C₂H₆O₂ and 60.0 g H₂O.

density = 1.0514 g/mL

volume of solution = = 95.1 mL = 0.0951 L

molarity = = 6.78 M

11.62 molality = = 10.7 m

11.63 molality = = 1.94 m

11.64 C₁₉H₂₁NO₃, 311.34 amu; 1.5 mg = 1.5 x 10^-3 g

1.3 x 10^-3 ; solve for kg of solvent.

kg of solvent = 0.0037 kg

Because the solution is very dilute, kg of solvent kg of solution.

g of solution = (0.0037 kg) = 3.7 g

11.65 C₁₂H₂₂O₁₁, 342.30 amu

32.5 g C₁₂H₂₂O₁₁ x = 0.0949 mol C₁₂H₂₂O₁₁

0.850 m = 0.850; kg of H₂O = = 0.112 kg

mass of H₂O = 0.112 kg x = 112 g H₂O

11.66 C₆H₁₂O₆, 180.16 amu; H₂O, 18.02 amu; Assume 1.00 L of solution.

mass of solution = (1000 mL)(1.0624 g/mL) = 1062.4 g

mass of solute = 0.944 mol x = 170.1 g C₆H₁₂O₆

mass of H₂O = 1062.4 g - 170.1 g = 892.3 g H₂O

mol C₆H₁₂O₆ = 0.944 mol; mol H₂O = 892.3 g x = 49.5 mol

(a) = 0.0187

(b) mass % = x 100% = 16.0%

(c) molality = = 1.06 m

11.67 C₁₂H₂₂O₁₁, 342.30 amu; Assume 1.00 L of solution.

mass of solution = (1000 mL)(1.0432 g/mL) = 1043.2 g

mass of solute = 0.335 mol C₁₂H₂₂O₁₁ x = 114.7 g C₁₂H₂₂O₁₁

mass of H₂O = 1043.2 g - 114.7 g = 928.5 g H₂O

mol C₁₂H₂₂O₁₁ = 0.335 mol; 928.5 g H₂O x = 51.53 mol H₂O

= 0.006 46

mass % C₁₂H₂₂O₁₁ = = 11.0 mass % C₁₂H₂₂O₁₁

molality = = 0.361 m

Solubility and Henry's Law

11.68 M = k P = (0.091 )(0.75 atm) = 0.068 M

11.69 M = k P; k = = 0.195 mol/(L atm)

P = 25.5 mm Hg x = 0.0336 atm

M = k P = (0.195 )(0.0336 atm) = 6.55 x 10^-3 M

11.70 M = k P

Calculate k: k = = 2.21 x 10^-3

Convert 4 mg/L to mol/L:

4 mg = 4 x 10^-3 g

O₂ molarity = = 1.25 x 10^-4 M

= 0.06 atm

11.71 k = 1.93 x 10^-3 mol/(L atm)

M = k P = [1.93 x 10^-3 mol/(L atm)]= 1.73 x 10^-4 mol/L

1.73 x 10^-4 mol/L x = 5.5 mg/L

11.72 [Xe] = 10 mmol/L = 0.010 M at STP

M = k P; k = = 0.010 mol/(L atm)

11.73 Assuming H₂O as the solvent, NH₃ does not obey Henry's law because NH₃ can both hydrogen bond and react with H₂O.

Colligative Properties

11.74 The difference in entropy between the solvent in a solution and a pure solvent is responsible for colligative properties.

11.75 Osmotic pressure is the amount of pressure that needs to be applied to cause osmosis to stop.

11.76

11.77 Molality is a temperature independent concentration unit. For freezing point depression and boiling point elevation, molality is used so that the solute concentration is independent of temperature changes. Molarity is temperature dependent. Molarity can be used for osmotic pressure because osmotic pressure is measured at a fixed temperature.

