# Chapter 12 Solutions.

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Chapter 12 Solutions

Classification of matter
Homogeneous (visibly indistinguishable) Heterogeneous (visibly distinguishable) (Solutions) Mixtures (multiple components) Pure Substances (one component) Matter Elements Compounds

Solution = Solute + Solvent

Vodka = ethanol + water Brass = copper + zinc

If solvent is water, the solution is called an aqueous solution.
Click to add notes If solvent is water, the solution is called an aqueous solution.

Beer Wine Liquor Ethanol Concentration

Four Concentrations (1) Unit: none (2) Unit: mol/L

Four Concentrations (3) Unit: none Unit: none

Four Concentrations (4) Unit: mol/kg

A solution contains 5.0 g of toluene (C7H8) and 225 g of
benzene (C6H6) and has a density of g/mL. Calculate the mass percent and mole fraction of C7H8, and the molarity and molality of the solution. 2.2 %, 0.018, 0.21 mol/L, 0.24 mol/kg Practice on Example 12.4 and compare your results with the answers.

Electrical Conductivity of Aqueous Solutions

strong acids strong bases salts strong electrolyte weak electrolyte nonelectrolyte weak acids weak bases solute many organic compounds

van’t Hoff factor nonelectrolyte: i = 1
Unit: none nonelectrolyte: i = 1 weak electrolyte: depends on degree of dissociation strong electrolyte: depends on chemical formula

NaCl MgBr2 MgSO4 FeCl3 NaOH Hexane Glucose

Four properties of solutions
(1) Boiling point elevation Solution compared to pure solvent water = solvent Boiling point = 100 °C water + sugar = solution Boiling point > 100 °C

Sugar Dissolved in Water to Make Candy Causes the Boiling Point to be Elevated

∆Tb = Tb,solution − Tb,solvent = i Kb m
i: van’t Hoff factor of solute m: molality Kb: boiling-point elevation constant Units Kb is characteristic of the solvent. Does not depend on solute.

i ― electrolyte or nonelectrolyte
Boiling point elevation can be used to find molar mass of solute. ∆Tb ― experiments i ― electrolyte or nonelectrolyte Kb ― table or reference book

A solution was prepared by dissolving 18.00 g glucose in 150.0 g
water. The resulting solution was found to have a boiling point of °C. Calculate the molar mass of glucose. Glucose is molecular solid that is present as individual molecules in solution. 180 g/mol

Four properties of solutions
(1) Boiling point elevation (2) Freezing point depression Solution compared to pure solvent water = solvent freezing point = 0 °C water + salt = solution freezing point < 0 °C

∆Tf = Tf,solvent − Tf,solution = i Kf m
i: van’t Hoff factor of solute m: molality Kf: freezing-point depression constant Units Kf is characteristic of the solvent. Does not depend on solute.

i ― electrolyte or nonelectrolyte
Freezing point depression can be used to find molar mass of solute. ∆Tf ― experiments i ― electrolyte or nonelectrolyte Kf ― table or reference book

A chemist is trying to identify a human hormone that controls
metabolism by determining its molar mass. A sample weighing 0.546 g was dissolved in 15.0 g benzene, and the freezing-point depression was determined to be °C. Calculate the molar mass of the hormone. 776 g/mol

Chapter 12, Unnumbered Figure 1, Page 535

The Addition of Antifreeze Lowers the Freezing Point of Water in a Car's Radiator
antifreeze = water + ethylene glycol

Experiment 23 this Thursday
Read the lab manual before coming to the lab Bring the manual, a copy of the data form, a pair of goggles, and an apron to the lab Wash your apron before Thursday Dress properly according to the syllabus

Four properties of solutions
(1) Boiling point elevation (2) Freezing point depression (3) Osmotic pressure

Osmotic Pressure

Π = iMRT Π ― osmotic pressuue M ― molarity R ― ideal gas constant T ― absolute temperature i ― van’t Hoff factor of solute

Π = iMRT Units Π ― atm M ― mol/L R ― atm·L·K−1·mol−1 T ― K i ― none

i ― electrolyte or nonelectrolyte
Osmotic pressure can be used to find molar mass of solute. Π ― experiments i ― electrolyte or nonelectrolyte R ― constant T ― experiments

To determine the molar mass of a certain protein, 1.00 x 10−3 g
of it was dissolved in enough water to make 1.00 mL of solution. The osmotic pressure of this solution was found to be 1.12 torr at 25.0 °C. Calculate the molar mass of the protein. 1.66 x 104 g/mol

