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# Chapter 18 Solutions - PowerPoint PPT Presentation

Chapter 18 Solutions . Milbank High School. Section 18.1 Properties of Solutions. OBJECTIVES: Identify the factors that determine the rate at which a solute dissolves. Section 18.1 Properties of Solutions. OBJECTIVES:

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### Chapter 18Solutions

Milbank High School

Section 18.1Properties of Solutions

• OBJECTIVES:

• Identify the factors that determine the rate at which a solute dissolves.

Section 18.1Properties of Solutions

• OBJECTIVES:

• Calculate the solubility of a gas in a liquid under various pressure conditions.

• Nature of the solute and the solvent

• Whether a substance will dissolve

• How much will dissolve

• Factors determining rate of solution...

• stirred or shaken (agitation)

• temperature is increased

• Why?

• In order to dissolve, the solvent molecules must come in contact with the solute.

• Stirring moves fresh solvent next to the solute.

• The solvent touches the surface of the solute.

• Smaller pieces increase the amount of surface area of the solute.

• Higher temperature makes the molecules of the solvent move around faster and contact the solute harder and more often.

• Speeds up dissolving.

• Usually increases the amount that will dissolve (exception is gases)

• Solubility- The maximum amount of substance that will dissolve at a specific temperature (g solute/100 g solvent)

• Saturated solution- Contains the maximum amount of solute dissolved

• Unsaturated solution- Can still dissolve more solute

• Supersaturated- solution that is holding more than it theoretically can; seed crystal will make it come out; Fig. 18.7, p.506

• Miscible means that two liquids can dissolve in each other

• water and antifreeze, water and ethanol

• Partially miscible- slightly

• water and ether

• Immiscible means they can’t

• oil and vinegar

• For solids in liquids, as the temperature goes up-the solubility usually goes up (Fig. 18.4, p.504)

• For gases in a liquid, as the temperature goes up-the solubility goes down (Fig. 18.5, p.505)

• For gases in a liquid, as the pressure goes up-the solubility goes up (Fig. 18.6, p.505)

• Henry’s Law - says the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid

• think of a bottle of soda pop, removing the lid releases pres.

• Equation: S1 S2

P1 P2

=

• Ever heard of seeding the clouds to make them produce rain?

• Clouds- mass of air supersaturated with water vapor

• Silver Iodide (AgI) crystals are dusted into the cloud

• The AgI attracts the water, forming droplets to attract others

Section 18.2Concentration of Solutions

• OBJECTIVES:

• Solve problems involving the molarity of a solution.

Section 18.2Concentration of Solutions

• OBJECTIVES:

• Describe how to prepare dilute solutions from more concentrated solutions of known molarity.

Section 18.2Concentration of Solutions

• OBJECTIVES:

• Explain what is meant by percent by volume [ % (v/v) ], and percent by mass [ % (m/v) ] solutions.

• a measure of the amount of solute dissolved in a given quantity of solvent

• Aconcentrated solution has a large amount of solute

• Adilute solution has a small amount of solute

• thus, only qualitative descriptions

• But, there are ways to express solution concentration quantitatively

• The number of moles of solute in 1 Liter of the solution.

• M = moles/Liter; such as 6.0 molar

• What is the molarity of a solution with 2.0 moles of NaCl in 250 mL of solution?

• Sample 18-2, page 510

• Pour in a small amount of solvent

• Then add the solute (to dissolve it)

• Carefully fill to final volume.

• Fig. 18-10, page 509

• Also: M x L = moles

• How many moles of NaCl are needed to make 6.0 L of a 0.75 M NaCl solution?

• 10.3 g of NaCl are dissolved in a small amount of water, then diluted to 250 mL. What is the concentration?

• How many grams of sugar are needed to make 125 mL of a 0.50 M C6H12O6 solution?

### Dilution

• The number of moles of solute doesn’t change if you add more solvent!

