notes one unit seven– chapter 13 solutions
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Pages 466-486. Notes One Unit Seven– Chapter 13 Solutions. Definitions Types of Mixtures Example Solutions Factors Affecting Solubility Like Dissolves Like Solubility of Solids Changes with Temperature Solubility of Gases Changes with Temperature Pressure Factor Molar Concentration - PowerPoint PPT PresentationTRANSCRIPT
Notes One Unit Seven– Chapter 13 Solutions
•Definitions•Types of Mixtures•Example Solutions•Factors Affecting Solubility•Like Dissolves Like•Solubility of Solids Changes with Temperature•Solubility of Gases Changes with Temperature•Pressure Factor•Molar Concentration•Finding Molarity From Mass and Volume•Finding Mass from Molarity and Volume•Finding Volume from Molarity and Mass
Pages 466-486
Definitions• Solutions are
homogeneous mixtures.
• Uniform throughout.• Solvent.• Determines the
state of solution• Is the largest
component.• Solute.• Other components
dissolved in solvent.
Common Mixtures
liquid liquid emulsion mayonnaisegas liquid liquid foam whipped cream
liquid gas aerosol hair spraysolid solid solid ruby glass
SOLUTE SOLVENT Type EXAMPLE
solid gas aerosol dust in air
liquid solid emulsion pearlgas solid solid foam Styrofoam
Solution Types
gas gas gas a i rgas liquid liquid soda popliquid liquid liquid antifreeze
solid liquid liquid seawatersolid solid solid brass
SOLUTE SOLVENT PHASE EXAMPLE
liquid solid solid filling
Factors Affecting Solubility• 1. Nature of Solute / Solvent1. Nature of Solute / Solvent.
• 2. Temperature Increase2. Temperature Increase
• i) Solid/Liquid…increase
• ii) gas…decreases
• 3. Pressure Factor -3. Pressure Factor -
• i) Solids/Liquids - Very little
• ii) gas - Increases.
• iii) squeezes gas into solution.
Like Dissolves Like
• Non-polar does not dissolve polar• Oil in H2O
• Polar dissolves Polar• C2H5OH in H2O
• Ionic compounds soluble in polar solvents• NaCl in H2O
Solubility of solids Changes with Temperature
• Does Solubility always increase for solids or gases?
• How many grams potassium chromate will dissolve 100g water at 70oC?
• 70g• How many grams lead(II)
nitrate will precipitate from 250g water as it cools from 70oC to 50oC?
19gx _____250g100g =48g
solids gases
101g
82g
Calculating Freezing Point Depression Mass• A solution containing 1.89 g of methanol in 51.96 g
of water freezes at -3.4oC. Calculate the molecular weight of methanol .
• 1.)Calculate Temperature Change• ΔTb =
• 2.)Calculate moles per Kilograms• ΔTf = Kf x m m = ΔTf /Kf
• m =0.500m/kg
• 3.)Calculate grams / kilograms• g =• g =23.0g/kg• fm=• 46.0g/m
0.000oC- 0.929oC= 0.929oC
0.929÷ 1.858oC/ m = m
7.67 g ÷ 0.3330kg
23.0 g/0.500m
Solubility of Gases Changes with Temperature
• a) Why are fish stressed, if the temperature of the water increases?
• How much does the solubility of oxygen change, for a 20oC to 60oC change?
• 0.90-0.60=0.30mg
0.60mg
0.90mg
Pressure FactorPressure FactorGreater pressure… more dissolved gas
Pressure FactorPressure Factor
Molar Concentration
• M=n/V
• MxV=n
• V=n/M
Finding Molarity From Mass and Volume• Calculate molarity for 25.5 g of NH3 in 600. mL solution.
• 1) Calculate Formula Mass:
• 2) Calculate the moles of solute:
• 3) Calculate the Moles/Liters Ratio• M = n / V • M = 1.50 moles / 0.600 L • M = 2.52 mol/L
25.5g ÷ 17.0g/m= 1.50m
NH
1x3x
14.0 =1.0 =
14.0 3.017.0g/m
E # Mass
Finding Volume from Molarity and Mass• How many milliliters of 2.50M solution can be made
using 25.5grams of NH3?