11.78 (a) CH₄N₂O, 60.06 amu; H₂O, 18.02 amu

10.0 g CH₄N₂O x = 0.167 mol CH₄N₂O

150.0 g H₂O x = 8.32 mol H₂O

= 0.980

P_soln = = (71.93 mm Hg)(0.980) = 70.5 mm Hg

(b) LiCl, 42.39 amu; 10.0 g LiCl x = 0.236 mol LiCl

LiCl dissociates into Li⁺(aq) and Cl^-(aq) in H₂O.

mol Li⁺ = mol Cl^- = mol LiCl = 0.236 mol

150.0 g H₂O x = 8.32 mol H₂O

= 0.946

P_soln = =(71.93 mm Hg)(0.946) = 68.0 mm Hg

11.79 C₆H₁₂O₆, 180.16 amu; CH₃OH, 32.04 amu

16.0 g C₆H₁₂O₆ x = 0.0888 mol C₆H₁₂O₆

80.0 g CH₃OH x = 2.50 mol CH₃OH

= 0.966

P_soln = = (140 mm Hg)(0.966) = 135 mm Hg

11.80 For H₂O, K_b = 0.51 ; 150.0 g = 0.1500 kg

(a) T_b = K_b m = = 0.57^oC

solution boiling point = 100.00^oC + T_b = 100.00^oC + 0.57^oC = 100.57^oC

(b) T_b = K_b m = = 1.6^oC

solution boiling point = 100.00^oC + T_b = 100.00^oC + 1.6^oC = 101.6^oC

11.81 For H₂O, K_f = 1.86 ; 150.0 g = 0.1500 kg

(a) T_f = K_f m = = 2.07^oC

solution freezing point = 0.00^oC - T_f = 0.00^oC - 2.07^oC = -2.07^oC

(b) T_f = K_f m = = 5.85^oC

solution freezing point = 0.00^oC - T_f = 0.00^oC - 5.85^oC = -5.85^oC

11.82 Acetone, C₃H₆O, 58.08 amu, = 285 mm Hg

Ethyl acetate, C₄H₈O₂, 88.11 amu, = 118 mm Hg

25.0 g C₃H₆O x = 0.430 mol C₃H₆O

25.0 g C₄H₈O₂ x = 0.284 mol C₄H₈O₂

= 0.602; = 0.398

P_soln =

P_soln = (285 mm Hg)(0.602) + (118 mm Hg)(0.398) = 219 mm Hg

11.83 CHCl₃, 119.38 amu, = 205 mm Hg; CH₂Cl₂, 84.93 amu, = 415 mm Hg

15.0 g CHCl₃ x = 0.126 mol CHCl₃

37.5 g CH₂Cl₂ x = 0.442 mol CH₂Cl₂

= 0.222; = 0.778

P_soln =

P_soln = (205 mm Hg)(0.222) + (415 mm Hg)(0.778) = 368 mm Hg

11.84 In the liquid, X_acetone = 0.602 and X_{ethyl acetate} = 0.398

In the vapor, P_Total = 219 mm Hg

P_acetone = X_acetone = (285 mm Hg)(0.602) = 172 mm Hg

P_{ethyl acetate} = X_{ethyl acetate} = (118 mm Hg)(0.398) = 47 mm Hg

= 0.785; = 0.215

11.85 In the liquid, = 0.222 and = 0.778

In the vapor, P_total = 368 mm Hg

= (205 mm Hg)(0.222) = 45.5 mm Hg

= (415 mm Hg)(0.778) = 323 mm Hg

= 0.124; = 0.876

11.86 C₉H₈O₄, 180.16 amu; 215 g = 0.215 kg

T_b = K_b m = 0.47^oC; K_b = = 3.6

11.87 C₆H₈O₆, 176.13 amu; 50.0 g = 0.0500 kg

T_f = K_f m = 1.33^oC; K_f = = 3.90

11.88 T_b = K_b m = 1.76^oC; m = = 0.573 m

11.89 C₆H₁₂O₆, 180.16 amu

For ethyl alcohol, K_b = 1.22 ; 285 g = 0.285 kg

T_b = K_b m = = 0.618^oC

solution boiling point = normal boiling point + T_b = 79.1^oC

normal boiling point = 79.1^oC - T_b = 79.1^oC - 0.618^oC = 78.5^oC

11.90 = MRT

(a) NaCl 58.44 amu; 350.0 mL = 0.3500 L

There are 2 moles of ions/mole of NaCl

= (2) = 13.0 atm

(b) CH₃CO₂Na, 82.03 amu; 55.0 mL = 0.0550 L

There are 2 moles of ions/mole of CH₃CO₂Na

= (2) = 65.2 atm

11.91 = MRT = (298 K) = 0.007 11 atm

= 0.007 11 atm x = 5.41 mm Hg

height of H₂O column = 5.41 mm Hg x = 73.2 mm

height of H₂O column = 73.2 mm x = 0.0732 m

11.92 = MRT; M = = 0.197 M

11.93 = MRT; M = = 0.30 M

Uses of Colligative Properties

11.94 Osmotic pressure is most often used for the determination of molecular mass because, of the four colligative properties, osmotic pressure gives the largest colligative property change per mole of solute.

