Chapter 11 Properties of Solutions

From Chapter 1: Classification of matter Matter Homogeneous (visibly indistinguishable) Heterogeneous (visibly distinguishable) Elements Compounds Mixtures (multiple components) Pure Substances (one component) (Solutions)

Solution = Solute + Solvent

Vodka = ethanol + waterBrass = copper + zinc

Liquor BeerWine Ethanol Concentration

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

Four Concentrations Unit: none Unit: mol/kg (3) (4)

A solution contains 5.0 g of toluene (C 7 H 8 ) and 225 g of benzene (C 6 H 6 ) and has a density of g/mL. Calculate the mass percent and mole fraction of C 7 H 8, and the molarity and molality of the solution. Practice on Sample Exercise 11.1 on page 486 and compare your results with the answers.

Electrical Conductivity of Aqueous Solutions

solute strong electrolyte weak electrolyte nonelectrolyte strong acids strong bases most salts weak acids weak bases many organic compounds Chapter 4

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

NaCl MgCl 2 MgSO 4 FeCl 3 HCl Glucose

Figure In an Aqueous Solution a Few Ions Aggregate, Forming Ion Pairs that Behave as a Unit

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

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

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

Table 11.5 Molal Boiling-Point Elevation Constants (K b ) and Freezing-Point Depression Constants (K f ) for Several Solvents

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

A solution was prepared by dissolving g glucose in 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. Sample exercise 11.8, page 505

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

Spreading Salt on a Highway

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

Table 11.5 Molal Boiling-Point Elevation Constants (K b ) and Freezing-Point Depression Constants (K f ) for Several Solvents

The Addition of Antifreeze Lowers the Freezing Point of Water in a Car's Radiator 0 °C 100 °C water < 0 °C> 100 °C water + antifreeze

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

A chemist is trying to identify a human hormone that controls metabolism by determining its molar mass. A sample weighing 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. Sample exercise 11.10, page 507

Table 11.5 Molal Boiling-Point Elevation Constants (K b ) and Freezing- Point Depression Constants (K f ) for Several Solvents

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 ― temperature

Π = iMRT Π ― atm M ― mol/L R ― atm·L·K −1 ·mol −1 T ― K Units

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. Sample exercise 11.11, page 509

What concentration of NaCl in water is needed to produce an aqueous solution isotonic with blood ( Π = 7.70 atm at 25 °C)? Sample exercise 11.12, page 510

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

Lowering Vapor Pressure Nonvolatile solute to volatile solvent

The Presence of a Nonvolatile Solute Lowers the Vapor Pressure of the Solvent

pure solvent Liquid Surface

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

Show that the packing efficiency for body centered cubic unit cell is 68 %. Pop Quiz: 0.5 extra point

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

Figure For a Solution that Obeys Raoult's Law, a Plot of P soln Versus X solvent, Give a Straight Line

Calculate the expected vapor pressure at 25 °C for a solution prepared by dissolving g of common table sugar (sucrose, molar mass = g/mol) in cm 3 of water. At 25 °C, the density of water is g/cm 3 and the vapor pressure is torr. Sample exercise 11.5, page 499

Predict the vapor pressure of a solution prepared by mixing 35.0 g solid Na 2 SO 4 (molar mass = 142 g/mol) with 175 g water at 25 °C. The vapor pressure of pure water at 25 °C is torr. Sample exercise 11.6, page 500

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 X A + X B = 1 0 1

A mixture of benzene (C 6 H 6 ) and toluene (C 7 H 8 ) 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?

Lowering vapor pressure can be used to find molar mass of solute. and ― experiments

At 25 °C a solution is prepared by dissolving g of common table sugar (sucrose, nonelectrolyte, nonvolatile) in cm 3 of water. The vapor pressure of this solution is torr. At 25 °C, the density of water is g/cm 3 and the vapor pressure is torr. Calculate the molar mass of sucrose. Modified sample exercise 11.5, page 499

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

A solution is prepared by mixing 5.81 g acetone (molar mass = 58.1 g/mol) and 11.0 g chloroform (molar mass = g/mol). At 35 °C, this solution has a total vapor pressure of 260. torr. Is this an ideal solution? The vapor pressure of pure acetone and pure chloroform at 35 °C are 345 torr and 293 torr, respectively. Sample exercise 11.7, page 503

What kind of solution is ideal?

10% P0P0 # of molecules in vapor = 100 x 1 x 10% = 10 χ pure solvent

10% 5% 15% # of molecules in vapor = 100 x 0.8 x 5% = 4 # of molecules in vapor = 100 x 0.8 x 15% = 12 # of molecules in vapor = 100 x 0.8 x 10% = 8 χ Raoult’s law: Deviate from Raoult’s law P0P0 solvent + solute

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

Figure a-c Vapor Pressure for a Solution of Two Volatile Liquids IdealNonideal

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

Rule of solubility: Like dissolves like Polarity

Henry’s Law: the amount of gas that dissolved in a solution is directly proportional to the pressure of the gas above the solution. C = kP

Figure 11.5 a-c Henry's Law

C = kP

What you must master in this chapter Four concentrations. Four colligative properties. A Quiz this week? Calculations associated with