• 1)Calculate formula mass:
• 2)Calculate the moles of solute:
• 3)Calculate Volume from Moles/Liters Ratio
• V=1.50m/2.50M=0.600L• 600.mL solution
NH
1x3x
14.0 =1.0 =
14.0 3.017.0g/m
E # Mass
25.5g ÷ 17.0g/m= 1.50m
Finding Mass from Molarity and Volume• How many grams of NH3 are in 600. mL solution at
2.50M?• 1) Calculate formula mass:
• 2) Calculate moles • M=n/V n=MxV
• n=1.50m• 3) Calculate mass• n=g/MW g=MWxn
• g=25.5g NH3
n= 2.50M x 0.600L
g=(17.0g/m)x(1.50m)
NH
1x3x
14.0 =1.0 =
14.0 3.017.0g/m
E # Mass
Notes Two Unit Seven– Chapter 13 Solutions
• Saturated versus Unsaturated• Colligative properties of water• Forming a Saturated Solution• How Does a Solution Form?• Colligative Properties• Vapor Pressure• Boiling and Freezing Point• BP Elevation and Freezing FP Depression• Calculating Freezing Point Depression Mass
Pages 487-501
Characteristics of Saturated Solutions
Soliddissolve
water
dissolve
precipitate
dissolve
precipitate
Unsaturated Unsaturated Saturated
DynamicEquilibrium
Cooling causes precipitation.Warming causes dissolving.
• As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them.
Solvation
Colligative Properties• Colligative properties depend on moles dissolved
particles.Vapor pressure lowering Boiling point elevationMelting point depressionOsmotic pressure
Vapor Pressure• vapor pressure of a solvent.
• vapor pressure of a solution.
Phase DiagramP
ress
ure
Temperature
solid
Liquid
gas
meltingfreezing
NFP NBP
1 atm
Triple point
Critical point
0.0oC 100.0oC
vaporizingcondensing
sublimationdepostion
One Molal Solution of WaterP
ress
ure
Temperature
solid
Liquid
gas
Kf Kb
1 atm
1.858oC 0.512oC
BP Elevation Constants (Kb)FP Depression Constants( Kf)
BP Elevation and FP Depression
Tb = Kb m
Kb=0.5120C/m
Tf = Kf m
Kf=1.8580C/m
Molarity versus Molality
Molality (m) = ________________moles of solutekilograms solvent
Molarity (M) = ________________moles of soluteliters of solution
Calculating Tf andTb• Calculate the freezing and boiling points of a
solution made using 1000.g antifreeze (C2H6O2) in 4450g water.
• 1) Calculate Moles
• 2) Calculate molality
• 3) Calculate Temperature Change • Δt=Kxm • ΔTf =• Tf = • ΔTb = • Tb =
1000.g ÷62.0g/mol = 16.1 moles
16.1 mole ÷ 4.45 Kg water =3.62m
(1.858oC/m) (3.62 m) = 6.73oC0.000oC- 6.73oC= -6.73oC(0.512oC/m) (3.62 m) =1.96oC
100.000oC + 1.96oC = 101.96oC
C
H
2x
6x
12.0 =
1.0 =
24.0
6.0
E # Mass
O 2x 16.0 = 32.0
62.0g/m
Calculating Boiling Point Elevation Mass• A solution containing 18.00 g of glucose in 150.0 g
of water boils at 100.34oC. Calculate the molecular weight of glucose.
• 1.)Calculate Temperature Change• ΔTb =
• 2.)Calculate moles per Kilograms• ΔTb = Kb x m m = ΔTb /Kb
• m =0.67m/kg
• 3.)Calculate grams / kilograms• g =• g =120 g/kg• MW=120 g/0.67m• 180g/m
100.34oC-100.00oC= 0.34oC
0.34÷ 0.512oC/ m = m
18.00 g ÷ 0.1500kg
Notes Three Unit Seven• Ice-cream Lab A Calculating Freezing Point
• Depression Mass
• Colligative Properties of Electrolytes
• Distillation
• Osmotic Pressure
• Dialysis
Pages 487-501
Ice-cream
Calculating Freezing Point Depression Mass
• 1.)Calculate Temperature Change
• 2.)Calculate moles per Kilograms
• 3.)Calculate grams / kilograms
Colligative Properties of Electrolytes• Colligative properties depend on the number of particles
dissolved.
• NaClNa+1+Cl-1 CH3OHAl2(SO4)32Al+3 + 3SO4
-2 C6H12O6
Distillation
Distillation
Osmotic Pressure• Hypertonic
• > 0.92% (9.g/L)
• Crenation
• Isotonic Saline
• = 0.92% (9.g/L)
• Hypotonic
• < 0.92% (9.g/L)
• Rupture
Dialysis
Kidney
Dialysis