11.95 C₆H₁₂O₆ does not dissociate in aqueous solution. LiCl and NaCl both dissociate into two solute particles per formula unit in aqueous solution. CaCl₂ dissociates into three solute particles per formula unit in aqueous solution. Assume that you have 1.00 g of each substance. Calculate the number of moles of solute particles in 1.00 g of each substance.

C₆H₁₂O₆, 180.2 amu; moles solute particles = 1.00 g x = 0.005 55 moles

LiCl, 42.4 amu; moles solute particles = 2 = 0.0472 moles

NaCl, 58.4 amu; moles solute particles = 2 = 0.0342 moles

CaCl₂, 111.0 amu; moles solute particles = 3 = 0.0270 moles

LiCl produces more solute particles/gram than any of the other three substances. LiCl would be the most efficient per unit mass.

11.96 = 407.2 mm Hg x = 0.5358 atm

= MRT; M = = 0.021 90 M = 0.021 90 mol/L

M = 0.021 90 mol/L = ; solve for the molar mass

molar mass of cellobiose = 342.5 g/mol; molecular mass = 342.5 amu

11.97 height of Hg column = 32.9 cm H₂O x = 2.43 cm Hg

= 2.43 cm Hg x = 0.0320 atm

= MRT; M =

M = 0.001 31 mol/L = ; solve for the molar mass.

molar mass of met-enkephalin = 573 g/mol; molecular mass = 573 amu

11.98 HCl is a strong electrolyte in H₂O and completely dissociates into two solute particles per each HCl.

HF is a weak electrolyte in H₂O. Only a few percent of the HF molecules dissociates into ions.

11.99 Na₂SO₄, 142.0 amu; m = = 0.50 m

T = K_b m = (0.50 m) = 0.26^oC

The experimental T is approximately 3 times that predicted by the equation above because Na₂SO₄ dissociates into three solute particles (2 Na⁺ and SO₄^2-) in aqueous solution.

11.100 First, determine the empirical formula:

Assume 100.0 g of -carotene.

10.51% H 10.51 g H x = 10.43 mol H

89.49% C 89.49 g C x = 7.45 mol C

C_7.45H_10.43; Divide each subscript by the smaller, 7.45.

C_{7.45 / 7.45}H_{10.43 / 7.45}

CH_1.4

Multiply each subscript by 5 to obtain integers. Empirical formula is C₅H₇, 67.1 amu.

Second, calculate the molecular mass:

T_f = K_f m; m =

m = 0.0310 mol/kg = ; solve for the molar mass.

molar mass of -carotene = 538 g/mol; molecular mass = 538 amu

Finally, determine the molecular formula:

Divide the molecular mass by the empirical formula mass.

; molecular formula is C_{(8 x 5)}H_{(8 x 7)}, or C₄₀H₅₆

11.101 First, determine the empirical formula:

Assume a 100.0 g sample of lysine.

49.29% C 49.29 g C x = 4.10 mol C

9.65% H 9.65 g H x = 9.57 mol H

19.16% N 19.16 g N x = 1.37 mol N

21.89% O 21.89 g O x = 1.37 mol O

C_4.10H_9.57N_1.37O_1.37; Divide each subscript by the smallest, 1.37.

C_{4.10 / 1.37}H_{9.57 / 1.37}N_{1.37 / 1.37}O_{1.37 / 1.37}

Empirical formula is C₃H₇NO, 73.09 amu

Second, calculate the molecular mass:

T_f = K_f m = 1.37 ^oC; m = = 0.171 m = 0.171 mol/kg

m = 0.171 mol/kg = ; solve for the molar mass.

molar mass of lysine = 146 g/mol; molecular mass = 146 amu

Finally, determine the molecular formula:

Divide the molecular mass by the empirical formula mass.