Practice on Example 12.10 and

Figure: 13-25

isotonic = isosmotic: same osmotic pressure
Chapter 12, Unnumbered Figure, Page 550 isotonic = isosmotic: same osmotic pressure

What molarity of NaCl in water is needed to produce an
aqueous solution isotonic with blood ( Π = 7.70 atm at 25 °C)? 0.158 mol/L

Four properties of solutions
(1) Boiling point elevation (2) Freezing point depression (3) Osmotic pressure (4) Lowering the vapor pressure

Nonvolatile solute to volatile solvent
Lowering Vapor Pressure

Liquid Surface

Liquid Surface pure solvent
Surface Molecules pure solvent number of solvent molecules above the liquid 100 x 100% x 5% = 5

Liquid Surface Liquid Surface pure solvent solvent + solute
number of solvent molecules above the liquid pure solvent: 100 x 100% x 5% = 5 solution: 100 x 80% x 5% = 4

Four Concentrations (3) Unit: none Unit: none

Raoult’s Law: Case 1 ― vapor pressure of solution
Nonvolatile solute in a Volatile solvent ― vapor pressure of solution ― vapor pressure of pure solvent ― mole fraction of solvent

For a Solution that Obeys Raoult's Law, a Plot of Psoln Versus Xsolvent, Give a Straight Line

Example 12.6 Calculate the vapor pressure at 25 °C of a solution containing 99.5 g of sucrose (C12H22O11) and 300 mL of water. The vapor pressure of pure water at 25 °C is 23.8 torr. Assume the density of water to be 1.00 g/mL. molar mass of sucrose = g/mol 23.4 torr

Predict the vapor pressure of a solution prepared by mixing
35.0 g solid Na2SO4 (molar mass = 142 g/mol) with 175 g water at 25 °C. The vapor pressure of pure water at 25 °C is 23.8 torr.

Liquid Surface solvent + solute
When you count the number of solute particles, use van’t Hoff factor i.

22.1 torr Predict the vapor pressure of a solution prepared by mixing
35.0 g solid Na2SO4 (molar mass = 142 g/mol) with 175 g water at 25 °C. The vapor pressure of pure water at 25 °C is 23.8 torr. 22.1 torr Try Example 12.12

aqueous solution ΔTb Figure: 11-21

Raoult’s Law: Case 2 Volatile solute in a Volatile solvent
Recall Dalton’s law of partial pressures

Vapor Pressure for a Solution of Two Volatile Liquids
XA + XB = 1

A mixture of benzene (C6H6) and toluene (C7H8) containing
1.0 mol of benzene and 2.0 mol of toluene. At 20 °C the vapor pressures of pure benzene and toluene are 75 torr and 22 torr, respectively. What is the vapor pressure of the mixture? What is the mole fraction of benzene in the vapor?

A solution that obeys Raoult’s Law is called an
ideal solution.

Chapter 12, Figure 12.15A Behavior of Nonideal Solutions

What kind of solution is ideal if
viewed at a molecular level?

pure solvent 10% P0 # of molecules in vapor = 100 x 1 x 10% = 10 χ

P0 χ Psln solvent + solute
15% solvent + solute 5% 10% P0 # of molecules in vapor = 100 x 0.8 x 10% = 8 χ Psln Raoult’s law: # of molecules in vapor = 100 x 0.8 x 5% = 4 Deviate from Raoult’s law # of molecules in vapor = 100 x 0.8 x 15% = 12

What kind of solution is ideal?
Solute-solute, solvent-solvent, and solute-solvent interactions are very similar.

Chapter 12, Figure 12.15 Behavior of Nonideal Solutions

Try Example 12.7

What kind of solution is ideal?
Solute-solute, solvent-solvent, and solute-solvent interactions are very similar. Comparison to ideal gas.

Four Colligative properties of solutions
(1) Boiling point elevation: ∆Tb = i Kb m (2) Freezing point depression: ∆Tf = i Kf m (3) Osmotic pressure: Π = iMRT (4) Lowering the vapor pressure: Colligative: depend on the quantity (number of particles, concentration) but not the kind or identity of the solute particles.

Quiz 3 Find van’t Hoff factors from formulas
Use osmotic pressure to find molar mass of a solute Raoult’s law, ideal solution