• The # moles before = the # moles after

• M1 x V1 = M2 x V2

• M1 and V1 are the starting concentration and volume.

• M2 and V2 are the final concentration and volume.

• Stock solutions are pre-made to known Molarity

• 2.0 L of a 0.88 M solution are diluted to 3.8 L. What is the new molarity?

• You have 150 mL of 6.0 M HCl. What volume of 1.3 M HCl can you make?

• Need 450 mL of 0.15 M NaOH. All you have available is a 2.0 M stock solution of NaOH. How do you make the required solution?

• Percent means parts per 100, so

• Percent by volume: = Volume of solute x 100% Volume of solution

• indicated %(v/v)

• What is the percent solution if 25 mL of CH3OH is diluted to 150 mL with water?

• Percent by mass: = Mass of solute(g) x 100% Volume of solution(mL)

• Indicated %(m/v)

• More commonly used

• 4.8 g of NaCl are dissolved in 82 mL of solution. What is the percent of the solution?

• How many grams of salt are there in 52 mL of a 6.3 % solution?

Section 18.3Colligative Properties of Solutions

• OBJECTIVES:

• Explain on a particle basis why a solution has a lower vapor pressure than the pure solvent of that solution.

Section 18.3Colligative Properties of Solutions

• OBJECTIVES:

• Explain on a particle basis why a solution has an elevated boiling point, and a depressed freezing point compared with the pure solvent.

### Colligative Properties

Depend only on the number of dissolved particles

Not on what kind of particle

• The bonds between molecules keep molecules from escaping.

• In a solution, some of the solvent is busy keeping the solute dissolved.

• Lowers the vapor pressure

• Electrolytes form ions when they are dissolved = more pieces.

• NaCl ® Na+ + Cl- (= 2 pieces)

• More pieces = bigger effect

• The vapor pressure determines the boiling point.

• Lower vapor pressure = higher boiling point.

• Salt water boils above 100ºC

• The number of dissolved particles determines how much, as well as the solvent itself.

• Solids form when molecules make an orderly pattern.

• The solute molecules break up the orderly pattern.

• Makes the freezing point lower.

• Salt water freezes below 0ºC

• How much depends on the number of solute particles dissolved.

Section 18.4Calculations Involving Colligative Properties

• OBJECTIVES:

• Calculate the molality and mole fraction of a solution.

Section 18.4Calculations Involving Colligative Properties

• OBJECTIVES:

• Calculate the molar mass of a molecular compound from the freezing point depression or boiling point elevation of a solution of the compound.

• a new unit for concentration

• m = Moles of solute kilogram of solvent

• m = Moles of solute 1000 g of solvent

• What is the molality of a solution with 9.3 mole of NaCl in 450 g of water?

• The size of the change in boiling point is determined by the molality.

• DTb= Kbx m x n

• DTb is the change in the boiling point

• Kb is a constant determined by the solvent (Table 18.2, page 523).

• m is the molality of the solution.

• n is the number of pieces it falls into when it dissolves.

• The size of the change in freezing point is also determined by molality.

• DTf= -Kfx m x n

• DTf is the change in freezing point

• Kf is a constant determined by the solvent (Table 18.3, page 524).

• m is the molality of the solution.

• n is the number of pieces it falls into when it dissolves.

• What is the boiling point of a solution made by dissolving 1.20 moles of NaCl in 750 g of water?

• What is the freezing point?

• What is the boiling point of a solution made by dissolving 1.20 moles of CaCl2 in 750 g of water?

• What is the freezing point?

• This is another way to express concentration

• It is the ratio of moles of solute to total number of moles of solute + solvent (Fig. 18-19, p.522)

na

na + nb

Sample 18-8, page 521

X =

• We can use changes in boiling and freezing to calculate the molar mass of a substance

• Find: 1) molality 2) moles, and then 3) molar mass

• Sample 18-10, page 524

• Note the key equations on page 527 to solve problems in this chapter.