= 2; molecular formula is C_{(2 x 3)}H_{(2 x 7)}N_{(2 x 1)}O_{(2 x 1)}, or C₆H₁₄N₂O₂

General Problems

11.102 K_f for snow (H₂O) is 1.86 . Reasonable amounts of salt are capable of lowering the freezing point (T_f) of the snow below an air temperature of -2^oC. Reasonable amounts of salt, however, are not capable of causing a T_f of more than 30^oC which would be required if it is to melt snow when the air temperature is -30^oC.

11.103 KBr, 119.00 amu; there are 2 ions/KBr; 125 g = 0.125 kg

T_b = 103.2 ^oC - 100.0 ^oC = 3.2 ^oC

T_b = K_b 2 m; m = = 3.137 mol/kg

m = 3.137 mol/kg = ; solve for the mass of KBr.

mass of KBr = 47 g

11.104 C₂H₆O₂, 62.07 amu; T_f = 22.0^oC

T_f = K_f m; m = = 11.8 m

m = 11.8

Solve for the mass of ethylene glycol. mass of ethylene glycol = 2600 g = 2.60 x 10³ g

11.105 The vapor pressure of toluene is lower than the vapor pressure of benzene at the same temperature. When 1 mL of toluene is added to 100 mL of benzene, the vapor pressure of the solution decreases, which means that the boiling point of the solution will increase. When 1 mL of benzene is added to 100 mL of toluene, the vapor pressure of the solution increases, which means that the boiling point of the solution will decrease.

11.106 When solid CaCl₂ is added to liquid water, the temperature rises because H_soln for CaCl₂ is exothermic.

When solid CaCl₂ is added to ice at 0^oC, some of the ice will melt (an endothermic process) and the temperature will fall because the CaCl₂ lowers the freezing point of an ice/water mixture.

11.107 AgCl, 143.32 amu; there are 2 ions/AgCl

= 2MRT

= 2(278 K) = 0.002 atm

11.108 C₁₀H₈, 128.17 amu; 150.0 g = 0.1500 kg; T_f = 0.35^oC

T_f = K_f m; m =

m = 0.0684

Solve for the mass of C₁₀H_8. mass of C₁₀H₈ = 1.3 g

11.109 Br₂, 159.81 amu; CCl₄, 153.82 amu

= 228.8 mm Hg

= 123.8 mm Hg

1.50 g Br₂ x = 9.39 x 10^-3 mol Br₂

145.0 g CCl₄ x = 0.943 mol CCl₄

= 0.009 86

= 0.990

P_soln =

P_soln = (228.8 mm Hg)(0.009 86) + (123.8 mm Hg)(0.990) = 125 mm Hg

11.110 NaCl, 58.44 amu; there are 2 ions/NaCl

A 3.5 mass % aqueous solution of NaCl contains 3.5 g NaCl and 96.5 g H₂O.

molality = = 0.62 m

T_f = K_f 2 m = = 2.3^oC

freezing point = 0.0^oC - T_f = 0.0^oC - 2.3^oC = -2.3^oC

T_b = K_b 2 m = = 0.63^oC

boiling point = 100.00^oC + T_b = 100.00^oC + 0.63^oC = 100.63^oC

11.111 (a) Assume a total mass of solution of 1000.0 g.

ppm = x 10⁶

For each ion: mass of solute ion =

Ion Mass Moles

Cl^- 19.0 g 0.536 mol

Na⁺ 10.5 g 0.457 mol

SO₄^2- 2.65 g 0.0276 mol

Mg²⁺ 1.35 g 0.0555 mol

Ca²⁺ 0.400 g 0.009 98 mol

K⁺ 0.380 g 0.009 72 mol

HCO₃^- 0.140 g 0.002 29 mol

Br^- 0.065 g 0.000 81 mol

Total 34.5 g 1.099 mol

Mass of H₂O = 1000.0 g - 34.5 g = 965.5 g H₂O = 0.9655 kg H₂O

molality = = 1.138 m

(b) Assume M = m for a dilute solution.

