Notes One Unit Seven– Chapter 13 Solutions

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