= MRT = (1.138 )(0.082 06 )(300 K) = 28.0 atm

11.112 (a) 90 mass % isopropyl alcohol =

Solve for the mass of H₂O.

mass of H₂O = = 1.2 g

mass of solution = 10.5 g + 1.2 g = 11.7 g

11.7 g of rubbing alcohol contains 10.5 g of isopropyl alcohol.

(b) C₃H₈O, 60.10 amu

mass C₃H₈O = (0.90)(50.0 g) = 45 g

45 g C₃H₈O x = 0.75 mol C₃H₈O

11.113 C₆H₁₂O₆, 180.16 amu; 50.0 mL = 0.0500 L; 17.5 mg = 17.5 x 10^-3 g

= MRT

T = = 312 K

11.114 First, determine the empirical formula.

3.47 mg = 3.47 x 10^-3 g sample

10.10 mg = 10.10 x 10^-3 g CO₂

2.76 mg = 2.76 x 10^-3 g H₂O

mass C = 10.10 x 10^-3 g CO₂ x = 2.76 x 10^-3 g C

mass H = 2.76 x 10^-3 g H₂O x = 3.09 x 10^-4 g H

mass O = 3.47 x 10^-3 g - 2.76 x 10^-3 g C - 3.09 x 10^-4 g H = 4.01 x 10^-4 g O

2.76 x 10^-3 g C x = 2.30 x 10^-4 mol C

3.09 x 10^-4 g H x = 3.07 x 10^-4 mol H

4.01 x 10^-4 g O x = 2.51 x 10^-5 mol O = 0.251 x 10^-4 mol O

To simplify the empirical formula, divide each mol quantity by 10^-4.

C_2.30H_3.07O_0.251; Divide all subscripts by the smallest, 0.251.

C_{2.30 / 0.251}H_{3.07 / 0.251}O_{0.251 / 0.251}

C_9.16H_12.23O, empirical formula is C₉H₁₂O (136 amu)

Second, determine the molecular mass.

7.55 mg = 7.55 x 10^-3 g estradiol; 0.500 g x = 5.00 x 10^-4 kg camphor

T_f = K_f m; m = = 0.0557 m

m =

mol estradiol = m x (kg solvent) = (0.0557 mol/kg)(5.00 x 10^-4 kg) = 2.79 x 10^-5 mol

molar mass = = 271 g/mol; molecular mass = 271 amu

Finally, determine the molecular formula:

Divide the molecular mass by the empirical formula mass.

= 2; molecular formula is C_{(2 x 9)}H_{(2 x 12)}O_{(2 x 1)}, or C₁₈H₂₄O₂

11.115 CCl₃CO₂H(aq) H⁺(aq) + CCl₃CO₂^-(aq)

1.00 - x x x

T = K_f m; m = = 1.36 m

1.36 = 1.00 - x + x + x = 1 + x; x = 0.36

36% of the acid molecules are dissociated.

11.116 (a) H₂SO₄, 98.08 amu; 2.238 mol H₂SO₄ x = 219.50 g H₂SO₄

mass of 2.238 m solution = 219.50 g H₂SO₄ + 1000 g H₂O = 1219.50 g

volume of 2.238 m solution = 1219.50 g x = 1084.68 mL = 1.0847 L

molarity of 2.238 m solution = = 2.063 M

The molarity of the H₂SO₄ solution is less than the molarity of the BaCl₂ solution. Because equal volumes of the two solutions are mixed, H₂SO₄ is the limiting reactant and the number of moles of H₂SO₄ determines the number of moles of BaSO₄ produced as the white precipitate.

(0.05000 L) x (2.063 mol H₂SO₄/L) x = 24.07 g BaSO₄

(b) More precipitate will form because of the excess BaCl₂ in the solution.

11.117 KCl, 74.55amu; KNO₃, 101.10 amu_; Ba(NO₃)₂, 261.34 amu

= MRT; M = = 0.040 07 M

0.040 07 M = ; mol ions = (0.040 07 mol/L)(0.500 L) = 0.020 035 mol ions

mass Cl = 1.000 g x 0.2092 = 0.2092 g Cl

mass KCl = = 0.440 g KCl

mol ions from KCl = 0.440 g KCl x = 0.0118 mol ions

mol ions from KNO₃ and Ba(NO₃)₂ = 0.020 035 - 0.0118 = 0.008 235 mol ions

Let x = mass KNO₃ and y = mass Ba(NO₃)₂

x + y = 1.000 g - 0.440 g = 0.560 g

= 0.008 235 mol ions

x = 0.560 - y

0.0198x + 0.0115y = 0.008 235

0.0198(0.560 - y) + 0.0115y = 0.008 235

0.011 09 - 0.0198y + 0.0115y = 0.008 235

0.002 855 = 0.0083y; y = = 0.3440; x = 0.560 - 0.3440 = 0.216

mass % KCl = = 44.0%

mass % KNO₃ = = 21.6%

mass % Ba(NO₃)₂ = = 34.4%

11.118 Let x = and y = and assume n_total = 1.00 mol

(14.5 mm Hg)x + (82.5 mm Hg)y = 39.4 mm Hg

(26.8 mm Hg)x + (140.3 mm Hg)y = 68.2 mm Hg

x =

+ 82.5y = 39.4

+ 82.5y = 39.4

36.90 - 75.91y + 82.5y = 39.4; 6.59 = 2.5; y = = 0.3794

x = = 0.5586

X_LiCl = 1 - - = 1 - 0.5586 - 0.3794 = 0.0620

The mole fraction equals the number of moles of each component because

n_total = 1.00 mol.

mass LiCl = 0.0620 mol LiCl x = 2.6 g LiCl

mass H₂O = 0.5588 mol H₂O x = 10.1 g H₂O

mass CH₃OH = 0.3794 mol CH₃OH x = 12.2 g CH₃OH

total mass = 2.6 g + 10.1 g + 12.2 g = 24.9 g

mass % LiCl = = 10%

mass % H₂O = = 41%

mass % CH₃OH = = 49%

Multi-Concept Problems

11.119 (a) 382.6 mL = 0.3826 L; 20.0^oC = 293.2 K

PV = nRT

= = 0.0158 mol H₂

(b) M + x HCl x/2 H₂ + MCl_x

moles HCl reacted = 0.0158 mol H₂ x = 0.0316 mol HCl

moles Cl reacted = moles HCl reacted = 0.0316 mol Cl

mass Cl = 0.0316 mol Cl x = 1.120 g Cl

mass MCl_x = mass M + mass Cl = 1.385 g + 1.120 g = 2.505 g MCl_x

(c) T_f = K_f · m; m == 1.90 m

(d) 25.0 g = 0.0250 kg

1.90 m = =

mol ions = (1.90 mol/kg)(0.0250 kg) = 0.0475 mol ions

(e) mol M = mol ions - mol Cl = 0.0475 mol - 0.0316 mol = 0.0159 mol M

= 2, the formula is MCl₂.

molar mass = = 157.5 g/mol; molecular mass = 157.5 amu

(f) atomic mass of M = 157.5 amu - 2(35.453 amu) = 86.6 amu; M = Sr

11.120 (a) 20.00 mL = 0.02000 L

mol NaOH = (0.02000 L)(2.00 mol/L) = 0.0400 mol NaOH

mol CO₂ = 0.0400 mol NaOH x = 0.0200 mol CO₂

mol C = 0.0200 mol CO₂ x = 0.0200 mol C

mass C = 0.0200 mol C x = 0.240 g C

mass H = mass of compound - mass of C = 0.270 g - 0.240 g = 0.030 g H

mol H = 0.030 g H x = 0.030 mol H

The mole ratio of C and H in the molecule is C_0.0200 H_0.030.

C_0.0200 H_0.030, divide both subscripts by the smaller of the two, 0.0200.

C_{0.0200 / 0.0200} H_{0.030 / 0.0200}

C₁H_1.5, multiply both subscripts by 2.

C_{(2 x 1)} H_{(2 x 1.5)}

C₂H₃ (27.05 amu) is the empirical formula.

(b) T_f = K_f m; m = = 0.050 m

50.0 g = 0.0500 kg

mol solute = (0.050 )(0.0500 kg) = 0.0025 mol

molar mass = = 108 g/mol; molecular mass = 108 amu

(c) To find the molecular formula, first divide the molecular mass by the mass of the empirical formula unit.

= 4

Multiply the subscripts in the empirical formula by the result of this division, 4.

C_{(4 x 2)} H_{(4 x 3)}

C₈H₁₂ is the molecular formula of the compound.